CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 5 | CHEM 107 - Essentials of General and Organic Chemistry

CHEM 107 - ESSENTIALS OF

GENERAL AND ORGANIC

CHEMISTRY - Calorimetry

Question Bank - Set 5

Liberty University

Question 1

Question

A 50 g aluminum block at 100

C is placed in 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 25

C. Calculate the specific

heat capacity of the aluminum block. The specific heat capacity of water is

cw= 4.18 J/g

Solution

Step 1: Calculate the heat gained by the water.

The heat gained by the water can be calculated using the formula:

Qwater =mwater ·cw·∆T

where - mwater = 200 g is the mass of water, - cw= 4.18 J/g

C is the specific

heat capacity of water, - ∆T=Tfinal −Tinitial = 25C−20C= 5Cis the change

in temperature.

Plugging in the values:

Qwater = 200 g ×4.18 J/g

C×5C= 4180 J

Step 2: Calculate the heat lost by the aluminum block.

The heat lost by the aluminum block is equal in magnitude to the heat

gained by the water:

Qwater =Qaluminum

Using the formula for heat:

Qaluminum =maluminum ·caluminum ·∆T

We know: - maluminum = 50 g is the mass of the aluminum block, - ∆T= 75C

(since the final temperature of the system is 25

C), - caluminum is the specific

heat capacity of aluminum (to be determined).

Plugging in the values:

4180 J = 50 g ×caluminum ×75C

Step 3: Solve for the specific heat capacity of aluminum.

caluminum =4180 J

3750 g

C= 1.114 J/g

Therefore, the specific heat capacity of the aluminum block is 1.114 J/g

Question 2

Question

A 50.0 g piece of aluminum at 120◦C is placed in 100.0 g of water at 25.0◦C.

Assuming no heat is lost to the surroundings, calculate the final temperature of

the system after thermal equilibrium is reached. The specific heat capacity of

aluminum is 0.900 J/g◦C and the specific heat capacity of water is 4.184 J/g◦C.

Solution

Step 1: Calculate the heat lost by aluminum and gained by water using the

formula:

q=m·c·∆T

where qis the heat energy, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature. For aluminum:

qAl = 50.0 g ×0.900 J/g◦C×(Tfinal −120◦C)

For water:

qH2O= 100.0 g ×4.184 J/g◦C×(Tfinal −25◦C)

Step 2: Since energy is conserved, the heat lost by aluminum is equal to the

heat gained by water:

qAl =−qH2O

Step 3: Substitute the expressions for qAl and qH2Ointo the energy conser-

vation equation and solve for Tfinal:

50.0 g×0.900 J/g◦C×(Tfinal −120◦C) = −100.0 g×4.184 J/g◦C×(Tfinal −25◦C)

Step 4: Solve for Tfinal:

50.0×0.900 ×(Tfinal −120) = −100.0×4.184 ×(Tfinal −25)

45(Tfinal −120) = −418.4(Tfinal −25)

45Tfinal −5400 = −418.4Tfinal + 10460

463.4Tfinal = 15860

Tfinal = 34.26◦C

Therefore, the final temperature of the system after thermal equilibrium is

reached is 34.26◦C.

Question 3

Question

A student conducts a calorimetry experiment by mixing 200 mL of water at

C with 100 mL of methanol at 60

C in a calorimeter. Assuming no heat is

lost to surroundings, calculate the final temperature of the system. The specific

heat capacity of water is 4.18 J/g

C and the specific heat capacity of methanol

is 2.51 J/g

C. The density of water is 1 g/mL and the density of methanol is

0.791 g/mL.

Solution

Step 1: Calculate the initial heat content of water and methanol. The heat

content (q) for a substance is given by the formula:

q=m×c×∆T

where: - mis the mass of the substance, - cis the specific heat capacity of

the substance, and - ∆Tis the change in temperature.

For water: - mwater = 200 mL ×1 g/mL = 200 g - cwater = 4.18 J/g

C -

∆Twater =Tfinal −25C

For methanol: - mmethanol = 100 mL ×0.791 g/mL = 79.1 g - cmethanol =

2.51 J/g

C - ∆Tmethanol =Tfinal −60C

Step 2: Since heat is conserved in the system (no heat lost to the surround-

ings), the total initial heat content must equal the total final heat content.

qwater +qmethanol = 0

mwater ×cwater ×∆Twater +mmethanol ×cmethanol ×∆Tmethanol = 0

Substitute in the known values and solve for Tfinal.

Question 4

Question

A 50 g piece of aluminum at 80

C is placed in 200 g of water at 20

C. The final

temperature of the system is 22.5

C. Assuming no heat is lost to the surround-

ings, calculate the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by the aluminum and the heat gained by the

water using the formula:

Q=mc∆T

where: - Qis the heat energy, - mis the mass of the substance, - cis the specific

heat capacity of the substance, and - ∆Tis the change in temperature.

For the aluminum:

Qaluminum =maluminumcaluminum∆Taluminum

Qaluminum = 50 ×caluminum ×(22.5−80)

Qaluminum =−3575caluminum

For the water:

Qwater =mwatercwater∆Twater

Qwater = 200 ×4.18 ×(22.5−20)

Qwater = 4388 J

Step 2: Since heat is conserved in this closed system, the heat lost by the

aluminum is equal to the heat gained by the water:

Qaluminum =Qwater

−3575caluminum = 4388

caluminum =4388

3575

caluminum ≈1.226 J/g

Therefore, the specific heat capacity of aluminum is approximately 1.226

J/g

Question 5

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 25

C. Assuming no heat

is lost to the surroundings, find the specific heat capacity of aluminum. The

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down from 80

to 25

C. The specific heat capacity of aluminum is 0.897 J/g

C. The formula to

calculate heat energy is Q=mc∆T, where: - Qis the heat energy, - mis the

mass, - cis the specific heat capacity, - ∆Tis the change in temperature.

Given: m= 50 g, c= 0.897 J/g

C, and ∆T= 80C−25C= 55C.

Using the formula, we have: Qaluminum = (50 g)(0.897 J/g

C)(55C).

Calculating Qaluminum, we get: Qaluminum = 2481.75 J.

Step 2: Calculate the heat gained by the water and the calorimeter. Using

the same formula as before with the given values: Qwater = (200 g)(4.18 J/g

C)(25C−

20C).

Calculating Qwater, we get: Qwater = 418 J.

Step 3: Since the total heat lost by the aluminum block is equal to the heat

gained by the water and the calorimeter (assuming no heat loss to surroundings),

we have: Qaluminum =Qwater +Qcalorimeter.

Substitute the values we found into the equation: 2481.75 J = 418 J +

Qcalorimeter.

Solving for Qcalorimeter, we get: Qcalorimeter = 2063.75 J.

Step 4: The heat gained by the calorimeter is due to the temperature change

of the water and calorimeter together. Let the specific heat capacity of the

calorimeter be C. Since the water and calorimeter have the same temperature

change, we have: Qcalorimeter = (200 g + mcalorimeter)C(25C−20C).

Given that the mass of water is 200 g, the specific heat capacity of water is

4.18 J/g

C, and the final temperature is 25C.

Substitute the known values into the equation: 2063.75 J = (200 g+mcalorimeter)4.18 J/g

C×

5C.

Solving for mcalorimeter, we find: mcalorimeter = 105 g.

Step 5: Calculate the specific heat capacity of aluminum. Given that the

mass of the aluminum block is 50 g and the specific heat capacity of the alu-

minum is 0.897 J/g

C, we have: Qaluminum = (50 g)(0.897 J/g

C)(55C).

Substitute the values we know: 2481.75 J = (50 g)(0.897 J/g

C)(55C).

Solving for the specific heat capacity of aluminum, we get: caluminum =

0.897 J/g

Therefore, the specific heat capacity of aluminum is 0.897

Question 6

Question

A 50 g piece of aluminum at 100◦C is dropped into 200 g of water at 20◦C in

a calorimeter. The final temperature of the system is 25◦C. Assuming all the

heat lost by the aluminum is gained by the water and the calorimeter, what is

the heat capacity of the calorimeter? The specific heat capacity of aluminum is

0.90 J/g◦C and that of water is 4.18 J/g◦C.

Solution

Step 1: Calculate the heat lost by the aluminum. The heat lost by the aluminum

can be calculated using the formula:

Qaluminum =maluminum ×caluminum ×∆T

where: - maluminum is the mass of aluminum (50 g) - caluminum is the specific

heat capacity of aluminum (0.90 J/g◦C) - ∆Tis the change in temperature

(final temperature - initial temperature)

Qaluminum = 50 g ×0.90 J/g◦

×(25 −100)◦CQaluminum = 50 ×0.90 ×(−75)

Qaluminum =−3375 J

Step 2: Calculate the heat gained by the water and the calorimeter. Since

the heat lost by the aluminum is equal to the heat gained by the water and the

calorimeter, we have:

Qaluminum =Qwater +Qcalorimeter

Substitute the values we know:

−3375 J = (200 g + Ccalorimeter)×4.18 J/g◦

×(25 −20)◦C

−3375 J = (200 + Ccalorimeter)×4.18 ×5

−3375 = (200 + Ccalorimeter)×20.9

−3375 = 4180 + 20.9Ccalorimeter

20.9Ccalorimeter =−7555

Ccalorimeter =−361.2 J/◦CJ/◦CJ/◦CJ/◦C

Therefore, the heat capacity of the calorimeter is -361.2 J/◦C.

Question 7

Question

A 50 gram piece of iron at 80

C is placed in 200 grams of water at 20

C. If the

final temperature of the system is 25

C, what is the specific heat capacity of

iron? Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat lost by the iron and the heat gained by the water

using the formula:

qiron =−qwater

where q=mc∆T(m= mass, c= specific heat capacity, ∆T= change in

temperature).

Step 2: Calculate the heat lost by the iron:

qiron =mironciron∆Tiron

qiron = (0.05 kg)(ciron)(80 −25)

Step 3: Calculate the heat gained by the water:

qwater =mwatercwater∆Twater

qwater = (0.2 kg)(4190 J/kg◦C)(25 −20)

Step 4: Set the two equations equal and solve for the specific heat capacity

of iron:

(0.05 kg)(ciron)(55) = (0.2 kg)(4190 J/kg◦C)(5)

ciron =(0.2 kg)(4190 J/kg◦C)(5)

(0.05 kg)(55)

ciron = 3821 J/kg◦C

Question 8

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 25

C, calculate the specific

heat capacity of aluminum. Assume that the specific heat capacity of water is

4.18 J/g

C and neglect any heat loss to the surroundings.

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down from

C to 25

C. The formula for heat transfer is given by:

q=mc∆T

Where: m= mass of the substance (Aluminum block) in grams = 50 g c=

specific heat capacity of the substance (Aluminum) in J/g

C ∆T= change in

temperature = 25C−80C=−55C(negative since it is cooling down)

Substitute the values into the formula:

qAluminum = 50 ×c×(−55)

Step 2: Calculate the heat gained by the water as it warms up from 20

C to

C. Since the water gains the same amount of heat that the aluminum loses,

we can write:

qAluminum =qWater

The formula for heat transfer in water is similar to that of the aluminum

block:

qWater =mc∆T

Where: m= mass of the substance (Water) in grams = 200 g c= specific heat

capacity of the substance (Water) = 4.18 J/g

C ∆T= change in temperature

= 25C−20C= 5C

Substitute the values into the formula:

50 ×c×(−55) = 200 ×4.18 ×5

Step 3: Solve the equation obtained in Step 2 to find the specific heat ca-

pacity of aluminum (c).

50c×(−55) = 200 ×4.18 ×5

c=200 ×4.18 ×5

50 ×55

c=4180

c≈76.0 J/g

Therefore, the specific heat capacity of aluminum is approximately 76.0

J/g

Question 9

Question

A 50 g lead bullet at 100

C is fired into a large block of ice at 0

C. If the bullet

comes to rest after melting some ice, what is the mass of ice melted? (Specific

heat of lead = 0.128 cal/g

C, heat of fusion of ice = 79.7 cal/g, specific heat of

ice = 0.5 cal/g

Solution

Step 1: Calculate the heat lost by the lead bullet as it cools down to the final

temperature. The specific heat formula Q=mc∆Tcan be used where Qis the

heat absorbed or released, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature.

Given: mlead = 50 g, clead = 0.128 cal/g

C, Ti-lead = 100C,Tf-lead =x

Heat lost by lead = −(50 g)(0.128 cal/g

C)(x−100) cal

Step 2: Calculate the heat required to melt the ice. The heat required for

a phase change (melting/freezing) is given by Q=mL where Qis the heat

absorbed or released, mis the mass, and Lis the latent heat of fusion.

Given: Lice = 79.7 cal/g, mice =y

Heat needed to melt ice = (y)(79.7) cal

Step 3: Calculate the heat gained by the melted ice as it warms up to its

final temperature. The specific heat formula can be used again.

Given: mice =y,cice = 0.5 cal/g

C, Ti-ice = 0C,Tf-ice =x

Heat gained by ice = (y)(0.5)(x−0) cal

Step 4: Setting up the conservation of energy equation. The heat lost by the

lead bullet equals the heat gained by the melted ice as well as the heat required

to melt the ice.

−(50)(0.128)(x−100) = (y)(79.7) + (y)(0.5)(x)

Step 5: Solve for y. Solving the equation found in Step 4 will give the mass

of ice melted. Let’s solve this equation.

−6.4x+ 640 = 79.7y+ 0.5yx

Since the initial temperature of the ice is 0

C, we can simplify this further

to only consider the ice reaching the final temperature:

−6.4x+ 640 = 79.7y+ 0.5yx

−6.4x+ 640 = y(79.7+0.5x)

Hence, the mass of ice melted is given by y=−6.4x+640

79.7+0.5x.

Question 10

Question

A student performs a calorimetry experiment by mixing 200 g of water at 20◦C

with 100 g of an unknown metal at 100◦C. The final temperature of the mixture

is 25◦C. If the specific heat capacity of water is 4.18 J/(g·◦C), determine the

specific heat capacity of the metal.

Solution

Step 1: Calculate the heat lost by the hot metal and gained by the cold water

using the formula:

Qmetal =−Qwater

where

Q=mc∆T

Step 2: Calculate the heat lost by the metal:

Qmetal =mmetalcmetal∆Tmetal

where - mmetal = 100 g (mass of metal) - cmetal (specific heat capacity of the

metal, to be determined) - ∆Tmetal =Tf−Tmetal = 25 −100 = −75 ◦C

Step 3: Calculate the heat gained by the water:

Qwater =mwatercwater∆Twater

where - mwater = 200 g (mass of water) - cwater = 4.18 J/(g·◦C) (specific heat

capacity of water) - ∆Twater =Tf−Twater = 25 −20 = 5 ◦C

Step 4: Set up the equation:

100 ×cmetal × −75 = −200 ×4.18 ×5

Step 5: Solve for cmetal:

cmetal =−200 ×4.18 ×5

100 × −75

Step 6: Calculate the specific heat capacity of the metal:

cmetal = 1.39 J/(g·◦C)

Therefore, the specific heat capacity of the metal is 1.39 J/(g·◦C).

Question 11

Question

A 50.0 g piece of aluminum at 80.0

C is dropped into 100.0 g of water at 20.0

Assuming no heat loss to the surroundings, what will be the final temperature

of the system? (Specific heat capacity of aluminum = 0.900 J/g

C, specific heat

capacity of water = 4.18 J/g

C, and the heat of fusion of water = 334 J/g)

Solution

Step 1: Calculate the heat lost by the aluminum as it cools down to the final

temperature. The heat lost is given by the formula:

Qlost, Al =m·c·∆T

where: - m= 50.0 g is the mass of the aluminum, - c= 0.900 J/g

C is the

specific heat capacity of aluminum, - ∆T=Tinitial −Tfinal, - Tinitial = 80.0Cis

the initial temperature of the aluminum, and - Tfinal is the final temperature of

the system.

Substitute the values:

Qlost, Al = 50.0 g ×0.900 J/g

C×(80.0−Tfinal)C

Step 2: Calculate the heat gained by the water as it warms up to the final

temperature. The heat gained by the water is given by the formula:

Qgain, water =m·c·∆T

where: - m= 100.0 g is the mass of the water, - c= 4.18 J/g

C is the specific

heat capacity of water, - ∆T=Tfinal −Tinitial, - Tinitial = 20.0Cis the initial

temperature of the water.

Substitute the values:

Qgain, water = 100.0 g ×4.18 J/g

C×(Tfinal −20.0)C

Step 3: Set up the equation based on the law of conservation of energy. The

heat lost by the aluminum must equal the heat gained by the water since no

energy is lost to the surroundings:

Qlost, Al =Qgain, water

Step 4: Solve the equation to find the final temperature of the system.

50.0 g ×0.900 J/g

C×(80.0−Tfinal)C= 100.0 g ×4.18 J/g

C×(Tfinal −20.0)C

Solve for Tfinal to find the final temperature of the system.

Question 12

Question

A student wants to determine the specific heat capacity of a metal block. The

student immerses the metal block, initially at a temperature of 200

C, into

a 1.5 kg of water at 20

C contained in a calorimeter. The final equilibrium

temperature of the system is found to be 40

C. If the specific heat capacity of

water is 4186 J/kg◦C, determine the specific heat capacity of the metal block.

Solution

Let’s denote the specific heat capacity of the metal block as cmand the mass

of the metal block as mm. We can use the principle of conservation of energy

to solve for cm.

Step 1: Calculate the heat lost by the metal block and the heat

gained by the water. The heat lost by the metal block is equal to the heat

gained by the water. Therefore, we can write:

mmcm∆T=mwcw∆T

where: mm= 1.5 kg (mass of water), cw= 4186 J/kg◦C (specific heat capacity

of water), ∆T= 20◦C (change in temperature).

Step 2: Calculate the change in temperature for the metal block.

The temperature of the metal block changes from 200

C to 40

C, thus ∆Tm=

40◦C−200◦C = −160◦C.

Step 3: Substitute known values into the heat transfer equation.

Substitute mm= 1.5 kg, cw= 4186 J/kg◦C, ∆T= 20◦C, and ∆Tm=−160◦C

into the heat transfer equation:

1.5cm(−160) = 1.5·4186 ·20

−240cm= 125580

cm=125580

−240

cm=−523.25 J/kg◦C

Step 4: Conclusion The specific heat capacity of the metal block is

−523.25 J/kg◦C. It is negative because heat transfer from the metal block causes

a temperature decrease of the metal block.

Question 13

Question

A 50.0 g sample of aluminum at 90.0

C is added to a calorimeter containing

100.0 g of water at 25.0

C. If the final temperature of both the aluminum and

water is 28.0

C, what is the specific heat capacity of aluminum? Assume the

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat absorbed by the water using the formula

qwater =mwater ×cwater ×∆T

where - mwater = 100.0 g is the mass of water, - cwater = 4.18 J/g

C is the specific

heat capacity of water, and - ∆T=Tfinal −Tinitial = 28.0C−25.0C= 3.0Cis

the change in temperature. Substitute the values into the formula to find qwater.

Step 2: Calculate the heat lost by the aluminum using the formula

qaluminum =−qwater

because heat lost by the aluminum is equal in magnitude but opposite in sign

to the heat gained by the water.

Step 3: Calculate the heat lost by the aluminum using its specific heat

capacity with the formula

qaluminum =maluminum ×caluminum ×∆T

where - maluminum = 50.0 g is the mass of aluminum, - caluminum is the specific

heat capacity of aluminum, and - ∆T=Tfinal −Tinitial = 28.0C−90.0C=

−62.0Cis the change in temperature.

Step 4: Set up and solve an equation to find caluminum using the relationships

between the heat gained by the water and the heat lost by the aluminum.

This equation can be written as

mwater ×cwater ×∆T=maluminum ×caluminum ×∆T

Substitute the known values and solve for caluminum.

Question 14

Question

A 50 g block of copper initially at 150◦C is dropped into 200 g of water at 20◦C.

The final temperature of the system is 22◦C. Assuming no heat is lost to the

surroundings, calculate the specific heat capacity of copper. The specific heat

capacity of water is 4.18 J/g◦C.

Solution

Step 1: Determine the heat transfer from the copper block to the water. The

heat lost by the copper block equals the heat gained by the water. Let qCu be

the heat lost by the copper block and qH2O be the heat gained by the water.

qCu =−qH2O

Step 2: Calculate the heat lost by the copper block. The heat lost by the

copper block is given by the formula:

qCu =−mc∆T

where mis the mass, cis the specific heat capacity, and ∆Tis the change

in temperature.

Substitute the given values: m= 50 g, c= specific heat capacity of copper,

∆T= (22 −150) ◦C = −128 ◦C.

qCu = 50 ·c·(−128) = −6400cJ

Step 3: Calculate the heat gained by the water. The heat gained by the

water is also given by the formula:

qH2O =mc∆T

Substitute the given values: m= 200 g, c= 4.18 J/g◦C, ∆T= (22 −20) ◦C

= 2 ◦C.

qH2O = 200 ·4.18 ·2 = 1672 J

Step 4: Set up the equation and solve for the specific heat capacity of copper.

Since qCu =−qH2O, we have

−6400c= 1672

Solving for c:

c=1672

6400 = 0.2615 J/g◦C

Therefore, the specific heat capacity of copper is 0.2615 J/g◦C.

Question 15

Question

A 200 g aluminum calorimeter initially at 20

C contains 500 g of water at 40

A 50 g iron ball at 100

C is placed into the calorimeter, causing the temperature

to stabilize at 30

C. Assuming no heat is lost to the surroundings, what is the

specific heat capacity of the iron ball? (Specific heat capacities: aluminum =

0.902 J/g

C, water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the iron ball as it cools to 30

The heat lost by the iron ball is equal to the heat gained by the aluminum

calorimeter, water, and the final mixture. Let Ciron be the specific heat capacity

of iron. The heat lost by the iron ball is given by:

Qiron =miron ·Ciron ·(Tf−Ti)

Qiron = 50 g ·Ciron ·(30 −100)

Qiron =−3500 ·Ciron J

Step 2: Calculate the heat gained by the water and aluminum calorimeter.

The heat gained by the water and aluminum calorimeter is given by:

Qwater =mwater ·Cwater ·(Tf−Ti)

Qaluminum =maluminum ·Caluminum ·(Tf−Ti)

We know that the total heat gained is equal to the heat lost by the iron ball:

Qwater +Qaluminum =−3500 ·Ciron J

Substitute the given values and solve for Ciron:

500 g ·4.18 J/g

C + 200 g ·0.902 J/g

C = −3500 ·Ciron

2090 + 180 ≈ −3500 ·Ciron

2270 ≈ −3500 ·Ciron

Ciron ≈ −0.648 J/g

Therefore, the specific heat capacity of the iron ball is approximately 0.648 J/g

Question 16

Question

A piece of iron with a mass of 150 grams at a temperature of 200◦C is placed

into 500 grams of water at 20◦C in a calorimeter. The final temperature of

the system is 25◦C. Assuming no heat is lost to the surroundings, calculate the

specific heat capacity of iron. (Specific heat capacity of water = 4.18 J/g◦C)

Solution

Step 1: Calculate the heat lost by the iron and the heat gained by the water

using the formula for heat transfer:

q=mc∆T

where qis the heat transferred, mis the mass, cis the specific heat capacity,

and ∆Tis the change in temperature.

For the iron:

qiron =mironciron∆Tiron

qiron = (0.150 kg)(ciron)(200 −25)

For the water:

qwater =mwatercwater∆Twater

qwater = (0.500 kg)(4.18 J/g◦C)(25 −20)

Step 2: Since the heat lost by the iron is equal to the heat gained by the

water (assuming no heat loss to the surroundings):

qiron =qwater

(0.150)(ciron)(175) = (0.500)(4.18)(5)

Step 3: Solve for the specific heat capacity of iron, ciron:

ciron =(0.500)(4.18)(5)

(0.150)(175)

ciron =10.45

26.25

ciron = 0.398 J/g◦C

Question 17

Question

A 50.0 g piece of aluminum at 90.0

C is placed in 100.0 g of water at 20.0

The final temperature of the mixture is 22.5

C. Assuming no heat is lost to the

surroundings, calculate the specific heat capacity of aluminum. Given: specific

heat capacity of water = 4.18 J/(g◦C).

Solution

Step 1: Calculate the heat lost by aluminum and the heat gained by water. The

heat lost by aluminum is equal to the heat gained by water.

Heat lost by aluminum = Heat gained by water

m1·c1·∆T1=m2·c2·∆T2

where: m1= 50.0 g (mass of aluminum), c1=? (specific heat capacity of alu-

minum), ∆T1=−67.5◦C (temperature change for aluminum), m2= 100.0 g

(mass of water), c2= 4.18 J/(g◦C) (specific heat capacity of water), ∆T2=

2.5◦C (temperature change for water).

Step 2: Substitute the values and solve for the specific heat capacity of

aluminum.

50.0 g ·c1·(−67.5◦C) = 100.0 g ·4.18 J/(g◦C) ·2.5◦C

c1=100.0 g ·4.18 J/(g◦C) ·2.5◦C

50.0 g ·(−67.5◦C)

c1=1045.0 J

−3375 g ·◦C

c1=−0.309 J/(g◦C)

Therefore, the specific heat capacity of aluminum is −0.309 J/(g◦C).

Question 18

Question

A 50.0 g piece of copper is heated to 95.0

C and then placed into 100.0 g of

water at 25.0

C. The final temperature of the system is 26.0

C. Assuming all

heat is transferred to the water, determine the specific heat of copper.

Solution

Step 1: Calculate the heat absorbed by the water The heat absorbed by the

water can be calculated using the formula:

qwater =mwater ·cwater ·∆T

where: - mwater is the mass of water (100.0 g), - cwater is the specific heat

capacity of water (4.18 J/g

C), and - ∆Tis the temperature change of the

water (final temperature - initial temperature).

Substitute the values into the formula:

qwater = (100.0 g) ·(4.18 J/g

C) ·(26.0−25.0)C

qwater = 100 ·4.18 ·1 = 418 J

Step 2: Calculate the heat lost by the copper The heat lost by the copper is

equal to the heat gained by the water, so:

qcopper =−qwater =−418 J

Step 3: Calculate the specific heat of copper The heat lost by the copper

can be calculated using the formula:

qcopper =mcopper ·ccopper ·∆T

where: - mcopper is the mass of copper (50.0 g), - ccopper is the specific heat

capacity of copper (unknown), and - ∆Tis the temperature change of the copper

(final temperature - initial temperature).

Substitute the values into the formula and solve for ccopper:

−418 = (50.0 g) ·ccopper ·(26.0−95.0)C

ccopper =−418

50 ·(26.0−95.0)

ccopper ≈0.39 J/g

Therefore, the specific heat of copper is approximately 0.39 J/g

Question 19

Question

A 50 g piece of copper at 95

C is placed in 200 g of water at 25

C in an insulated

container. Assuming no heat is lost to the surroundings, what will be the final

temperature of the system? (Specific heat capacity of copper is 0.385 J/g

C and

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat gained by the water.

The heat gained by the water can be calculated using the formula:

Qwater =mwater ×cwater ×∆T

where: - mwater = 200 g (mass of water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T=Tfinal −Tinitial.

The initial temperature of water, Tinitial = 25C.

Step 2: Calculate the heat lost by the copper.

The heat lost by the copper can be calculated using the formula:

Qcopper =mcopper ×ccopper ×∆T

where: - mcopper = 50 g (mass of copper), - ccopper = 0.385 J/g

C (specific heat

capacity of copper), - ∆T=Tfinal −Tinitial.

The initial temperature of copper, Tinitial = 95C.

Step 3: Set up the energy conservation equation.

Since no heat is lost to the surroundings, the heat gained by the water must

be equal to the heat lost by the copper.

Qwater =Qcopper

Step 4: Solve for the final temperature.

Substitute the expressions for Qwater and Qcopper into the energy conserva-

tion equation and solve for the final temperature, Tfinal.

mwater ×cwater ×∆T=mcopper ×ccopper ×∆T

200 ×4.18 ×(Tfinal −25) = 50 ×0.385 ×(95 −Tfinal)

Solve this equation to find the final temperature, Tfinal.

Question 20

Question

A 50 g piece of copper is heated to 100

C and then placed in 200 g of water at

C. The final temperature of the system is 25

C. Assuming no heat is lost to

the surroundings, calculate the specific heat capacity of copper. (Specific heat

capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper piece as it cools down from

100

C to 25

C. The heat lost is given by the formula: Q=mc∆T, where: -

mis the mass of the copper piece (in grams), - cis the specific heat capacity

of copper (in J/g

C), - ∆Tis the change in temperature (final temperature -

initial temperature).

Given: - mcopper = 50 g, - Tinitial = 100

C, - Tfinal = 25

Substitute these values into the formula: Qcopper = (50 g)(c)(25

Step 2: Calculate the heat gained by the water as it heats up from 20

to 25

C. The heat gained is given by the same formula: Q=mc∆T, where:

-mis the mass of the water (in grams), - cis the specific heat capacity of

water (in J/g

C), - ∆Tis the change in temperature (final temperature - initial

temperature).

Given: - mwater = 200 g, - cwater = 4.18 J/g

C, - Tinitial = 20

C, - Tfinal =

Substitute these values into the formula: Qwater = (200 g)(4.18 J/g

C)(5

Step 3: Since heat lost by the copper equals heat gained by the water (as-

suming no heat loss to the surroundings), we can set the two equations equal

to each other: (50 g)(c)(25

C) = (200 g)(4.18 J/g

C)(5

Step 4: Solve for the specific heat capacity of copper, c:c=(200 g)(4.18 J/g

C)(5

(50 g)(25

c=4180 J

1250 g

c= 3.344 J/g

Therefore, the specific heat capacity of copper is 3.344 J/g

Question 21

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in

a calorimeter. The specific heat capacity of aluminum is 0.90 J/g

C and the

specific heat capacity of water is 4.18 J/g

C. If the final temperature of the

system is 24.5

C, determine the heat capacity of the calorimeter.

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down to the

final temperature of the system. The formula for heat transfer is given by

Q=mc∆T,

where Qis the heat transfer, mis the mass of the substance, cis the specific

heat capacity, and ∆Tis the change in temperature.

The heat lost by the aluminum block is calculated as follows:

QAluminum = (50 g)(0.90 J/g

C)(80 −24.5) = 2075 J.

Step 2: Calculate the heat gained by the water and the calorimeter. The

heat gained by the water and calorimeter is given by

Qwater+calorimeter = (50 g + 200 g)(4.18 J/g

C)(24.5−20) = 3503 J.

Step 3: Set up an equation to determine the heat capacity of the calorimeter.

Since the heat lost by the aluminum block is equal to the heat gained by the

water and calorimeter (law of conservation of energy), we have

QAluminum =Qwater+calorimeter.

Step 4: Solve for the heat capacity of the calorimeter.

2075 J = 3503 J + Ccalorimeter,

where Ccalorimeter is the heat capacity of the calorimeter.

Therefore, the heat capacity of the calorimeter is

Ccalorimeter = 2075 J −3503 J = −1428 J.

The negative sign indicates that the calorimeter releases 1428 J of heat,

which makes sense since the calorimeter would lose heat to the surroundings

during the experiment.

Question 22

Question

A 50 g piece of aluminum at 80

C is dropped into 200 g of water at 20

in a calorimeter. The final temperature of the system is 25

C. Calculate the

specific heat capacity of aluminum. Assume the specific heat capacity of water

is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the aluminum piece. The formula for calcu-

lating heat transfer is:

Q=mc∆T

where: Q= heat transfer (in J) m= mass of the substance (in g) c= specific

heat capacity of the substance (in J/g

C) ∆T= change in temperature (in

Given: maluminum = 50 g cwater = 0.903 J/g

C ∆Taluminum = (80 −25)

C =

Substitute the values into the formula:

Qaluminum = 50 g ×caluminum ×55

Step 2: Calculate the heat gained by the water. Since the system is isolated,

the heat lost by the aluminum piece is equal to the heat gained by the water.

Therefore:

Qaluminum =Qwater

Using the same formula and the given values:

Qwater = 200 g ×4.18 J/g

C×(25 −20)

Step 3: Set the two equations equal to each other:

50 g ×caluminum ×55

C = 200 g ×4.18 J/g

C×5

Step 4: Solve for caluminum.

caluminum =200 g ×4.18 J/g

C×5

50 g ×55

Question 23

Question

A 50 g piece of aluminum at 80◦C is dropped into 200 g of water at 20◦C in a

calorimeter. The final temperature of the system is 25◦C. Assuming the specific

heat capacity of water is 4.18 J/g◦C and that of aluminum is 0.897 J/g◦C,

calculate the specific heat capacity of the calorimeter.

Solution

Step 1: Calculate the heat gained by water: The heat gained by the water is

given by the equation

Q=mc∆T

where: - mis the mass of water (200 g), - cis the specific heat capacity of water

(4.18 J/g◦C), - ∆Tis the change in temperature of the water.

The change in temperature of the water is

∆T=Tf−Ti= 25◦C−20◦C= 5◦C.

Substitute the values in:

Q= (200 g)(4.18 J/g◦C)(5◦C)

Q= 4180 J

Step 2: Calculate the heat lost by the aluminum: The heat lost by the

aluminum is given by the equation

Q=mc∆T

where: - mis the mass of aluminum (50 g), - cis the specific heat capacity of

aluminum (0.897 J/g◦C), - ∆Tis the change in temperature of the aluminum.

The change in temperature of the aluminum is

∆T=Tf−Ti= 25◦C−80◦C=−55◦C.

Notice that the negative sign indicates that the aluminum is losing heat

energy.

Substitute the values in:

Q= (50 g)(0.897 J/g◦C)(−55◦C)

Q=−2456.25 J

Step 3: Calculate the heat absorbed by the calorimeter (water + aluminum):

Since we assume no heat is lost to the surroundings, the heat lost by the alu-

minum is equal to the heat gained by the water and the calorimeter:

Qcalorimeter =−Qaluminum =Qwater

Qcalorimeter = 4180 J

Step 4: Calculate the heat capacity of the calorimeter: We know the heat

capacity formula is

Q=mc∆T

where: - mis the mass of the calorimeter, - cis the specific heat capacity of the

calorimeter, - ∆Tis the change in temperature of the calorimeter.

Substitute the values in:

4180 J = (m)(c)(5◦C)

We are looking for c, the specific heat capacity of the calorimeter.

Solving for c, we get

c=4180 J

(m)(5◦C)

Since we don’t have the mass of the calorimeter, further calculation is not

possible based on the information provided.

Question 24

Question

A piece of metal of mass 120.0 g at a temperature of 200.0

C is placed in 250.0

g of water at 20.0

C. The final equilibrium temperature of the system is 26.5

Assume the specific heat capacities of the metal and water are constant at 0.385

J/g

C and 4.18 J/g

C, respectively. Determine the specific heat capacity of the

metal.

Solution

Step 1: Calculate the heat gained by the metal as it cools down to the final

temperature. The formula for heat gained or lost is given by:

q=mc∆T

Where: - qis the heat gained or lost - mis the mass of the object - cis the

specific heat capacity of the substance - ∆Tis the change in temperature

Given that the metal is cooling down, the heat gained by the metal is equal

to the heat lost by the water. Thus, we can set up the equation:

mmetalcmetal∆Tmetal =−mwatercwater∆Twater

Substitute the values to find cmetal:

120.0 g ·cmetal ·(26.5C−26.5C) = −250.0 g ·4.18 J/g

C·(26.5C−20.0C)

120.0 g ·cmetal ·0 = −250.0 g ·4.18 J/g

C·6.5C

0 = −250.0·4.18 ·6.5

0 = −6505.0

Since our calculated value is negative, there seems to be an error in our

calculation. Let’s go back and verify our computations.

Question 25

Question

A 50 g piece of aluminum at 100

C is placed in 150 g of water at 20

C. If the

final temperature of the system is 25

C, calculate the specific heat capacity of

aluminum. (Specific heat capacity of water = 4.18 J/g

Solution

Step 1: First, we find the heat absorbed by the aluminum and the heat released

by the water. The heat lost by the aluminum equals the heat gained by the

water since no heat is lost to the surroundings. Let the specific heat capacity

of aluminum be cAl.

Step 2: The heat absorbed by the aluminum can be calculated using the

formula:

qAl =mAl ·cAl ·∆TAl

where

mAl = 50 g

∆TAl = (25 −100) = −75

Therefore,

qAl = 50 g ·cAl ·(−75)

Step 3: The heat released by the water can be calculated using the formula:

qH2O =mH2O ·cH2O ·∆TH2O

where

mH2O = 150 g

cH2O = 4.18 J/g

∆TH2O = (25 −20) = 5

Therefore,

qH2O = 150 g ·4.18 J/g

C·5

Step 4: Since the heat lost by the aluminum equals the heat gained by the

water:

qAl =qH2O

50 g ·cAl ·(−75) = 150 g ·4.18 J/g

C·5

Step 5: Solve for cAl to find the specific heat capacity of aluminum.

50 g ·cAl ·(−75) = 150 g ·4.18 J/g

C·5

cAl =150 g ·4.18 J/g

C·5

50 g ·(−75)

cAl =3135 J

−3750 g

Step 6: Calculate the specific heat capacity of aluminum.

cAl =−0.836 J/g

Therefore, the specific heat capacity of aluminum is −0.836 J/g

Question 26

Question

A student conducts an experiment to determine the specific heat capacity of

a metal by immersing it in a container of water. The metal is heated to a

high temperature and then quickly transferred to the water, causing the water’s

temperature to rise. The student records the initial temperature of the water

as 20

C and the final temperature as 32

C. The mass of the metal is 0.2 kg and

its initial temperature is 100

C. The mass of the water is 0.5 kg and its specific

heat capacity is 4186 J/kg ·K. If the final temperature of the metal and water

is 30

C, what is the specific heat capacity of the metal?

Solution

Step 1: Calculate the heat absorbed by the water. The heat absorbed by the

water can be calculated using the formula:

Qwater =mc∆T

where: m= mass of water = 0.5 kg, c= specific heat capacity of water =

4186 J/kg ·K, ∆T= change in temperature = final temperature - initial tem-

perature = 32C−20C= 12C.

Substitute the values into the formula:

Qwater = 0.5 kg ×4186 J/kg ·K×12C= 25116 J

Step 2: Calculate the heat lost by the metal. The heat lost by the metal can

be calculated using the formula:

Qmetal =mc∆T

where: m= mass of metal = 0.2 kg, c= specific heat capacity of the metal

(to be determined), ∆T= change in temperature = initial temperature - final

temperature = 100C−30C= 70C.

Substitute the values into the formula:

Qmetal = 0.2 kg ×c×70C

Step 3: Set up the heat balance equation. Since the system is isolated and

there is no heat lost to the surroundings, the heat lost by the metal is equal to

the heat gained by the water.

Qwater =Qmetal

0.2 kg ×c×70C= 25116 J

Step 4: Solve for the specific heat capacity of the metal.

0.2 kg ×c×70C= 25116 J

c=25116 J

0.2 kg ×70C

c= 179.4 J/kg ·K

Therefore, the specific heat capacity of the metal is 179.4 J/kg ·K.

Question 27

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C in a calorimeter.

If the final temperature of the system is 25

C, what is the specific heat capacity

of the iron? Assume the specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat gained or lost by the water. Given: Mass of water,

mw= 200 g Initial temperature of water, Tin = 20

C Final temperature of the

system, Tfinal = 25

C Specific heat capacity of water, Cw= 4.18 J/g

The heat gained by the water is given by the formula:

Qwater =mw·Cw·∆T

where ∆T=Tfinal −Tin.

Substitute the given values:

∆T= 25

C−20

C=5

Qwater = 200 g ·4.18 J/g

C·5

Step 2: Calculate the heat gained or lost by the iron. Given: Mass of iron,

miron = 50 g Initial temperature of iron, Tin = 80

C Final temperature of the

system, Tfinal = 25

The heat lost by the iron is equal to the heat gained by the water (since this

is a closed system). Therefore:

Qwater =Qiron

mw·Cw·∆T=miron ·Ciron ·∆Tiron

Substitute the known values:

200 g ·4.18 J/g

C·5

C = 50 g ·Ciron ·(80

C−25

Step 3: Solve for the specific heat capacity of iron.

200 ·4.18 ·5 = 50 ·Ciron ·55

Ciron =200 ·4.18 ·5

50 ·55

Calculating the specific heat capacity of iron:

Ciron =4180

110 = 38 J/g

Therefore, the specific heat capacity of the iron is 38 J/g

Question 28

Question

A piece of copper of mass 200 g at a temperature of 100

C is placed in 400 g of

water at 20

C. Assuming no heat is lost to the surroundings, calculate the final

temperature of the mixture. The specific heat capacity of copper is 0.385 J/g

and the specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper when it cools down to the final

temperature.

The heat lost by the copper can be calculated using the formula:

Qcopper =mc∆T

where: - mis the mass of the copper (200 g), - cis the specific heat capacity

of copper (0.385 J/g

C), - ∆Tis the temperature change of the copper.

The temperature change of the copper can be calculated as follows:

∆T=Tfinal −Tinitial

∆T=Tfinal −100C

Step 2: Calculate the heat gained by the water when it warms up to the

final temperature.

The heat gained by the water can be calculated using the formula:

Qwater =mc∆T

where: - mis the mass of the water (400 g), - cis the specific heat capacity

of water (4.18 J/g

C), - ∆Tis the temperature change of the water.

The temperature change of the water can be calculated as follows:

∆T=Tfinal −Tinitial

∆T=Tfinal −20C

Step 3: Set up the equation based on the conservation of energy.

Since there is no heat loss to the surroundings, the heat lost by the copper

must equal the heat gained by the water. Thus, we have:

Qcopper =Qwater

Step 4: Solve for the final temperature, Tfinal.

Substitute the expressions for Qcopper and Qwater into the equation from

Step 3 and solve for Tfinal.

mc∆Tcopper =mc∆Twater

Finally, solve for Tfinal.

Question 29

Question

A 150 g aluminum block at 175◦C is placed into a container of water at 20◦C.

The water has a mass of 200 g. The final temperature of the system is 25◦C.

Assuming no heat is lost to the surroundings, calculate the specific heat capacity

of aluminum. (Specific heat of water = 4.18 J/g◦C)

Solution

Step 1: Calculate the heat absorbed by the aluminum block as it cools down:

The heat absorbed by the aluminum block can be calculated using the formula:

qAl =mAl ·cAl ·∆T

Where: - mAl = mass of aluminum block = 150 g - cAl = specific heat capacity of

aluminum (to be determined) - ∆T= change in temperature of aluminum block

= 25◦C−175◦C = −150◦C (negative sign indicates a decrease in temperature)

Substitute the given values into the formula:

qAl = 150 g ×cAl ×(−150◦C)

Step 2: Calculate the heat released by the cooling water as it heats up: The

heat released by the water can be calculated using the same formula but with

the specific heat capacity of water:

qwater =mwater ×cwater ×∆T

Where: - mwater = mass of water = 200 g - cwater = specific heat capacity of

water = 4.18 J/g◦C-∆T= change in temperature of water = 25◦C−20◦C =

5◦C

Substitute the given values into the formula:

qwater = 200 g ×4.18 J/g◦C×5◦C

Step 3: Since the system is isolated and assuming no heat is lost to the

surroundings, the heat absorbed by the aluminum block is equal to the heat

released by the water:

qAl =qwater

Step 4: Solve for the specific heat capacity of aluminum: Set the two heat

values equal to each other and solve for cAl:

150 g ×cAl ×(−150◦C) = 200 g ×4.18 J/g◦C×5◦C

cAl =200 g ×4.18 J/g◦C×5◦C

150 g ×(−150◦C)

cAl =2090 J

−22500 g◦C

cAl ≈ −0.093 J/g◦C

The specific heat capacity of aluminum is approximately −0.093 J/g◦C.

Question 30

Question

A 50 g block of copper at 100◦C is dropped into 200 g of water at 20◦C contained

in a 100 g aluminum calorimeter cup. The final temperature of the system is

30◦C. Given the specific heat capacities of copper, water, and aluminum as

0.385 J/g

C, 4.18 J/g

C, and 0.897 J/g

C, respectively, calculate the initial

temperature of the water before the copper block was dropped into it.

Solution

Step 1: Calculate the heat absorbed by the copper block. The heat absorbed

by the copper block can be calculated using the formula:

Q=mc∆T

where: - mis the mass of the copper block (50 g), - cis the specific heat capacity

of copper (0.385 J/g

C), and - ∆Tis the change in temperature of the copper

block (30 −100 = −70

C).

Substitute the given values into the formula to calculate Q:

Q= 50 ×0.385 ×(−70)

Step 2: Calculate the heat lost by the water. The heat lost by the water can

be calculated using the formula:

Q=mc∆T

where: - mis the mass of the water (200 g), - cis the specific heat capacity

of water (4.18 J/g

C), and - ∆Tis the change in temperature of the water

(30 −x

C, where xis the initial temperature of the water).

Substitute the given values into the formula to calculate Q.

Step 3: Calculate the heat absorbed by the aluminum calorimeter cup. The

heat absorbed by the aluminum calorimeter cup can be calculated using the

formula:

Q=mc∆T

where: - mis the mass of the aluminum calorimeter cup (100 g), - cis the

specific heat capacity of aluminum (0.897 J/g

C), and - ∆Tis the change in

temperature of the aluminum calorimeter cup (30 −20 = 10

C).

Substitute the given values into the formula to calculate Q.

Step 4: Apply the principle of conservation of energy. According to the

principle of conservation of energy, the heat lost by the copper block is equal to

the sum of the heat gained by the water and the aluminum calorimeter cup:

Qcopper =Qwater +Qcup

Substitute the calculated values of Qcopper,Qwater, and Qcup into the equa-

tion and solve for x, the initial temperature of the water.

Question 31

Question

A 50 g aluminum block at 80

C is placed in 200 g of water at 20

C in a calorime-

ter. The final temperature of the system is 30

C. Assuming no heat is lost to

the surroundings and that the specific heat capacity of water is 4186 J/kg

calculate the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by the aluminum block

Heat lost by aluminum = m×c×∆T

where m= 50 g = 0.05 kg (mass of aluminum block), c(specific heat capacity

of aluminum),

∆T=Tf−Ti= 30 −80 = −50

Therefore,

Heat lost by aluminum = 0.05 ×c×(−50)

Step 2: Calculate the heat gained by water

Heat gained by water = m×cw×∆T

where m= 200 g = 0.2 kg (mass of water), cw= 4186 J/kg

C (specific heat

capacity of water),

∆T=Tf−Ti= 30 −20 = 10

Therefore,

Heat gained by water = 0.2×4186 ×10

Step 3: Since there is no heat loss to the surroundings, the heat lost by the

aluminum block is equal to the heat gained by the water.

0.05 ×c×(−50) = 0.2×4186 ×10

Step 4: Solve for the specific heat capacity of aluminum, c

0.05 ×c×(−50) = 0.2×4186 ×10

c=0.2×4186 ×10

0.05 ×(−50)

Question 32

Question

A piece of aluminum of mass 250 g at a temperature of 90

C is placed in 500 g of

water at 20

C in a calorimeter of negligible heat capacity. The final temperature

of the system is 23

C. Assuming no heat loss to the surroundings, calculate the

specific heat capacity of aluminum. (Specific heat capacity of water = 4.18

J/g

Solution

Step 1: Calculate the heat gained by the water. Given: mwater = 500 g

∆Twater = 23−20 = 3

Ccwater = 4.18 J/g

C Using the formula Q=mc∆T, the

heat gained by the water can be calculated as: Qwater = (500 g)×(4.18 J/g

C)×

Step 2: Calculate the heat lost by the aluminum. Given: maluminum =

250 g ∆Taluminum = 23 −90 = −67

C (negative as aluminum is losing heat)

Let caluminum be the specific heat capacity of aluminum. Using the formula

Q=mc∆T, the heat lost by the aluminum can be calculated as: Qaluminum =

(250 g) ×caluminum ×(−67)

Step 3: Since no heat is lost to the surroundings, the heat gained by the

water is equal to the heat lost by the aluminum. Therefore, we have: Qwater =

Qaluminum

Step 4: Equate Qwater and Qaluminum and solve for caluminum to find the

specific heat capacity of aluminum. This is left as an exercise for the reader.

Question 33

Question

A 50 g piece of copper at 150

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the mixture is 25

C. Assuming no heat

is lost to the surroundings, calculate the specific heat capacity of copper. The

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper and the heat gained by the water.

The formula for heat transfer in calorimetry is given by:

Q=mc∆T

where: Q= heat energy transferred, m= mass of the material, c= specific

heat capacity of the material, ∆T= change in temperature.

Let’s first calculate the heat lost by the copper piece: Given: mcopper = 50 g,

cwater = 0.387 J/g

C (specific heat capacity of copper), Tinitial = 150C,Tfinal =

25C.

The change in temperature for copper (∆Tcopper) is:

∆Tcopper =Tfinal −Tinitial = 25C−150C=−125C

(The negative sign indicates the decrease in temperature).

Now, calculate the heat lost by the copper:

Qcopper =mcopper ·ccopper ·∆Tcopper

Step 2: Calculate the heat gained by the water.

Given: mwater = 200 g, cwater = 4.18 J/g

C (specific heat capacity of water),

Tinitial = 20C,Tfinal = 25C.

The change in temperature for water (∆Twater) is:

∆Twater =Tfinal −Tinitial = 25C−20C= 5C

Now, calculate the heat gained by the water:

Qwater =mwater ·cwater ·∆Twater

Step 3: Set up the equation for heat conservation. Since no heat is lost to

the surroundings, the heat lost by the copper should be equal to the heat gained

by the water:

Qcopper =Qwater

Step 4: Solve for the specific heat capacity of copper. Equating the two

equations:

mcopper ·ccopper ·∆Tcopper =mwater ·cwater ·∆Twater

Substitute the known values and solve for ccopper to find the specific heat

capacity of copper.

Question 34

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 25

C, calculate the specific

heat capacity of aluminum. Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat absorbed by the aluminum block. The heat absorbed

by the aluminum block can be calculated using the formula:

Qaluminum =mc∆T

Where: - mis the mass of the aluminum block (50 g) - cis the specific heat

capacity of aluminum - ∆Tis the temperature change of the aluminum block

(final temperature - initial temperature)

Given that the initial temperature of the aluminum block is 80

C and the

final temperature is 25

C, we have:

∆T= 25C−80C=−55C

Substitute the known values into the formula:

Qaluminum = (50 g)(c)(−55C)

Step 2: Calculate the heat released by the aluminum block to the water.

The heat released by the aluminum block is equal to the heat absorbed by the

water, which can be calculated using the formula:

Qwater =mc∆T

Where: - mis the mass of the water (200 g) - cis the specific heat capacity of

water - ∆Tis the temperature change of the water (final temperature - initial

temperature)

Given that the initial temperature of the water is 20

C and the final tem-

perature is 25

C, we have:

∆T= 25C−20C= 5C

Substitute the known values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Step 3: Set up the equation for heat transfer. Since the total heat gained by

the water is equal to the total heat lost by the aluminum block, we can set up

the equation:

Qaluminum =Qwater

Step 4: Solve for the specific heat capacity of aluminum. Equating Qaluminum

and Qwater and solving for c:

(50 g)(c)(−55C) = (200 g)(4.18 J/g

C)(5C)

c=(200 g)(4.18 J/g

C)(5C)

(50 g)(−55C)

Question 35

Question

A 50.0 g sample of aluminum at 95.0

C is added to 100.0 g of water at 25.0

The final temperature of the system is 30.0

C. Assuming no heat is lost to the

surroundings and the specific heat capacity of water is 4.18 J/g

C, determine

the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by the aluminum and gained by the water.

The heat lost by the aluminum is equal to the heat gained by the water:

(m·c·∆T)aluminum = (m·c·∆T)water

where maluminum = 50.0 g (mass of aluminum), cwater = 4.18 J/g

C (specific

heat capacity of water), Tinitial, aluminum = 95.0

C (initial temperature of alu-

minum), Tfinal = 30.0

C (final temperature of the system), mwater = 100.0 g

(mass of water), Tinitial, water = 25.0

C (initial temperature of water).

Therefore, the equation becomes:

(50.0 g ·caluminum ·(30.0

C−95.0

C)) = (100.0 g ·4.18 J/g

C·(30.0

C−25.0

C))

Step 2: Solve for caluminum.

Begin by simplifying both sides of the equation:

6750 ·caluminum = 20.9·100

caluminum =20.9·100

6750 = 0.310 J/g

Therefore, the specific heat capacity of aluminum is 0.310 J/g

Question 5

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 25

C. Assuming no heat

is lost to the surroundings, find the specific heat capacity of aluminum. The

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down from 80

to 25

C. The specific heat capacity of aluminum is 0.897 J/g

C. The formula to

calculate heat energy is Q=mc∆T, where: - Qis the heat energy, - mis the

mass, - cis the specific heat capacity, - ∆Tis the change in temperature.

Given: m= 50 g, c= 0.897 J/g

C, and ∆T= 80C−25C= 55C.

Using the formula, we have: Qaluminum = (50 g)(0.897 J/g

C)(55C).

Calculating Qaluminum, we get: Qaluminum = 2481.75 J.

Step 2: Calculate the heat gained by the water and the calorimeter. Using

the same formula as before with the given values: Qwater = (200 g)(4.18 J/g

C)(25C−

20C).

Calculating Qwater, we get: Qwater = 418 J.

Step 3: Since the total heat lost by the aluminum block is equal to the heat

gained by the water and the calorimeter (assuming no heat loss to surroundings),

we have: Qaluminum =Qwater +Qcalorimeter.

Substitute the values we found into the equation: 2481.75 J = 418 J +

Qcalorimeter.

Solving for Qcalorimeter, we get: Qcalorimeter = 2063.75 J.

Step 4: The heat gained by the calorimeter is due to the temperature change

of the water and calorimeter together. Let the specific heat capacity of the

calorimeter be C. Since the water and calorimeter have the same temperature

change, we have: Qcalorimeter = (200 g + mcalorimeter)C(25C−20C).

Given that the mass of water is 200 g, the specific heat capacity of water is

4.18 J/g

C, and the final temperature is 25C.

Substitute the known values into the equation: 2063.75 J = (200 g+mcalorimeter)4.18 J/g

C×

5C.

Solving for mcalorimeter, we find: mcalorimeter = 105 g.

Step 5: Calculate the specific heat capacity of aluminum. Given that the

mass of the aluminum block is 50 g and the specific heat capacity of the alu-

minum is 0.897 J/g

C, we have: Qaluminum = (50 g)(0.897 J/g

C)(55C).

Substitute the values we know: 2481.75 J = (50 g)(0.897 J/g

C)(55C).

Solving for the specific heat capacity of aluminum, we get: caluminum =

0.897 J/g

Therefore, the specific heat capacity of aluminum is 0.897

Question 6

Question

A 50 g piece of aluminum at 100◦C is dropped into 200 g of water at 20◦C in

a calorimeter. The final temperature of the system is 25◦C. Assuming all the

heat lost by the aluminum is gained by the water and the calorimeter, what is

the heat capacity of the calorimeter? The specific heat capacity of aluminum is

0.90 J/g◦C and that of water is 4.18 J/g◦C.

Solution

Step 1: Calculate the heat lost by the aluminum. The heat lost by the aluminum

can be calculated using the formula:

Qaluminum =maluminum ×caluminum ×∆T

where: - maluminum is the mass of aluminum (50 g) - caluminum is the specific

heat capacity of aluminum (0.90 J/g◦C) - ∆Tis the change in temperature

(final temperature - initial temperature)

Qaluminum = 50 g ×0.90 J/g◦

×(25 −100)◦CQaluminum = 50 ×0.90 ×(−75)

Qaluminum =−3375 J

Step 2: Calculate the heat gained by the water and the calorimeter. Since

the heat lost by the aluminum is equal to the heat gained by the water and the

calorimeter, we have:

Qaluminum =Qwater +Qcalorimeter

Substitute the values we know:

−3375 J = (200 g + Ccalorimeter)×4.18 J/g◦

×(25 −20)◦C

−3375 J = (200 + Ccalorimeter)×4.18 ×5

−3375 = (200 + Ccalorimeter)×20.9

−3375 = 4180 + 20.9Ccalorimeter

20.9Ccalorimeter =−7555

Ccalorimeter =−361.2 J/◦CJ/◦CJ/◦CJ/◦C

Therefore, the heat capacity of the calorimeter is -361.2 J/◦C.

Question 7

Question

A 50 gram piece of iron at 80

C is placed in 200 grams of water at 20

C. If the

final temperature of the system is 25

C, what is the specific heat capacity of

iron? Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat lost by the iron and the heat gained by the water

using the formula:

qiron =−qwater

where q=mc∆T(m= mass, c= specific heat capacity, ∆T= change in

temperature).

Step 2: Calculate the heat lost by the iron:

qiron =mironciron∆Tiron

qiron = (0.05 kg)(ciron)(80 −25)

Step 3: Calculate the heat gained by the water:

qwater =mwatercwater∆Twater

qwater = (0.2 kg)(4190 J/kg◦C)(25 −20)

Step 4: Set the two equations equal and solve for the specific heat capacity

of iron:

(0.05 kg)(ciron)(55) = (0.2 kg)(4190 J/kg◦C)(5)

ciron =(0.2 kg)(4190 J/kg◦C)(5)

(0.05 kg)(55)

ciron = 3821 J/kg◦C

Question 8

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 25

C, calculate the specific

heat capacity of aluminum. Assume that the specific heat capacity of water is

4.18 J/g

C and neglect any heat loss to the surroundings.

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down from

C to 25

C. The formula for heat transfer is given by:

q=mc∆T

Where: m= mass of the substance (Aluminum block) in grams = 50 g c=

specific heat capacity of the substance (Aluminum) in J/g

C ∆T= change in

temperature = 25C−80C=−55C(negative since it is cooling down)

Substitute the values into the formula:

qAluminum = 50 ×c×(−55)

Step 2: Calculate the heat gained by the water as it warms up from 20

C to

C. Since the water gains the same amount of heat that the aluminum loses,

we can write:

qAluminum =qWater

The formula for heat transfer in water is similar to that of the aluminum

block:

qWater =mc∆T

Where: m= mass of the substance (Water) in grams = 200 g c= specific heat

capacity of the substance (Water) = 4.18 J/g

C ∆T= change in temperature

= 25C−20C= 5C

Substitute the values into the formula:

50 ×c×(−55) = 200 ×4.18 ×5

Step 3: Solve the equation obtained in Step 2 to find the specific heat ca-

pacity of aluminum (c).

50c×(−55) = 200 ×4.18 ×5

c=200 ×4.18 ×5

50 ×55

c=4180

c≈76.0 J/g

Therefore, the specific heat capacity of aluminum is approximately 76.0

J/g

Question 9

Question

A 50 g lead bullet at 100

C is fired into a large block of ice at 0

C. If the bullet

comes to rest after melting some ice, what is the mass of ice melted? (Specific

heat of lead = 0.128 cal/g

C, heat of fusion of ice = 79.7 cal/g, specific heat of

ice = 0.5 cal/g

Solution

Step 1: Calculate the heat lost by the lead bullet as it cools down to the final

temperature. The specific heat formula Q=mc∆Tcan be used where Qis the

heat absorbed or released, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature.

Given: mlead = 50 g, clead = 0.128 cal/g

C, Ti-lead = 100C,Tf-lead =x

Heat lost by lead = −(50 g)(0.128 cal/g

C)(x−100) cal

Step 2: Calculate the heat required to melt the ice. The heat required for

a phase change (melting/freezing) is given by Q=mL where Qis the heat

absorbed or released, mis the mass, and Lis the latent heat of fusion.

Given: Lice = 79.7 cal/g, mice =y

Heat needed to melt ice = (y)(79.7) cal

Step 3: Calculate the heat gained by the melted ice as it warms up to its

final temperature. The specific heat formula can be used again.

Given: mice =y,cice = 0.5 cal/g

C, Ti-ice = 0C,Tf-ice =x

Heat gained by ice = (y)(0.5)(x−0) cal

Step 4: Setting up the conservation of energy equation. The heat lost by the

lead bullet equals the heat gained by the melted ice as well as the heat required

to melt the ice.

−(50)(0.128)(x−100) = (y)(79.7) + (y)(0.5)(x)

Step 5: Solve for y. Solving the equation found in Step 4 will give the mass

of ice melted. Let’s solve this equation.

−6.4x+ 640 = 79.7y+ 0.5yx

Since the initial temperature of the ice is 0

C, we can simplify this further

to only consider the ice reaching the final temperature:

−6.4x+ 640 = 79.7y+ 0.5yx

−6.4x+ 640 = y(79.7+0.5x)

Hence, the mass of ice melted is given by y=−6.4x+640

79.7+0.5x.

Question 10

Question

A student performs a calorimetry experiment by mixing 200 g of water at 20◦C

with 100 g of an unknown metal at 100◦C. The final temperature of the mixture

is 25◦C. If the specific heat capacity of water is 4.18 J/(g·◦C), determine the

specific heat capacity of the metal.

Solution

Step 1: Calculate the heat lost by the hot metal and gained by the cold water

using the formula:

Qmetal =−Qwater

where

Q=mc∆T

Step 2: Calculate the heat lost by the metal:

Qmetal =mmetalcmetal∆Tmetal

where - mmetal = 100 g (mass of metal) - cmetal (specific heat capacity of the

metal, to be determined) - ∆Tmetal =Tf−Tmetal = 25 −100 = −75 ◦C

Step 3: Calculate the heat gained by the water:

Qwater =mwatercwater∆Twater

where - mwater = 200 g (mass of water) - cwater = 4.18 J/(g·◦C) (specific heat

capacity of water) - ∆Twater =Tf−Twater = 25 −20 = 5 ◦C

Step 4: Set up the equation:

100 ×cmetal × −75 = −200 ×4.18 ×5

Step 5: Solve for cmetal:

cmetal =−200 ×4.18 ×5

100 × −75

Step 6: Calculate the specific heat capacity of the metal:

cmetal = 1.39 J/(g·◦C)

Therefore, the specific heat capacity of the metal is 1.39 J/(g·◦C).

Question 11

Question

A 50.0 g piece of aluminum at 80.0

C is dropped into 100.0 g of water at 20.0

Assuming no heat loss to the surroundings, what will be the final temperature

of the system? (Specific heat capacity of aluminum = 0.900 J/g

C, specific heat

capacity of water = 4.18 J/g

C, and the heat of fusion of water = 334 J/g)

Solution

Step 1: Calculate the heat lost by the aluminum as it cools down to the final

temperature. The heat lost is given by the formula:

Qlost, Al =m·c·∆T

where: - m= 50.0 g is the mass of the aluminum, - c= 0.900 J/g

C is the

specific heat capacity of aluminum, - ∆T=Tinitial −Tfinal, - Tinitial = 80.0Cis

the initial temperature of the aluminum, and - Tfinal is the final temperature of

the system.

Substitute the values:

Qlost, Al = 50.0 g ×0.900 J/g

C×(80.0−Tfinal)C

Step 2: Calculate the heat gained by the water as it warms up to the final

temperature. The heat gained by the water is given by the formula:

Qgain, water =m·c·∆T

where: - m= 100.0 g is the mass of the water, - c= 4.18 J/g

C is the specific

heat capacity of water, - ∆T=Tfinal −Tinitial, - Tinitial = 20.0Cis the initial

temperature of the water.

Substitute the values:

Qgain, water = 100.0 g ×4.18 J/g

C×(Tfinal −20.0)C

Step 3: Set up the equation based on the law of conservation of energy. The

heat lost by the aluminum must equal the heat gained by the water since no

energy is lost to the surroundings:

Qlost, Al =Qgain, water

Step 4: Solve the equation to find the final temperature of the system.

50.0 g ×0.900 J/g

C×(80.0−Tfinal)C= 100.0 g ×4.18 J/g

C×(Tfinal −20.0)C

Solve for Tfinal to find the final temperature of the system.

Question 12

Question

A student wants to determine the specific heat capacity of a metal block. The

student immerses the metal block, initially at a temperature of 200

C, into

a 1.5 kg of water at 20

C contained in a calorimeter. The final equilibrium

temperature of the system is found to be 40

C. If the specific heat capacity of

water is 4186 J/kg◦C, determine the specific heat capacity of the metal block.

Solution

Let’s denote the specific heat capacity of the metal block as cmand the mass

of the metal block as mm. We can use the principle of conservation of energy

to solve for cm.

Step 1: Calculate the heat lost by the metal block and the heat

gained by the water. The heat lost by the metal block is equal to the heat

gained by the water. Therefore, we can write:

mmcm∆T=mwcw∆T

where: mm= 1.5 kg (mass of water), cw= 4186 J/kg◦C (specific heat capacity

of water), ∆T= 20◦C (change in temperature).

Step 2: Calculate the change in temperature for the metal block.

The temperature of the metal block changes from 200

C to 40

C, thus ∆Tm=

40◦C−200◦C = −160◦C.

Step 3: Substitute known values into the heat transfer equation.

Substitute mm= 1.5 kg, cw= 4186 J/kg◦C, ∆T= 20◦C, and ∆Tm=−160◦C

into the heat transfer equation:

1.5cm(−160) = 1.5·4186 ·20

−240cm= 125580

cm=125580

−240

cm=−523.25 J/kg◦C

Step 4: Conclusion The specific heat capacity of the metal block is

−523.25 J/kg◦C. It is negative because heat transfer from the metal block causes

a temperature decrease of the metal block.

Question 13

Question

A 50.0 g sample of aluminum at 90.0

C is added to a calorimeter containing

100.0 g of water at 25.0

C. If the final temperature of both the aluminum and

water is 28.0

C, what is the specific heat capacity of aluminum? Assume the

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat absorbed by the water using the formula

qwater =mwater ×cwater ×∆T

where - mwater = 100.0 g is the mass of water, - cwater = 4.18 J/g

C is the specific

heat capacity of water, and - ∆T=Tfinal −Tinitial = 28.0C−25.0C= 3.0Cis

the change in temperature. Substitute the values into the formula to find qwater.

Step 2: Calculate the heat lost by the aluminum using the formula

qaluminum =−qwater

because heat lost by the aluminum is equal in magnitude but opposite in sign

to the heat gained by the water.

Step 3: Calculate the heat lost by the aluminum using its specific heat

capacity with the formula

qaluminum =maluminum ×caluminum ×∆T

where - maluminum = 50.0 g is the mass of aluminum, - caluminum is the specific

heat capacity of aluminum, and - ∆T=Tfinal −Tinitial = 28.0C−90.0C=

−62.0Cis the change in temperature.

Step 4: Set up and solve an equation to find caluminum using the relationships

between the heat gained by the water and the heat lost by the aluminum.

This equation can be written as

mwater ×cwater ×∆T=maluminum ×caluminum ×∆T

Substitute the known values and solve for caluminum.

Question 14

Question

A 50 g block of copper initially at 150◦C is dropped into 200 g of water at 20◦C.

The final temperature of the system is 22◦C. Assuming no heat is lost to the

surroundings, calculate the specific heat capacity of copper. The specific heat

capacity of water is 4.18 J/g◦C.

Solution

Step 1: Determine the heat transfer from the copper block to the water. The

heat lost by the copper block equals the heat gained by the water. Let qCu be

the heat lost by the copper block and qH2O be the heat gained by the water.

qCu =−qH2O

Step 2: Calculate the heat lost by the copper block. The heat lost by the

copper block is given by the formula:

qCu =−mc∆T

where mis the mass, cis the specific heat capacity, and ∆Tis the change

in temperature.

Substitute the given values: m= 50 g, c= specific heat capacity of copper,

∆T= (22 −150) ◦C = −128 ◦C.

qCu = 50 ·c·(−128) = −6400cJ

Step 3: Calculate the heat gained by the water. The heat gained by the

water is also given by the formula:

qH2O =mc∆T

Substitute the given values: m= 200 g, c= 4.18 J/g◦C, ∆T= (22 −20) ◦C

= 2 ◦C.

qH2O = 200 ·4.18 ·2 = 1672 J

Step 4: Set up the equation and solve for the specific heat capacity of copper.

Since qCu =−qH2O, we have

−6400c= 1672

Solving for c:

c=1672

6400 = 0.2615 J/g◦C

Therefore, the specific heat capacity of copper is 0.2615 J/g◦C.

Question 15

Question

A 200 g aluminum calorimeter initially at 20

C contains 500 g of water at 40

A 50 g iron ball at 100

C is placed into the calorimeter, causing the temperature

to stabilize at 30

C. Assuming no heat is lost to the surroundings, what is the

specific heat capacity of the iron ball? (Specific heat capacities: aluminum =

0.902 J/g

C, water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the iron ball as it cools to 30

The heat lost by the iron ball is equal to the heat gained by the aluminum

calorimeter, water, and the final mixture. Let Ciron be the specific heat capacity

of iron. The heat lost by the iron ball is given by:

Qiron =miron ·Ciron ·(Tf−Ti)

Qiron = 50 g ·Ciron ·(30 −100)

Qiron =−3500 ·Ciron J

Step 2: Calculate the heat gained by the water and aluminum calorimeter.

The heat gained by the water and aluminum calorimeter is given by:

Qwater =mwater ·Cwater ·(Tf−Ti)

Qaluminum =maluminum ·Caluminum ·(Tf−Ti)

We know that the total heat gained is equal to the heat lost by the iron ball:

Qwater +Qaluminum =−3500 ·Ciron J

Substitute the given values and solve for Ciron:

500 g ·4.18 J/g

C + 200 g ·0.902 J/g

C = −3500 ·Ciron

2090 + 180 ≈ −3500 ·Ciron

2270 ≈ −3500 ·Ciron

Ciron ≈ −0.648 J/g

Therefore, the specific heat capacity of the iron ball is approximately 0.648 J/g

Question 16

Question

A piece of iron with a mass of 150 grams at a temperature of 200◦C is placed

into 500 grams of water at 20◦C in a calorimeter. The final temperature of

the system is 25◦C. Assuming no heat is lost to the surroundings, calculate the

specific heat capacity of iron. (Specific heat capacity of water = 4.18 J/g◦C)

Solution

Step 1: Calculate the heat lost by the iron and the heat gained by the water

using the formula for heat transfer:

q=mc∆T

where qis the heat transferred, mis the mass, cis the specific heat capacity,

and ∆Tis the change in temperature.

For the iron:

qiron =mironciron∆Tiron

qiron = (0.150 kg)(ciron)(200 −25)

For the water:

qwater =mwatercwater∆Twater

qwater = (0.500 kg)(4.18 J/g◦C)(25 −20)

Step 2: Since the heat lost by the iron is equal to the heat gained by the

water (assuming no heat loss to the surroundings):

qiron =qwater

(0.150)(ciron)(175) = (0.500)(4.18)(5)

Step 3: Solve for the specific heat capacity of iron, ciron:

ciron =(0.500)(4.18)(5)

(0.150)(175)

ciron =10.45

26.25

ciron = 0.398 J/g◦C

Question 17

Question

A 50.0 g piece of aluminum at 90.0

C is placed in 100.0 g of water at 20.0

The final temperature of the mixture is 22.5

C. Assuming no heat is lost to the

surroundings, calculate the specific heat capacity of aluminum. Given: specific

heat capacity of water = 4.18 J/(g◦C).

Solution

Step 1: Calculate the heat lost by aluminum and the heat gained by water. The

heat lost by aluminum is equal to the heat gained by water.

Heat lost by aluminum = Heat gained by water

m1·c1·∆T1=m2·c2·∆T2

where: m1= 50.0 g (mass of aluminum), c1=? (specific heat capacity of alu-

minum), ∆T1=−67.5◦C (temperature change for aluminum), m2= 100.0 g

(mass of water), c2= 4.18 J/(g◦C) (specific heat capacity of water), ∆T2=

2.5◦C (temperature change for water).

Step 2: Substitute the values and solve for the specific heat capacity of

aluminum.

50.0 g ·c1·(−67.5◦C) = 100.0 g ·4.18 J/(g◦C) ·2.5◦C

c1=100.0 g ·4.18 J/(g◦C) ·2.5◦C

50.0 g ·(−67.5◦C)

c1=1045.0 J

−3375 g ·◦C

c1=−0.309 J/(g◦C)

Therefore, the specific heat capacity of aluminum is −0.309 J/(g◦C).

Question 18

Question

A 50.0 g piece of copper is heated to 95.0

C and then placed into 100.0 g of

water at 25.0

C. The final temperature of the system is 26.0

C. Assuming all

heat is transferred to the water, determine the specific heat of copper.

Solution

Step 1: Calculate the heat absorbed by the water The heat absorbed by the

water can be calculated using the formula:

qwater =mwater ·cwater ·∆T

where: - mwater is the mass of water (100.0 g), - cwater is the specific heat

capacity of water (4.18 J/g

C), and - ∆Tis the temperature change of the

water (final temperature - initial temperature).

Substitute the values into the formula:

qwater = (100.0 g) ·(4.18 J/g

C) ·(26.0−25.0)C

qwater = 100 ·4.18 ·1 = 418 J

Step 2: Calculate the heat lost by the copper The heat lost by the copper is

equal to the heat gained by the water, so:

qcopper =−qwater =−418 J

Step 3: Calculate the specific heat of copper The heat lost by the copper

can be calculated using the formula:

qcopper =mcopper ·ccopper ·∆T

where: - mcopper is the mass of copper (50.0 g), - ccopper is the specific heat

capacity of copper (unknown), and - ∆Tis the temperature change of the copper

(final temperature - initial temperature).

Substitute the values into the formula and solve for ccopper:

−418 = (50.0 g) ·ccopper ·(26.0−95.0)C

ccopper =−418

50 ·(26.0−95.0)

ccopper ≈0.39 J/g

Therefore, the specific heat of copper is approximately 0.39 J/g

Question 19

Question

A 50 g piece of copper at 95

C is placed in 200 g of water at 25

C in an insulated

container. Assuming no heat is lost to the surroundings, what will be the final

temperature of the system? (Specific heat capacity of copper is 0.385 J/g

C and

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat gained by the water.

The heat gained by the water can be calculated using the formula:

Qwater =mwater ×cwater ×∆T

where: - mwater = 200 g (mass of water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T=Tfinal −Tinitial.

The initial temperature of water, Tinitial = 25C.

Step 2: Calculate the heat lost by the copper.

The heat lost by the copper can be calculated using the formula:

Qcopper =mcopper ×ccopper ×∆T

where: - mcopper = 50 g (mass of copper), - ccopper = 0.385 J/g

C (specific heat

capacity of copper), - ∆T=Tfinal −Tinitial.

The initial temperature of copper, Tinitial = 95C.

Step 3: Set up the energy conservation equation.

Since no heat is lost to the surroundings, the heat gained by the water must

be equal to the heat lost by the copper.

Qwater =Qcopper

Step 4: Solve for the final temperature.

Substitute the expressions for Qwater and Qcopper into the energy conserva-

tion equation and solve for the final temperature, Tfinal.

mwater ×cwater ×∆T=mcopper ×ccopper ×∆T

200 ×4.18 ×(Tfinal −25) = 50 ×0.385 ×(95 −Tfinal)

Solve this equation to find the final temperature, Tfinal.

Question 20

Question

A 50 g piece of copper is heated to 100

C and then placed in 200 g of water at

C. The final temperature of the system is 25

C. Assuming no heat is lost to

the surroundings, calculate the specific heat capacity of copper. (Specific heat

capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper piece as it cools down from

100

C to 25

C. The heat lost is given by the formula: Q=mc∆T, where: -

mis the mass of the copper piece (in grams), - cis the specific heat capacity

of copper (in J/g

C), - ∆Tis the change in temperature (final temperature -

initial temperature).

Given: - mcopper = 50 g, - Tinitial = 100

C, - Tfinal = 25

Substitute these values into the formula: Qcopper = (50 g)(c)(25

Step 2: Calculate the heat gained by the water as it heats up from 20

to 25

C. The heat gained is given by the same formula: Q=mc∆T, where:

-mis the mass of the water (in grams), - cis the specific heat capacity of

water (in J/g

C), - ∆Tis the change in temperature (final temperature - initial

temperature).

Given: - mwater = 200 g, - cwater = 4.18 J/g

C, - Tinitial = 20

C, - Tfinal =

Substitute these values into the formula: Qwater = (200 g)(4.18 J/g

C)(5

Step 3: Since heat lost by the copper equals heat gained by the water (as-

suming no heat loss to the surroundings), we can set the two equations equal

to each other: (50 g)(c)(25

C) = (200 g)(4.18 J/g

C)(5

Step 4: Solve for the specific heat capacity of copper, c:c=(200 g)(4.18 J/g

C)(5

(50 g)(25

c=4180 J

1250 g

c= 3.344 J/g

Therefore, the specific heat capacity of copper is 3.344 J/g

Question 21

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in

a calorimeter. The specific heat capacity of aluminum is 0.90 J/g

C and the

specific heat capacity of water is 4.18 J/g

C. If the final temperature of the

system is 24.5

C, determine the heat capacity of the calorimeter.

Solution

Step 1: Calculate the heat lost by the aluminum block as it cools down to the

final temperature of the system. The formula for heat transfer is given by

Q=mc∆T,

where Qis the heat transfer, mis the mass of the substance, cis the specific

heat capacity, and ∆Tis the change in temperature.

The heat lost by the aluminum block is calculated as follows:

QAluminum = (50 g)(0.90 J/g

C)(80 −24.5) = 2075 J.

Step 2: Calculate the heat gained by the water and the calorimeter. The

heat gained by the water and calorimeter is given by

Qwater+calorimeter = (50 g + 200 g)(4.18 J/g

C)(24.5−20) = 3503 J.

Step 3: Set up an equation to determine the heat capacity of the calorimeter.

Since the heat lost by the aluminum block is equal to the heat gained by the

water and calorimeter (law of conservation of energy), we have

QAluminum =Qwater+calorimeter.

Step 4: Solve for the heat capacity of the calorimeter.

2075 J = 3503 J + Ccalorimeter,

where Ccalorimeter is the heat capacity of the calorimeter.

Therefore, the heat capacity of the calorimeter is

Ccalorimeter = 2075 J −3503 J = −1428 J.

The negative sign indicates that the calorimeter releases 1428 J of heat,

which makes sense since the calorimeter would lose heat to the surroundings

during the experiment.

Question 22

Question

A 50 g piece of aluminum at 80

C is dropped into 200 g of water at 20

in a calorimeter. The final temperature of the system is 25

C. Calculate the

specific heat capacity of aluminum. Assume the specific heat capacity of water

is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the aluminum piece. The formula for calcu-

lating heat transfer is:

Q=mc∆T

where: Q= heat transfer (in J) m= mass of the substance (in g) c= specific

heat capacity of the substance (in J/g

C) ∆T= change in temperature (in

Given: maluminum = 50 g cwater = 0.903 J/g

C ∆Taluminum = (80 −25)

C =

Substitute the values into the formula:

Qaluminum = 50 g ×caluminum ×55

Step 2: Calculate the heat gained by the water. Since the system is isolated,

the heat lost by the aluminum piece is equal to the heat gained by the water.

Therefore:

Qaluminum =Qwater

Using the same formula and the given values:

Qwater = 200 g ×4.18 J/g

C×(25 −20)

Step 3: Set the two equations equal to each other:

50 g ×caluminum ×55

C = 200 g ×4.18 J/g

C×5

Step 4: Solve for caluminum.

caluminum =200 g ×4.18 J/g

C×5

50 g ×55

Question 23

Question

A 50 g piece of aluminum at 80◦C is dropped into 200 g of water at 20◦C in a

calorimeter. The final temperature of the system is 25◦C. Assuming the specific

heat capacity of water is 4.18 J/g◦C and that of aluminum is 0.897 J/g◦C,

calculate the specific heat capacity of the calorimeter.

Solution

Step 1: Calculate the heat gained by water: The heat gained by the water is

given by the equation

Q=mc∆T

where: - mis the mass of water (200 g), - cis the specific heat capacity of water

(4.18 J/g◦C), - ∆Tis the change in temperature of the water.

The change in temperature of the water is

∆T=Tf−Ti= 25◦C−20◦C= 5◦C.

Substitute the values in:

Q= (200 g)(4.18 J/g◦C)(5◦C)

Q= 4180 J

Step 2: Calculate the heat lost by the aluminum: The heat lost by the

aluminum is given by the equation

Q=mc∆T

where: - mis the mass of aluminum (50 g), - cis the specific heat capacity of

aluminum (0.897 J/g◦C), - ∆Tis the change in temperature of the aluminum.

The change in temperature of the aluminum is

∆T=Tf−Ti= 25◦C−80◦C=−55◦C.

Notice that the negative sign indicates that the aluminum is losing heat

energy.

Substitute the values in:

Q= (50 g)(0.897 J/g◦C)(−55◦C)

Q=−2456.25 J

Step 3: Calculate the heat absorbed by the calorimeter (water + aluminum):

Since we assume no heat is lost to the surroundings, the heat lost by the alu-

minum is equal to the heat gained by the water and the calorimeter:

Qcalorimeter =−Qaluminum =Qwater

Qcalorimeter = 4180 J

Step 4: Calculate the heat capacity of the calorimeter: We know the heat

capacity formula is

Q=mc∆T

where: - mis the mass of the calorimeter, - cis the specific heat capacity of the

calorimeter, - ∆Tis the change in temperature of the calorimeter.

Substitute the values in:

4180 J = (m)(c)(5◦C)

We are looking for c, the specific heat capacity of the calorimeter.

Solving for c, we get

c=4180 J

(m)(5◦C)

Since we don’t have the mass of the calorimeter, further calculation is not

possible based on the information provided.

Question 24

Question

A piece of metal of mass 120.0 g at a temperature of 200.0

C is placed in 250.0

g of water at 20.0

C. The final equilibrium temperature of the system is 26.5

Assume the specific heat capacities of the metal and water are constant at 0.385

J/g

C and 4.18 J/g

C, respectively. Determine the specific heat capacity of the

metal.

Solution

Step 1: Calculate the heat gained by the metal as it cools down to the final

temperature. The formula for heat gained or lost is given by:

q=mc∆T

Where: - qis the heat gained or lost - mis the mass of the object - cis the

specific heat capacity of the substance - ∆Tis the change in temperature

Given that the metal is cooling down, the heat gained by the metal is equal

to the heat lost by the water. Thus, we can set up the equation:

mmetalcmetal∆Tmetal =−mwatercwater∆Twater

Substitute the values to find cmetal:

120.0 g ·cmetal ·(26.5C−26.5C) = −250.0 g ·4.18 J/g

C·(26.5C−20.0C)

120.0 g ·cmetal ·0 = −250.0 g ·4.18 J/g

C·6.5C

0 = −250.0·4.18 ·6.5

0 = −6505.0

Since our calculated value is negative, there seems to be an error in our

calculation. Let’s go back and verify our computations.

Question 25

Question

A 50 g piece of aluminum at 100

C is placed in 150 g of water at 20

C. If the

final temperature of the system is 25

C, calculate the specific heat capacity of

aluminum. (Specific heat capacity of water = 4.18 J/g

Solution

Step 1: First, we find the heat absorbed by the aluminum and the heat released

by the water. The heat lost by the aluminum equals the heat gained by the

water since no heat is lost to the surroundings. Let the specific heat capacity

of aluminum be cAl.

Step 2: The heat absorbed by the aluminum can be calculated using the

formula:

qAl =mAl ·cAl ·∆TAl

where

mAl = 50 g

∆TAl = (25 −100) = −75

Therefore,

qAl = 50 g ·cAl ·(−75)

Step 3: The heat released by the water can be calculated using the formula:

qH2O =mH2O ·cH2O ·∆TH2O

where

mH2O = 150 g

cH2O = 4.18 J/g

∆TH2O = (25 −20) = 5

Therefore,

qH2O = 150 g ·4.18 J/g

C·5

Step 4: Since the heat lost by the aluminum equals the heat gained by the

water:

qAl =qH2O

50 g ·cAl ·(−75) = 150 g ·4.18 J/g

C·5

Step 5: Solve for cAl to find the specific heat capacity of aluminum.

50 g ·cAl ·(−75) = 150 g ·4.18 J/g

C·5

cAl =150 g ·4.18 J/g

C·5

50 g ·(−75)

cAl =3135 J

−3750 g

Step 6: Calculate the specific heat capacity of aluminum.

cAl =−0.836 J/g

Therefore, the specific heat capacity of aluminum is −0.836 J/g

Question 26

Question

A student conducts an experiment to determine the specific heat capacity of

a metal by immersing it in a container of water. The metal is heated to a

high temperature and then quickly transferred to the water, causing the water’s

temperature to rise. The student records the initial temperature of the water

as 20

C and the final temperature as 32

C. The mass of the metal is 0.2 kg and

its initial temperature is 100

C. The mass of the water is 0.5 kg and its specific

heat capacity is 4186 J/kg ·K. If the final temperature of the metal and water

is 30

C, what is the specific heat capacity of the metal?

Solution

Step 1: Calculate the heat absorbed by the water. The heat absorbed by the

water can be calculated using the formula:

Qwater =mc∆T

where: m= mass of water = 0.5 kg, c= specific heat capacity of water =

4186 J/kg ·K, ∆T= change in temperature = final temperature - initial tem-

perature = 32C−20C= 12C.

Substitute the values into the formula:

Qwater = 0.5 kg ×4186 J/kg ·K×12C= 25116 J

Step 2: Calculate the heat lost by the metal. The heat lost by the metal can

be calculated using the formula:

Qmetal =mc∆T

where: m= mass of metal = 0.2 kg, c= specific heat capacity of the metal

(to be determined), ∆T= change in temperature = initial temperature - final

temperature = 100C−30C= 70C.

Substitute the values into the formula:

Qmetal = 0.2 kg ×c×70C

Step 3: Set up the heat balance equation. Since the system is isolated and

there is no heat lost to the surroundings, the heat lost by the metal is equal to

the heat gained by the water.

Qwater =Qmetal

0.2 kg ×c×70C= 25116 J

Step 4: Solve for the specific heat capacity of the metal.

0.2 kg ×c×70C= 25116 J

c=25116 J

0.2 kg ×70C

c= 179.4 J/kg ·K

Therefore, the specific heat capacity of the metal is 179.4 J/kg ·K.

Question 27

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C in a calorimeter.

If the final temperature of the system is 25

C, what is the specific heat capacity

of the iron? Assume the specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat gained or lost by the water. Given: Mass of water,

mw= 200 g Initial temperature of water, Tin = 20

C Final temperature of the

system, Tfinal = 25

C Specific heat capacity of water, Cw= 4.18 J/g

The heat gained by the water is given by the formula:

Qwater =mw·Cw·∆T

where ∆T=Tfinal −Tin.

Substitute the given values:

∆T= 25

C−20

C=5

Qwater = 200 g ·4.18 J/g

C·5

Step 2: Calculate the heat gained or lost by the iron. Given: Mass of iron,

miron = 50 g Initial temperature of iron, Tin = 80

C Final temperature of the

system, Tfinal = 25

The heat lost by the iron is equal to the heat gained by the water (since this

is a closed system). Therefore:

Qwater =Qiron

mw·Cw·∆T=miron ·Ciron ·∆Tiron

Substitute the known values:

200 g ·4.18 J/g

C·5

C = 50 g ·Ciron ·(80

C−25

Step 3: Solve for the specific heat capacity of iron.

200 ·4.18 ·5 = 50 ·Ciron ·55

Ciron =200 ·4.18 ·5

50 ·55

Calculating the specific heat capacity of iron:

Ciron =4180

110 = 38 J/g

Therefore, the specific heat capacity of the iron is 38 J/g

Question 28

Question

A piece of copper of mass 200 g at a temperature of 100

C is placed in 400 g of

water at 20

C. Assuming no heat is lost to the surroundings, calculate the final

temperature of the mixture. The specific heat capacity of copper is 0.385 J/g

and the specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper when it cools down to the final

temperature.

The heat lost by the copper can be calculated using the formula:

Qcopper =mc∆T

where: - mis the mass of the copper (200 g), - cis the specific heat capacity

of copper (0.385 J/g

C), - ∆Tis the temperature change of the copper.

The temperature change of the copper can be calculated as follows:

∆T=Tfinal −Tinitial

∆T=Tfinal −100C

Step 2: Calculate the heat gained by the water when it warms up to the

final temperature.

The heat gained by the water can be calculated using the formula:

Qwater =mc∆T

where: - mis the mass of the water (400 g), - cis the specific heat capacity

of water (4.18 J/g

C), - ∆Tis the temperature change of the water.

The temperature change of the water can be calculated as follows:

∆T=Tfinal −Tinitial

∆T=Tfinal −20C

Step 3: Set up the equation based on the conservation of energy.

Since there is no heat loss to the surroundings, the heat lost by the copper

must equal the heat gained by the water. Thus, we have:

Qcopper =Qwater

Step 4: Solve for the final temperature, Tfinal.

Substitute the expressions for Qcopper and Qwater into the equation from

Step 3 and solve for Tfinal.

mc∆Tcopper =mc∆Twater

Finally, solve for Tfinal.

Question 29

Question

A 150 g aluminum block at 175◦C is placed into a container of water at 20◦C.

The water has a mass of 200 g. The final temperature of the system is 25◦C.

Assuming no heat is lost to the surroundings, calculate the specific heat capacity

of aluminum. (Specific heat of water = 4.18 J/g◦C)

Solution

Step 1: Calculate the heat absorbed by the aluminum block as it cools down:

The heat absorbed by the aluminum block can be calculated using the formula:

qAl =mAl ·cAl ·∆T

Where: - mAl = mass of aluminum block = 150 g - cAl = specific heat capacity of

aluminum (to be determined) - ∆T= change in temperature of aluminum block

= 25◦C−175◦C = −150◦C (negative sign indicates a decrease in temperature)

Substitute the given values into the formula:

qAl = 150 g ×cAl ×(−150◦C)

Step 2: Calculate the heat released by the cooling water as it heats up: The

heat released by the water can be calculated using the same formula but with

the specific heat capacity of water:

qwater =mwater ×cwater ×∆T

Where: - mwater = mass of water = 200 g - cwater = specific heat capacity of

water = 4.18 J/g◦C-∆T= change in temperature of water = 25◦C−20◦C =

5◦C

Substitute the given values into the formula:

qwater = 200 g ×4.18 J/g◦C×5◦C

Step 3: Since the system is isolated and assuming no heat is lost to the

surroundings, the heat absorbed by the aluminum block is equal to the heat

released by the water:

qAl =qwater

Step 4: Solve for the specific heat capacity of aluminum: Set the two heat

values equal to each other and solve for cAl:

150 g ×cAl ×(−150◦C) = 200 g ×4.18 J/g◦C×5◦C

cAl =200 g ×4.18 J/g◦C×5◦C

150 g ×(−150◦C)

cAl =2090 J

−22500 g◦C

cAl ≈ −0.093 J/g◦C

The specific heat capacity of aluminum is approximately −0.093 J/g◦C.

Question 30

Question

A 50 g block of copper at 100◦C is dropped into 200 g of water at 20◦C contained

in a 100 g aluminum calorimeter cup. The final temperature of the system is

30◦C. Given the specific heat capacities of copper, water, and aluminum as

0.385 J/g

C, 4.18 J/g

C, and 0.897 J/g

C, respectively, calculate the initial

temperature of the water before the copper block was dropped into it.

Solution

Step 1: Calculate the heat absorbed by the copper block. The heat absorbed

by the copper block can be calculated using the formula:

Q=mc∆T

where: - mis the mass of the copper block (50 g), - cis the specific heat capacity

of copper (0.385 J/g

C), and - ∆Tis the change in temperature of the copper

block (30 −100 = −70

C).

Substitute the given values into the formula to calculate Q:

Q= 50 ×0.385 ×(−70)

Step 2: Calculate the heat lost by the water. The heat lost by the water can

be calculated using the formula:

Q=mc∆T

where: - mis the mass of the water (200 g), - cis the specific heat capacity

of water (4.18 J/g

C), and - ∆Tis the change in temperature of the water

(30 −x

C, where xis the initial temperature of the water).

Substitute the given values into the formula to calculate Q.

Step 3: Calculate the heat absorbed by the aluminum calorimeter cup. The

heat absorbed by the aluminum calorimeter cup can be calculated using the

formula:

Q=mc∆T

where: - mis the mass of the aluminum calorimeter cup (100 g), - cis the

specific heat capacity of aluminum (0.897 J/g

C), and - ∆Tis the change in

temperature of the aluminum calorimeter cup (30 −20 = 10

C).

Substitute the given values into the formula to calculate Q.

Step 4: Apply the principle of conservation of energy. According to the

principle of conservation of energy, the heat lost by the copper block is equal to

the sum of the heat gained by the water and the aluminum calorimeter cup:

Qcopper =Qwater +Qcup

Substitute the calculated values of Qcopper,Qwater, and Qcup into the equa-

tion and solve for x, the initial temperature of the water.

Question 31

Question

A 50 g aluminum block at 80

C is placed in 200 g of water at 20

C in a calorime-

ter. The final temperature of the system is 30

C. Assuming no heat is lost to

the surroundings and that the specific heat capacity of water is 4186 J/kg

calculate the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by the aluminum block

Heat lost by aluminum = m×c×∆T

where m= 50 g = 0.05 kg (mass of aluminum block), c(specific heat capacity

of aluminum),

∆T=Tf−Ti= 30 −80 = −50

Therefore,

Heat lost by aluminum = 0.05 ×c×(−50)

Step 2: Calculate the heat gained by water

Heat gained by water = m×cw×∆T

where m= 200 g = 0.2 kg (mass of water), cw= 4186 J/kg

C (specific heat

capacity of water),

∆T=Tf−Ti= 30 −20 = 10

Therefore,

Heat gained by water = 0.2×4186 ×10

Step 3: Since there is no heat loss to the surroundings, the heat lost by the

aluminum block is equal to the heat gained by the water.

0.05 ×c×(−50) = 0.2×4186 ×10

Step 4: Solve for the specific heat capacity of aluminum, c

0.05 ×c×(−50) = 0.2×4186 ×10

c=0.2×4186 ×10

0.05 ×(−50)

Question 32

Question

A piece of aluminum of mass 250 g at a temperature of 90

C is placed in 500 g of

water at 20

C in a calorimeter of negligible heat capacity. The final temperature

of the system is 23

C. Assuming no heat loss to the surroundings, calculate the

specific heat capacity of aluminum. (Specific heat capacity of water = 4.18

J/g

Solution

Step 1: Calculate the heat gained by the water. Given: mwater = 500 g

∆Twater = 23−20 = 3

Ccwater = 4.18 J/g

C Using the formula Q=mc∆T, the

heat gained by the water can be calculated as: Qwater = (500 g)×(4.18 J/g

C)×

Step 2: Calculate the heat lost by the aluminum. Given: maluminum =

250 g ∆Taluminum = 23 −90 = −67

C (negative as aluminum is losing heat)

Let caluminum be the specific heat capacity of aluminum. Using the formula

Q=mc∆T, the heat lost by the aluminum can be calculated as: Qaluminum =

(250 g) ×caluminum ×(−67)

Step 3: Since no heat is lost to the surroundings, the heat gained by the

water is equal to the heat lost by the aluminum. Therefore, we have: Qwater =

Qaluminum

Step 4: Equate Qwater and Qaluminum and solve for caluminum to find the

specific heat capacity of aluminum. This is left as an exercise for the reader.

Question 33

Question

A 50 g piece of copper at 150

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the mixture is 25

C. Assuming no heat

is lost to the surroundings, calculate the specific heat capacity of copper. The

specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper and the heat gained by the water.

The formula for heat transfer in calorimetry is given by:

Q=mc∆T

where: Q= heat energy transferred, m= mass of the material, c= specific

heat capacity of the material, ∆T= change in temperature.

Let’s first calculate the heat lost by the copper piece: Given: mcopper = 50 g,

cwater = 0.387 J/g

C (specific heat capacity of copper), Tinitial = 150C,Tfinal =

25C.

The change in temperature for copper (∆Tcopper) is:

∆Tcopper =Tfinal −Tinitial = 25C−150C=−125C

(The negative sign indicates the decrease in temperature).

Now, calculate the heat lost by the copper:

Qcopper =mcopper ·ccopper ·∆Tcopper

Step 2: Calculate the heat gained by the water.

Given: mwater = 200 g, cwater = 4.18 J/g

C (specific heat capacity of water),

Tinitial = 20C,Tfinal = 25C.

The change in temperature for water (∆Twater) is:

∆Twater =Tfinal −Tinitial = 25C−20C= 5C

Now, calculate the heat gained by the water:

Qwater =mwater ·cwater ·∆Twater

Step 3: Set up the equation for heat conservation. Since no heat is lost to

the surroundings, the heat lost by the copper should be equal to the heat gained

by the water:

Qcopper =Qwater

Step 4: Solve for the specific heat capacity of copper. Equating the two

equations:

mcopper ·ccopper ·∆Tcopper =mwater ·cwater ·∆Twater

Substitute the known values and solve for ccopper to find the specific heat

capacity of copper.

Question 34

Question

A 50 g aluminum block at 80

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 25

C, calculate the specific

heat capacity of aluminum. Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat absorbed by the aluminum block. The heat absorbed

by the aluminum block can be calculated using the formula:

Qaluminum =mc∆T

Where: - mis the mass of the aluminum block (50 g) - cis the specific heat

capacity of aluminum - ∆Tis the temperature change of the aluminum block

(final temperature - initial temperature)

Given that the initial temperature of the aluminum block is 80

C and the

final temperature is 25

C, we have:

∆T= 25C−80C=−55C

Substitute the known values into the formula:

Qaluminum = (50 g)(c)(−55C)

Step 2: Calculate the heat released by the aluminum block to the water.

The heat released by the aluminum block is equal to the heat absorbed by the

water, which can be calculated using the formula:

Qwater =mc∆T

Where: - mis the mass of the water (200 g) - cis the specific heat capacity of

water - ∆Tis the temperature change of the water (final temperature - initial

temperature)

Given that the initial temperature of the water is 20

C and the final tem-

perature is 25

C, we have:

∆T= 25C−20C= 5C

Substitute the known values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Step 3: Set up the equation for heat transfer. Since the total heat gained by

the water is equal to the total heat lost by the aluminum block, we can set up

the equation:

Qaluminum =Qwater

Step 4: Solve for the specific heat capacity of aluminum. Equating Qaluminum

and Qwater and solving for c:

(50 g)(c)(−55C) = (200 g)(4.18 J/g

C)(5C)

c=(200 g)(4.18 J/g

C)(5C)

(50 g)(−55C)

Question 35

Question

A 50.0 g sample of aluminum at 95.0

C is added to 100.0 g of water at 25.0

The final temperature of the system is 30.0

C. Assuming no heat is lost to the

surroundings and the specific heat capacity of water is 4.18 J/g

C, determine

the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by the aluminum and gained by the water.

The heat lost by the aluminum is equal to the heat gained by the water:

(m·c·∆T)aluminum = (m·c·∆T)water

where maluminum = 50.0 g (mass of aluminum), cwater = 4.18 J/g

C (specific

heat capacity of water), Tinitial, aluminum = 95.0

C (initial temperature of alu-

minum), Tfinal = 30.0

C (final temperature of the system), mwater = 100.0 g

(mass of water), Tinitial, water = 25.0

C (initial temperature of water).

Therefore, the equation becomes:

(50.0 g ·caluminum ·(30.0

C−95.0

C)) = (100.0 g ·4.18 J/g

C·(30.0

C−25.0

C))

Step 2: Solve for caluminum.

Begin by simplifying both sides of the equation:

6750 ·caluminum = 20.9·100

caluminum =20.9·100

6750 = 0.310 J/g

Therefore, the specific heat capacity of aluminum is 0.310 J/g

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 5

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