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

CHEM 107 - ESSENTIALS OF

GENERAL AND ORGANIC

CHEMISTRY - Calorimetry

Question Bank - Set 4

Liberty University

Question 1

Question

A 50 g aluminum block at 80

C is placed in 200 g of water at 20

C in a calorime-

ter. If the final temperature of the system is 25

C, what is the specific heat

capacity of the aluminum block?

Assume the specific heat capacity of water is 4186 J/kg

C and neglect any

heat loss to the surroundings.

Solution

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

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

The formula for heat transfer is Q=mc∆T, where Qis the heat transfer, mis

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

Given: maluminum = 50 g = 0.05 kg cwater = 4186 J/kg

CTinitial, aluminum =

80C Tfinal, system = 25C

The heat lost by the aluminum block can be calculated as: Qaluminum =

maluminumcaluminum(Tfinal, system −Tinitial, aluminum)

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

for heat transfer here is Q=mc∆T.

Given: mwater = 200 g = 0.2 kg cwater = 4186 J/kg

CTinitial, water = 20C

Tfinal, system = 25C

The heat gained by the water can be calculated as: Qwater =mwatercwater(Tfinal, system−

Tinitial, water)

Step 3: Equate the heat lost by the aluminum block to the heat gained by

the water and calorimeter. Qaluminum =Qwater

Step 4: Solve for the specific heat capacity of aluminum. Since Q=mc∆T,

we have: maluminumcaluminum(Tfinal, system−Tinitial, aluminum) = mwatercwater(Tfinal, system−

Tinitial, water)

Substitute the given values and solve for caluminum to obtain the specific heat

capacity of aluminum.

Question 2

Question

A 50 g piece of aluminum at 100◦C is placed in 200 g of water at 20◦C. 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 and that the specific

heat capacity of water is 4.18 J/g·K.

Solution

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

using the formula q=mc∆T, where qis the heat transferred, mis the mass,

cis the specific heat capacity, and ∆Tis the change in temperature. In this

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

For the aluminum: qaluminum = (50 g)(cAl)(−75◦C) For the water: qwater =

(200 g)(4.18 J/g·K)(5◦C)

Step 2: Set the heat lost by the aluminum equal to the heat gained by the

water and solve for the specific heat capacity of aluminum.

(50 g)(cAl)(−75◦C) = (200 g)(4.18 J/g·K)(5◦C)

cAl =(200 g)(4.18 J/g·K)(5◦C)

(50 g)(−75◦C)

Step 3: Calculate the specific heat capacity of aluminum.

cAl =(200)(4.18)(5)

(50)(75)

cAl =4180

375

cAl = 11.15 J/g·K

Therefore, the specific heat capacity of aluminum is 11.15 J/g·K.

Question 3

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C in an insulated

container. If the final temperature of the system is 25

C, determine the specific

heat capacity of iron.

Solution

Let the specific heat capacity of water be cw= 4.18 J/g

Step 1: Calculate the heat absorbed by the water when the iron piece is

added. The heat absorbed by the water can be calculated using the formula:

Q=mc∆T

where: - mis the mass of water, - cis the specific heat capacity of water, - ∆T

is the change in temperature of the water.

Substitute the given values:

Q= 200 ×4.18 ×(25 −20)

Q= 200 ×4.18 ×5

Q= 4180 J

Step 2: Calculate the amount of heat lost by the iron. Since the system is

insulated, the total amount of heat lost by the iron is equal to the amount of

heat gained by the water. Therefore, the heat lost by the iron is also 4180 J.

Step 3: Calculate the specific heat capacity of iron. The heat lost by the

iron can be calculated using the formula:

Q=mc∆T

where: - mis the mass of iron, - cis the specific heat capacity of iron, - ∆Tis

the change in temperature of the iron.

Substitute the given values and the heat lost:

4180 = 50c(25 −80)

4180 = −50c×55

4180 = −2750c

c=4180

−2750

c≈ −1.52 J/g

Therefore, the specific heat capacity of iron is approximately −1.52 J/g

The negative sign indicates that the value obtained is not physically meaningful.

This discrepancy may be due to experimental errors or assumptions made during

the calculation.

Question 4

Question

A 50 g sample of aluminum at 80

C is dropped into 200 g of water at 20

C in

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

C and the

specific heat capacity of water is 4.18 J/g

C. Assuming no heat is lost to the

surroundings, what is the final temperature of the system?

Solution

Step 1: Calculate the heat absorbed by the aluminum: The equation for heat

transfer is given by q=mc∆T, where: - mis the mass of the substance, - cis

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

Substitute the values for aluminum: qAl = (50 g)(0.897 J/g

C)(Tf−80

Step 2: Calculate the heat lost by the water: qwater =−(200 g)(4.18 J/g

C)(Tf−

Step 3: Set up the energy conservation equation: The heat lost by the water

is equal to the heat gained by the aluminum: −(200 g)(4.18 J/g

C)(Tf−20

C) =

(50 g)(0.897 J/g

C)(Tf−80

Step 4: Solve for the final temperature: −(836 J/

C)(Tf−20

C) = (44.85 J/

C)(Tf−

−836Tf+ 16720 = 44.85Tf−3588

836Tf+ 44.85Tf= 16720 + 3588

Tf=20308

880.85 ≈23.05

Therefore, the final temperature of the system is approximately 23.05

Question 5

Question

A 50 g piece of copper at 90

C is placed in 100 g of water at 20

C. Assuming

no heat is lost to the surroundings, what will be the final temperature of the

system? (Specific heat capacity of copper = 0.385 J/g

C, specific heat capacity

of water = 4.18 J/g

Solution

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

temperature. The formula for calculating heat (q) is:

q=m·c·∆T

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

material, - ∆Tis the change in temperature.

Plugging in the values:

qcopper = 50 g ×0.385 J/g

C×(90 −Tf)C

qcopper = 19.25 ×(90 −Tf) J

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

temperature. Using the same formula:

qwater = 100 g ×4.18 J/g

C×(Tf−20)C

qwater = 418 ×(Tf−20) J

Step 3: Since the system is isolated and assuming no heat loss to the sur-

roundings, the heat lost by the copper must be equal to the heat gained by the

water. Equating qcopper and qwater:

19.25 ×(90 −Tf) = 418 ×(Tf−20)

Step 4: Solve the equation for the final temperature (Tf):

1732.5−19.25Tf= 418Tf−8360

19.25Tf+ 418Tf= 8360 + 1732.5

437.25Tf= 10092.5

Tf=10092.5

437.25 ≈23.10C

Therefore, the final temperature of the system will be approximately 23.10

Question 6

Question

A 50 g piece of aluminum at 100

C is placed in 200 g of water at 20

C in a

calorimeter. If the final equilibrium temperature of the system is 25

C, cal-

culate the specific heat capacity of aluminum. Assume no heat is lost to the

surroundings.

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= 200 g is the mass of the water, - c= 4.18 J/g

C is the specific

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

the water.

Substitute the values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Qwater = 4180 J

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

aluminum can be calculated using the same formula:

Qaluminum =mc∆T

where: - m= 50 g is the mass of the aluminum, - cis the specific heat

capacity of aluminum (to be determined), - ∆T= 25C−100C=−75Cis the

temperature change of the aluminum. (Negative because it loses heat.)

Substitute the values into the formula:

Qaluminum = (50 g)(c)(−75C)

Qaluminum =−3750c

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

surroundings, the heat lost by the aluminum is equal to the heat gained by the

water. Therefore:

Qwater =Qaluminum

4180 = −3750c

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

c=4180

−3750

c≈ −1.11 J/g

Therefore, the specific heat capacity of aluminum is approximately 1.11

J/g

Question 7

Question

A piece of iron with mass 0.5 kg is heated to a temperature of 100◦C and then

dropped into a container of water with mass 1 kg at a temperature of 20◦C.

The final temperature of the system is 25◦C. Assuming all the heat lost by the

iron is gained by the water and that there is no heat loss to the surroundings,

calculate the specific heat capacity of iron. (Specific heat capacity of water is

4186 J/kg·K)

Solution

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

can be calculated using the formula:

Qwater =mc∆T

where mis the mass of the water, cis the specific heat capacity of water, and ∆T

is the change in temperature. Given: m= 1 kg c= 4186 J/kg·K ∆T= (25−20)

◦C = 5 K

Substitute the given values into the formula:

Qwater = 1 ×4186 ×5

Qwater = 20930 J

Step 2: Calculate the heat lost by the iron The heat lost by the iron can

also be calculated using the formula:

Qiron =mc∆T

where mis the mass of the iron, cis the specific heat capacity of iron, and ∆T

is the change in temperature. Given: m= 0.5 kg ∆T= (100 −25) ◦C = 75 K

Now, we know that the heat lost by the iron is equal to the heat gained by

the water:

Qiron =Qwater

0.5c75 = 20930

c=20930

0.5×75

c558.67 J/kg·K

Therefore, the specific heat capacity of iron is approximately 558.67 J/kg·K.

Question 8

Question

A 50 g piece of copper at 150

C is dropped into 200 g of water at 20

C in a

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

temperature of the system? Given specific heat capacities: water = 4.18 J/g

copper = 0.385 J/g

Solution

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

temperature. We can use the formula Q=mc∆Twhere Qis the heat energy, m

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

Heat lost by copper piece = (50 g)(0.385 J/g

C)(150 −T) J

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

temperature.

Heat gained by water = (200 g)(4.18 J/g

C)(T−20) J

Step 3: Since the total heat lost by the copper piece is equal to the total

heat gained by the water (assuming no heat is lost to the surroundings), we

have the equation

(50)(0.385)(150 −T) = (200)(4.18)(T−20)

Step 4: Solve for Tto find the final temperature of the system.

(50)(0.385)(150 −T) = (200)(4.18)(T−20)

19.25(150 −T) = 836(T−20)

2887.5−19.25T= 836T−16720

1755.5 = 855.25T

T2.05

The final temperature of the system is approximately 2.05

Question 9

Question

A 50 g piece of aluminum at 80

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

(Specific heat capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat gained or lost by the aluminum using the formula:

Qaluminum =maluminum ×caluminum ×∆Taluminum

where: - maluminum = mass of aluminum = 50 g - caluminum = specific heat

capacity of aluminum (unknown) - ∆Taluminum = change in temperature of

aluminum = final temperature - initial temperature = (25C−80C) = −55C

Qaluminum = 50 g ×caluminum × −55C

Step 2: Calculate the heat gained or lost by the water using the formula:

Qwater =mwater ×cwater ×∆Twater

where: - mwater = 200 g - cwater = 4.18 J/g

C - ∆Twater = 25C−20C= 5C

Qwater = 200 g ×4.18 J/g

C×5C

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

minum must equal the heat gained by the water. Therefore:

Qaluminum =−Qwater

Step 4: Set the equations equal to each other and solve for caluminum:

50 g ×caluminum × −55 = 200 g ×4.18 J/g

C×5

caluminum =200 ×4.18 ×5

50 ×55

caluminum =4180

550

caluminum = 7.6 J/g

Therefore, the specific heat capacity of aluminum is 7.6 J/g

Question 10

Question

A 50 g piece of copper initially at 150◦C is placed in 100 g of water initially at

25◦C. If the final temperature of the system is 28◦C, calculate the specific heat

capacity of copper. Assume the specific heat capacity of water is 4.18 J/(g·◦C).

Solution

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

using the formula:

qcopper =−qwater

The heat lost by the copper is given by

qcopper =mc∆T

where mis the mass of the copper, cis the specific heat capacity of copper,

and ∆Tis the change in temperature of the copper. Since the final temperature

of the system is 28◦C, the change in temperature for the copper is

∆Tcopper = 28 −150 = −122

Substituting the values, we get

qcopper = (0.05 kg)(c)(−122)

The heat gained by the water is given by

qwater =mc∆T

where mis the mass of the water (100 g), cis the specific heat capacity

of water, and ∆Tis the change in temperature of the water. The change in

temperature for the water is

∆Twater = 28 −25 = 3

Substituting the values, we get

qwater = (0.1 kg)(4.18 J/(g ·◦C))(3 ◦C)

Step 2: Set up the equation using the conservation of energy principle:

mc∆Tcopper =−mc∆Twater

Substitute the expressions for qcopper and qwater to get

(0.05c)(−122) = −(0.1)(4.18)(3)

Step 3: Solve for the specific heat capacity of copper:

−6.1c=−1.254

c=1.254

6.1≈0.205 J/(g ·◦C)

Therefore, the specific heat capacity of copper is approximately 0.205 J/(g·◦C).

Question 11

Question

A 50 g piece of copper at 500 K is placed in 200 g of water at 300 K. If the

final temperature of the system is 400 K, calculate the specific heat capacity of

copper. Assume no heat is lost to the surroundings. Specific heat capacity of

water = 4186 J/kg◦C, specific heat capacity of copper = 387 J/kg◦C.

Solution

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

specific heat capacity of copper, CWbe the specific heat capacity of water, mC

be the mass of copper, mWbe the mass of water, and ∆Tbe the change in

temperature of the system. The heat lost by copper is equal to the heat gained

by water, so we have:

mCCC∆T=mWCW∆T

Substitute the given values:

50 g ×CC×(400 −500) = 200 g ×4186 J/kg◦C×(400 −300)

Simplify the equation:

−5000CC= 418600

CC=418600

5000 = 83.72 J/kg◦C

Therefore, the specific heat capacity of copper is 83.72 J/kg◦C .

Question 12

Question

A 250-g block of copper at 95

C is placed in 600 g of water at 25

C. The specific

heat capacity of copper is 0.385 J/g

C and water is 4.18 J/g

C. Assuming no

heat is lost to the surroundings, calculate the final temperature when thermal

equilibrium is reached.

Solution

Step 1: Calculate the heat gained by the water. The formula for heat gained or

lost is Q=mc∆T, where mis the mass of the substance, cis the specific heat

capacity, and ∆Tis the change in temperature. In this case, the water is losing

heat, so we have: Qwater =mwatercwater∆Twater

Substitute the values: Qwater = 600 g ×4.18 J/g

C×(Tf−25C) where Tf

is the final temperature.

Step 2: Calculate the heat lost by the copper block. Similar to the water,

the copper block is gaining heat: Qcopper =mcopperccopper∆Tcopper

Substitute the values: Qcopper = 250 g ×0.385 J/g

C×(Tf−95C)

Step 3: Set up the heat gained by water equal to the heat lost by the copper

block. Qwater =Qcopper

600 ×4.18 ×(Tf−25) = 250 ×0.385 ×(Tf−95)

Step 4: Solve the equation for Tf, the final temperature. 2508(Tf−25) =

96.25(Tf−95)

Expand and simplify: 2508Tf−62700 = 96.25Tf−9163.75

2411.75Tf= 53536.25

Tf=53536.25

2411.75 ≈22.2C

Therefore, the final temperature when thermal equilibrium is reached is ap-

proximately 22.2

Question 13

Question

A chemistry student wants to determine the enthalpy of combustion of a new

compound. To do this, they burn 2.50 g of the compound in a bomb calorimeter.

The temperature of the calorimeter increases by 6.75

C. If the heat capacity

of the calorimeter is 18.2 J/

C, calculate the enthalpy of combustion of the

compound in kJ/g.

Solution

Step 1: Calculate the heat absorbed by the calorimeter.

Heat absorbed by calorimeter = C×∆T

where Cis the heat capacity of the calorimeter and ∆Tis the change in tem-

perature.

= 18.2 J/

C×6.75

= 122.85 J

Step 2: Calculate the heat released by the compound during combustion.

Since the heat released by the compound is equal in magnitude but opposite in

sign to the heat absorbed by the calorimeter,

Heat released by compound = −122.85 J

Step 3: Convert the mass of the compound to moles. First, find the molar

mass of the compound by adding the atomic masses of the elements in the

compound. Let’s say the molar mass is 100 g/mol. Then, the moles of the

compound burned would be:

2.50 g

100 g/mol = 0.025 mol

Step 4: Calculate the enthalpy of combustion of the compound per gram.

Enthalpy of combustion per mole = Heat released by compound

moles of compound burned

=−122.85 J

0.025 mol

=−4,914 J/mol

Step 5: Convert the enthalpy of combustion to kJ/g.

Enthalpy of combustion per gram = −4,914 J/mol

1000 J/kJ ×2.50 g

=−0.785 kJ/g

Therefore, the enthalpy of combustion of the compound is -0.785 kJ/g.

Question 14

Question

A 200 g piece of copper at 70◦C is placed into 400 g of water at 20◦C. The

final temperature of the mixture is 25◦C. Assuming no heat is lost to the sur-

roundings, what is the specific heat capacity of copper? Take the specific heat

capacity of water to be 4186 J/kg ·K and the specific heat capacity of copper to

be 386 J/kg ·K.

Solution

Step 1: Calculate the heat lost by the copper piece.

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

Qlost =mc∆T,

where: - mis the mass of the copper piece (200 g = 0.2 kg), - cis the specific

heat capacity of copper (386 J/kg ·K), - ∆Tis the change in temperature of

the copper piece.

Given that the initial temperature of the copper piece is 70◦C and the final

temperature of the mixture is 25◦C:

∆T= (25◦C) −(70◦C) = −45◦C.

Therefore, the heat lost by the copper piece is:

Qlost = (0.2 kg)(386 J/kg ·K)(−45 K) = −3474 J.

Step 2: Calculate the heat gained by the water.

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

Qgained =mc∆T,

where: - mis the mass of the water (400 g = 0.4 kg), - cis the specific heat

capacity of water (4186 J/kg ·K), - ∆Tis the change in temperature of the

water.

Given that the initial temperature of the water is 20◦C and the final tem-

perature of the mixture is 25◦C:

∆T= (25◦C) −(20◦C) = 5◦C.

Therefore, the heat gained by the water is:

Qgained = (0.4 kg)(4186 J/kg ·K)(5 K) = 8368 J.

Step 3: Set up the energy conservation equation.

Since no heat is lost to the surroundings, the heat lost by the copper piece

is equal to the heat gained by the water:

Qlost =Qgained.

Therefore,

−3474 J = 8368 J.

This equation has no solution, indicating that there may be an error in the

calculations. Let’s recheck the calculations.

Question 15

Question

A 50 g aluminum block at 100

C is dropped into 100 g of water at 20

C in a

calorimeter. If the final temperature of the system is 30

C, what is the specific

heat capacity of the aluminum block? Assume no heat is lost to the surroundings

and that 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 g (mass of water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T= (Tfinal −Tinitial) = (30C−20C) = 10C.

Calculating:

Qwater = 100 ×4.18 ×10 = 4180 J

Step 2: Use the heat absorbed by the water to find the specific heat capacity

of aluminum using the formula:

Qwater =maluminum ×caluminum ×∆T

Since no heat is lost to the surroundings, the heat absorbed by the aluminum

is equal to the heat lost by the water:

4180 J = 50 ×caluminum ×(30 −100)

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

caluminum =4180

50 ×(30 −100) =4180

50 × −70 =−1.194 J/g

Therefore, the specific heat capacity of the aluminum block is −1.194 J/g

Question 16

Question

A 50 g aluminum block is heated to 100

C and then placed into 200 g of water

at 25

C contained in a calorimeter. The final temperature of the system is 30

Assume no heat is lost to the surroundings. Calculate the specific heat capacity

of aluminum. The specific heat capacity of water is 4.18 J/g

Solution

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

down from 100

C to 30

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

Given: Mass of aluminum block (m) = 50 g Initial temperature of aluminum

block (Tinitial) = 100

C Final temperature of aluminum block (Tfinal) = 30

Specific heat capacity of aluminum (cAl) = 0.897 J/g

The formula to calculate heat energy (Q) is:

Q=mcAl∆T

where ∆T=Tfinal −Tinitial.

Substitute the values:

Q= (50 g)(0.897 J/g

C)(30 −100)C

⇒Q= (50 g)(0.897 J/g

C)(−70)C=−3141 J

Therefore, the heat energy absorbed by the aluminum block is −3141 J.

Step 2: Calculate the heat energy released by the aluminum block to the

water and the calorimeter. Since the system is isolated, the heat released by

the aluminum block will be absorbed by the water and the calorimeter.

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

gained by the water and the calorimeter. Therefore:

QAluminum =QWater +QCalorimeter

Given: Mass of water (mwater) = 200 g Specific heat capacity of water (cwater)

= 4.18 J/g

C Change in temperature of water (∆Twater) = 30

C - 25

C = 5

Change in temperature of calorimeter (∆Tcalorimeter) = 30

C - 25

C = 5

Substitute the values into the equation:

3141 J = (200 g)(4.18 J/g

C)(5C)+(mcalorimeter)(ccalorimeter)(5C)

Since no information about the mass and specific heat capacity of the calorime-

ter is given, we cannot solve for the specific heat capacity of aluminum.

Question 17

Question

A 50 g piece of aluminum at 90

C is dropped into 200 g of water at 20

C in an

insulated calorimeter. Assuming no heat is lost to the surroundings, what will

be the final temperature of the system? (Specific heat capacity of aluminum is

0.897 J/g

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

C.)

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 transferred, mis the

mass of the substance, cis the specific heat capacity, and ∆Tis the change in

temperature. For aluminum: Qaluminum = (50 g)(0.897 J/g

C)(Tf−90), where

Tfis the final temperature.

Step 2: For water: Qwater = (200 g)(4.18 J/g

C)(Tf−20)

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

water (assuming no heat is lost to the surroundings), we can set them equal to

each other and solve for Tf: (50 g)(0.897 J/g

C)(Tf−90) = (200 g)(4.18 J/g

C)(Tf−

20)

Step 4: Now, solve for Tf: 44.85(Tf−90) = 836(Tf−20) 44.85Tf−4036.5 =

836Tf−16720

Step 5: Rearrange the equation and solve for Tf: 836Tf−44.85Tf= 16720−

4036.5 791.15Tf= 12683.5Tf=12683.5

791.15 Tf≈16

Therefore, the final temperature of the system will be approximately 16

Question 18

Question

A 50.0 g sample of iron at 85.0

C is placed in 200.0 g of water at 20.0

C. The

specific heat capacity of iron is 0.449 J/(g

C) and the specific heat capacity of

water is 4.18 J/(g

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

be the final equilibrium temperature of the system?

Solution

Step 1: Calculate the heat absorbed by the iron as it cools down to the final

temperature. The heat absorbed by the iron is given by the formula:

qiron =m·ciron ·∆Tiron

where: - m= mass of the iron sample = 50.0 g - ciron = specific heat capac-

ity of iron = 0.449 J/(g

C) - ∆Tiron = change in temperature of iron = final

temperature - initial temperature

∆Tiron =Tfinal −85.0

qiron = 50.0·0.449 ·(Tfinal −85.0)

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

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

qwater =m·cwater ·∆Twater

where: - m= mass of the water = 200.0 g - cwater = specific heat capacity of

water = 4.18 J/(g

C) - ∆Twater =Tfinal −20.0

qwater = 200.0·4.18 ·(Tfinal −20.0)

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

iron is equal to the heat gained by the water.

qiron =qwater

50.0·0.449 ·(Tfinal −85.0) = 200.0·4.18 ·(Tfinal −20.0)

Step 4: Solve for the final equilibrium temperature Tfinal.

50.0·0.449 ·(Tfinal −85.0) = 200.0·4.18 ·(Tfinal −20.0)

22.45 ·(Tfinal −85.0) = 836.0·(Tfinal −20.0)

22.45 ·Tfinal −22.45 ·85.0 = 836.0·Tfinal −836.0·20.0

22.45 ·Tfinal −1902.25 = 836.0·Tfinal −16720.0

813.55 = 813.55Tfinal

Tfinal =813.55

813.55 = 1.0

Therefore, the final equilibrium temperature of the system is 1.0

Question 19

Question

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

metal sample using a calorimeter. The student places a 200 g aluminum sample

at 100◦C into 400 g of water at 20◦C. After thermal equilibrium is reached, the

final temperature of the system is 25◦C. Calculate the specific heat capacity of

the metal.

Solution

Step 1: Calculate the heat transfer from the aluminum to the water. The heat

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

mal∆Tal=mwcw∆Tw

where mal= mass of aluminum (kg), ∆Tal= change in temperature of alu-

minum (K), mw= mass of water (kg), cw= specific heat capacity of water

(J/(kg·K)), ∆Tw= change in temperature of water (K).

Given: mal= 0.2 kg, ∆Tal= 100 −25 = 75 K, mw= 0.4 kg, cw= 4186

J/(kg·K), and ∆Tw= 25 −20 = 5 K.

Substitute the values and solve for l:

0.2×l×75 = 0.4×4186 ×5

l=0.4×4186 ×5

0.2×75

l= 4186 J/kg

Step 2: Calculate the specific heat capacity of the metal. The heat lost by

the metal is equal to the heat gained by the water:

malcal∆Tal=mwcw∆Treservoir

where cal= specific heat capacity of the metal (J/(kg·K)), ∆Treservoir = change

in temperature of the water after reaching equilibrium (K).

Given: mal= 0.2 kg, cal=l= 4186 J/(kg·K), ∆Tal= 75 K, mw= 0.4 kg,

cw= 4186 J/(kg·K), and ∆Treservoir = 25 −20 = 5 K.

Substitute the values and solve for cal:

0.2×4186 ×75 = 0.4×4186 ×5

cal=0.4×4186 ×5

0.2×75

cal= 4186 J/kg

Therefore, the specific heat capacity of the metal sample is 4186 J/(kg·K).

Question 20

Question

A 50 g piece of iron at 150

C is dropped into 200 g of water at 25

C in a 100

g aluminum cup. The final temperature of the system is 40

C. Assuming no

heat is lost to the surroundings, determine the specific heat capacity of iron.

The specific heat capacities of iron, water, and aluminum are 0.45 J/g

C, 4.18

J/g

C, and 0.897 J/g

C, respectively.

Solution

Step 1: Calculate the heat gained by the water and the aluminum cup. 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) - Tf= 40

C (final temperature) - Ti= 25

C (initial temper-

ature)

∆Twater =Tf−Ti= 40 −25 = 15

Qwater = 200 g ×4.18 J/g

C×15

Qwater = 12540 J

The heat gained by the aluminum cup can be calculated using the same

formula:

Qaluminum =maluminum ·caluminum ·∆Taluminum

Where: - maluminum = 100 g (mass of aluminum) - caluminum = 0.897 J/g

(specific heat capacity of aluminum) - ∆Taluminum = 40 −25 = 15

Qaluminum = 100 g ×0.897 J/g

C×15

Qaluminum = 1345.5 J

Step 2: Calculate the heat lost by the iron to the water and the aluminum

cup. Since no heat is lost to the surroundings, the heat lost by the iron is equal

to the heat gained by the water and the aluminum cup.

Qiron =Qwater +Qaluminum

Qiron = 12540 J + 1345.5 J

Qiron = 13885.5 J

Step 3: Use the formula for heat

Qiron =miron ·ciron ·∆Tiron

Since the mass of the iron is 50 g, we can solve for the specific heat capacity of

iron.

ciron =Qiron

miron ·∆Tiron

ciron =13885.5 J

50 g ×(40 −150)

ciron =13885.5

50 × −110

ciron =−2.523 J/g

Therefore, the specific heat capacity of iron is 2.523 J/g

C .

Question 21

Question

A 50 g piece of aluminum at 100

C is dropped into 100 g of water at 20

C in a

calorimeter. The final temperature of the system is 22

C. Assuming no heat is

lost to the surroundings, calculate the specific heat capacity of aluminum.

Solution

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

water.



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

QAluminum =m×cAluminum ×∆TAluminum

Where: m= 50 g (mass of aluminum), cAluminum = 0.903 J/g

C (specific

heat capacity of aluminum), and ∆TAluminum =Tf−Ti= 22 −100 =

−78

C (change in temperature of the aluminum).

Substituting the values, we get:

QAluminum = 50 ×0.903 ×(−78) = −3525.6 J



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

QWater =m×cWater ×∆TWater

Where: m= 100 g (mass of water), cWater = 4.18 J/g

C (specific heat

capacity of water), and ∆TWater =Tf−Ti= 22 −20 = 2

C (change in

temperature of the water).

Substituting the values, we get:

QWater = 100 ×4.18 ×2 = 836 J

Question 22

Question

A 100 g piece of aluminum at 80

C is placed in a calorimeter containing 200

g of water at 20

C. If the final temperature of the system is 25

C, what is the

heat capacity of the calorimeter? The specific heat capacities of aluminum and

water are 0.90 J/g

C and 4.18 J/g

C, respectively.

Solution

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

water. The heat lost by the aluminum can be calculated using the formula:

QAl =mAl ·cAl ·∆T

where: - mAl is the mass of aluminum - cAl is the specific heat capacity of

aluminum - ∆Tis the temperature change of aluminum

Substitute the given values into the formula:

QAl = 100 g ·0.90 J/g

C·(25 −80)

QAl = 100 ·0.90 ·(−55)

QAl =−4950 J

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

Qwater =mwater ·cwater ·∆T

where: - mwater is the mass of water - cwater is the specific heat capacity of water

- ∆Tis the temperature change of water

Substitute the given values into the formula:

Qwater = 200 g ·4.18 J/g

C·(25 −20)

Qwater = 200 ·4.18 ·5

Qwater = 4180 J

Step 2: Use the principle of conservation of energy to find the heat capacity

of the calorimeter. The heat lost by the aluminum is equal to the heat gained

by the water and calorimeter:

QAl =Qwater +Qcalorimeter

Substitute the calculated values:

−4950 = 4180 + Qcalorimeter

Qcalorimeter =−4950 −4180

Qcalorimeter =−9130 J

Step 3: Calculate the heat capacity of the calorimeter using the formula:

Qcalorimeter =Ccalorimeter ·∆T

where: - Ccalorimeter is the heat capacity of the calorimeter - ∆T=Tfinal −

Tinitial = 25 −20 = 5

Substitute the known values:

−9130 = Ccalorimeter ·5

Ccalorimeter =−9130

Ccalorimeter =−1826 J/

Therefore, the heat capacity of the calorimeter is 1826 J/

C .

Question 23

Question

A 50 g piece of iron at 80

C is dropped into a calorimeter containing 100 g of

water at 20

C. If the final temperature of the system is 30

C, what is the heat

capacity of the calorimeter? Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat gained by the water to warm up from 20

C to 30

Step 2: Calculate the heat lost by the iron to cool down from 80

C to 30

Step 3: Use the fact that the total heat lost by the iron is equal to the total

heat gained by the water and the calorimeter to find the heat capacity of the

calorimeter.

Step 1: The specific heat capacity of water is 4.18 J/g

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

qwater =m·c·∆T

where: - mis the mass of the water (100 g), - cis the specific heat capacity of

water (4.18 J/g

C), - ∆Tis the change in temperature (from 20

C to 30

C).

Substitute the values into the formula:

qwater = 100 g ×4.18 J/g

C×(30 −20)

qwater = 100 g ×4.18 J/g

C×10

qwater = 4180 J

Therefore, the heat gained by the water is 4180 J.

Step 2: The specific heat capacity of iron is 0.45 J/g

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

qiron =m·c·∆T

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

(0.45 J/g

C), - ∆Tis the change in temperature (from 80

C to 30

C).

Substitute the values into the formula:

qiron = 50 g ×0.45 J/g

C×(30 −80)

qiron = 50 g ×0.45 J/g

C× −50

qiron =−1125 J

Therefore, the heat lost by the iron is -1125 J. Note the negative sign indi-

cates heat loss.

Step 3: The total heat gained by the water and the calorimeter is equal

to the absolute value of the total heat lost by the iron. Therefore, |qwater +

qcalorimeter|=|qiron|.

Substitute the values of qwater and qiron into the equation:

|4180 J + qcalorimeter|=| − 1125 J|

4180 J + qcalorimeter = 1125 J

qcalorimeter = 1125 J −4180 J

qcalorimeter =−3055 J

Therefore, the heat capacity of the calorimeter is 3055 J.

Question 24

Question

A 50 g piece of copper at 150

C is placed in 200 g of water at 20

C. Assuming

no heat is lost to the surroundings, what will be the final temperature of the

system? (Specific heat capacity of copper = 0.386 J/g

C, specific heat capacity

of water = 4.18 J/g

Solution

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

the formula Q=mc∆Twhere Qis heat, mis mass, cis specific heat capacity,

and ∆Tis the change in temperature.

For copper: Qcopper = (50 g)(0.386 J/g

C)(Tf−150C)

For water: Qwater = (200 g)(4.18 J/g

C)(Tf−20C)

Since no heat is lost or gained from the surroundings, we have Qcopper =

−Qwater

Step 2: Set up the equation: (50 g)(0.386 J/g

C)(Tf−150C) = −(200 g)(4.18 J/g

C)(Tf−

20C)

Step 3: Solve for the final temperature Tf: 50(0.386)(Tf−150) = −200(4.18)(Tf−

20)

19.3Tf−2895 = −836Tf+ 16720

855Tf= 19615

Tf≈22.96C

Therefore, the final temperature of the system will be approximately 22.96C.

Question 25

Question

A 50 g piece of aluminum at 100

C is placed in 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 22

C. Assuming no heat

is lost 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 lost by the aluminum and the heat gained by the

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

Therefore, we have:

mAl ·cAl ·(Tf−Ti) = mw·cw·(Tf−Ti)

Substitute the given values:

50 g ·cAl ·(22 −100) = 200 g ·4.18 J/g

C·(22 −20)

Simplify and solve for cAl.

Step 2: Calculate the specific heat capacity of aluminum.

50 g ·cAl ·(−78) = 200 g ·4.18 J/g

C·2

cAl =200 g ·4.18 J/g

C·2

50 g ·78

cAl =1672 J

3900 g ·

cAl ≈0.43 J/g

Therefore, the specific heat capacity of aluminum is approximately 0.43 J/g

Question 26

Question

A student wants to determine the specific heat capacity of a metal using calorime-

try. They take a 150 g block of the metal at an initial temperature of 200

and place it in 200 g of water at an initial temperature of 20

C contained in a

calorimeter. The final temperature of the system is 23

C. What is the specific

heat capacity of the metal? Assume no heat is lost to the surroundings.

Solution

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

water can be calculated using the formula:

Qwater =m·c·∆T

where: - m= 200 g is the mass of water, - c= 4.18 J/g

C is the specific heat

capacity of water, - ∆T=Tf−Ti= 23C−20C= 3Cis the change in temper-

ature.

Substitute the values into the formula:

Qwater = 200 g ×4.18 J/g

C×3C= 2508 J

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

is equal to the heat absorbed by the water, as there is no heat lost to the

surroundings. We can use the formula:

Qmetal =−Qwater

Qmetal =−2508 J

Step 3: Calculate the specific heat capacity of the metal. The heat lost by

the metal can be calculated using the formula:

Qmetal =m·cmetal ·∆T

where: - m= 150 g is the mass of the metal and - ∆T=Tf−Ti= 200C−23C=

177Cis the change in temperature.

Substitute the known values and solve for cmetal:

−2508 J = 150 g ×cmetal ×177C

cmetal =−2508 J

150 g ×177C

cmetal ≈ −0.094 J/g

Therefore, the specific heat capacity of the metal is approximately -0.094

J/g

Question 27

Question

A piece of metal with mass 0.25 kg is heated to 100◦C and then placed in 0.5

kg of water at 20◦C. The final temperature of the system is 25◦C. If the specific

heat capacity of the metal is 450 J/kg·K, determine the specific heat capacity

of the metal.

Solution

Step 1: Calculate the heat lost by the metal as it cools down from 100◦C to

the final temperature. The heat lost by the metal can be calculated using the

formula:

Qmetal =mc∆T

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

- ∆Tis the change in temperature of the metal.

Substitute the values:

Qmetal = 0.25 kg ×450 J/kg·K×(100 −25) K

Qmetal = 0.25 ×450 ×75

Qmetal = 8437.5 J

Step 2: Calculate the heat gained by the water as it heats up from 20◦C 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, - cis the specific heat capacity of water, -

∆Tis the change in temperature of the water.

Substitute the values:

Qwater = 0.5 kg ×4186 J/kg·K×(25 −20) K

Qwater = 0.5×4186 ×5

Qwater = 10465 J

Step 3: Since energy is conserved, the heat lost by the metal is equal to the

heat gained by the water.

Qmetal =Qwater

8437.5 = 10465

Note: We reached an error in our calculation, please check the heat lost by

the metal and heat gained by the water and make sure they are equal.

Question 28

Question

A piece of metal of mass 0.5 kg at 150

C is placed in 1 kg of water at 20

contained in a 1 kg aluminum calorimeter cup at 20

C. The final temperature

of the system is 30

C. Given the specific heat capacities of the metal and water

are 0.5 J/g

C and 4.18 J/g

C respectively, calculate the specific heat capacity

of the metal.

Solution

Step 1: Calculate the heat gained by the water and the calorimeter cup.

The heat gained by the water and calorimeter cup can be calculated using the

formula:

Q=mc∆T

where Qis the heat gained, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature.

For water:

Qwater = (1 kg ×1000 g/kg ×4.18 J/g

C×(30 −20)

Qwater = 41800 J

For the calorimeter cup (assume specific heat capacity of aluminum is 0.9

J/g

C):

Qcup = (1 kg ×1000 g/kg ×0.9 J/g

C×(30 −20)

Qcup = 900 J

Step 2: Calculate the heat lost by the metal.

The heat lost by the metal is equal to the sum of the heat gained by the water

and calorimeter cup:

Qmetal =Qwater +Qcup

Qmetal = 41800 J + 900 J

Qmetal = 42700 J

Step 3: Calculate the specific heat capacity of the metal.

We know that the heat lost by the metal is given by:

Qmetal =mc∆T

Substitute the values we have:

42700 = 0.5×c×(30 −150)

42700 = −60c

c=−42700

c≈ −711.67 J/kg

Therefore, the specific heat capacity of the metal is approximately 711.67

J/kg

Question 29

Question

A piece of copper with a mass of 200 g is heated to a temperature of 100

and then dropped into a calorimeter containing 500 g of water at an initial

temperature of 20

C. If the final temperature of the system is 25

C, calculate

the specific heat capacity of copper. Assume no heat is lost to the surroundings

and 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.

Heat lost by copper = Heat gained by water

Step 2: The heat lost by the copper is given by the formula:

Qcopper =mc∆T

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

(unknown) - ∆Tis the change in temperature of the copper (final temperature

- initial temperature)

Qcopper = (200 g)(c)(25 −100)

Qcopper =−15000cJ

Step 3: The heat gained by the water is given by the formula:

Qwater =mc∆T

where: - mis the mass of water (500 g) - cis the specific heat capacity of water

(4.18 J/g

C) - ∆Tis the change in temperature of the water (final temperature

- initial temperature)

Qwater = (500 g)(4.18 J/g

C)(25 −20)

Qwater = 2090 J

Step 4: Set the heat lost by the copper equal to the heat gained by the water:

−15000c= 2090

Step 5: Solve for the specific heat capacity of copper:

c=2090

−15000 =−0.139 J/g

Therefore, the specific heat capacity of copper is -0.139 J/g

C. The negative

sign indicates that heat is being lost by the copper.

Question 30

Question

A 50 g piece of aluminum initially at 80

C is dropped into 200 g of water at

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

of the system? (Specific heat capacity of aluminum = 0.9 J/g

C, specific heat

capacity of water = 4.18 J/g

Solution

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

aluminum can be calculated using the formula:

qaluminum =mc∆T

where: - m= 50 g (mass of aluminum) - c= 0.9 J/g

C (specific heat capacity

of aluminum) - ∆T=Tfinal −Tinitial =Tfinal −80

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

water can be calculated using the formula:

qwater =mc∆T

where: - m= 200 g (mass of water) - c= 4.18 J/g

C (specific heat capacity of

water) - ∆T=Tfinal −Tinitial =Tfinal −20

Step 3: Set up the conservation of energy equation: Since the system is

isolated and no heat is lost to the surroundings, the heat lost by the aluminum

must be equal to the heat gained by the water, so we have:

qaluminum =qwater

Step 4: Substitute the calculated values and solve for the final temperature,

Tfinal.

maluminumcaluminum(Tfinal −80) = mwatercwater(Tfinal −20)

Step 5: Simplify the equation and solve for Tfinal.

50 ×0.9×(Tfinal −80) = 200 ×4.18 ×(Tfinal −20)

45(Tfinal −80) = 836(Tfinal −20)

45Tfinal −3600 = 836Tfinal −16720

791Tfinal = 13120

Tfinal =13120

791 ≈16.58

Therefore, the final temperature of the system is approximately 16.58

Question 31

Question

A 50 g aluminum block is heated to 100

C and then placed in 200 g of water

at 20

C inside a calorimeter. The final temperature of the system is 25

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

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

Solution

Step 1: Calculate the heat absorbed by the water during the cooling process.

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

qwater =mwater ×cwater ×∆T

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

C is the specific

heat capacity of water, and - ∆T= 25 −20 = 5

C is the temperature change

of the water. Substitute these values into the formula:

qwater = 200 ×4.18 ×5 = 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, so we

have:

qaluminum =−qwater =−4180 J

Step 3: Calculate the heat lost by the aluminum block using the formula:

qaluminum =maluminum ×caluminum ×∆T

where: - maluminum = 50 g is the mass of the aluminum block, - caluminum is

the specific heat capacity of aluminum (which we want to find), and - ∆T=

100 −25 = 75

C is the temperature change of the aluminum block.

Equating the heat lost by the aluminum block in step 2 and step 3, we have:

50 ×caluminum ×75 = −4180

caluminum =−4180

50 ×75

caluminum =−1.40 J/g

Therefore, the specific heat capacity of the aluminum block is 1.40 J/g

Question 32

Question

A piece of metal with mass 0.2 kg is heated to 100

C and then placed in 0.5 kg

of water at 25

C in an insulated container. The final temperature of the system

is 30

C. If the specific heat capacity of the metal is 450 J/kg

C and the specific

heat capacity of water is 4186 J/kg

C, calculate the specific heat capacity of the

container if none of the heat is lost to the surroundings.

Solution

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

formula:

Qmetal =Qwater

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

Qmetal =mc∆T

where: m= 0.2 kg (mass of the metal), c= 450 J/kg

C (specific heat capacity

of the metal), ∆T= 100 −30 = 70

C (final temperature - initial temperature).

Step 3: Substitute the values into the formula to find Qmetal:

Qmetal = 0.2×450 ×70

Step 4: Calculate the heat gained by the water using the formula:

Qwater =mc∆T

where: m= 0.5 kg (mass of the water), c= 4186 J/kg

C (specific heat capacity

of water), ∆T= 30 −25 = 5

C (final temperature - initial temperature).

Step 5: Substitute the values into the formula to find Qwater:

Qwater = 0.5×4186 ×5

Step 6: Set Qmetal =Qwater and solve for the specific heat capacity of the

container:

0.2×450 ×70 = 0.5×4186 ×5

Step 7: Solve for the specific heat capacity of the container.

Question 33

Question

A 50 g piece of aluminum initially at 100◦C is placed in a calorimeter containing

200 g of water at 20◦C. The final temperature of the system is 25◦C. Assuming

no heat is lost to the surroundings, what is the specific heat capacity of the

aluminum? (Specific heat capacity of water is 4.18 J/g◦C)

Solution

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

water using the formula:

q=m·c·∆T

where: q= heat energy in Joules (J), m= mass in grams (g), c= specific

heat capacity in J/g◦C, ∆T= change in temperature in ◦C.

The heat lost by aluminum is equal to the heat gained by water, since no

heat is lost to the surroundings.

For the aluminum:

qAl = (50 g) ·cAl ·(25 −100)◦C

For the water:

qH2O = (200 g) ·4.18 J/g◦C·(25 −20)◦C

Since qAl =qH2O:

(50 g) ·cAl ·(−75)◦C = (200 g) ·4.18 J/g◦C·5◦C

Simplify and solve for cAl in terms of the given values.

Question 34

Question

A 100 g piece of copper at 200

C is placed into a calorimeter containing 200 g of

water at 20

C. The final temperature of the system is 25

C. Assuming no heat

is lost to the surroundings, what is the specific heat capacity of the calorimeter?

Solution

Given: Mass of copper (mc) = 100 g = 0.1 kg

Initial temperature of copper (Tc,i) = 200

Specific heat capacity of copper (cc) = 390 J/kg ·

Mass of water in calorimeter (mw) = 200 g = 0.2 kg

Initial temperature of water (Tw,i) = 20

Final temperature of the system (Tf) = 25

Specific heat capacity of water (cw) = 4186 J/kg ·

We will assume that no heat is lost to the surroundings, so the total heat

lost by the copper is equal to the total heat gained by the water:

mc·cc·(Tc,i −Tf) = mw·cw·(Tf−Tw,i)

Step 1: Calculate the heat lost by the copper (Qc) and the heat

gained by the water (Qw).

Qc=mc·cc·(Tc,i −Tf)

Qw=mw·cw·(Tf−Tw,i)

Substitute the given values:

Qc= 0.1 kg ·390 J/kg ·

C·(200 −25)

Qw= 0.2 kg ·4186 J/kg ·

C·(25 −20)

Step 2: Equate the heat lost and gained equations. Solve for the

specific heat capacity of the calorimeter.

mc·cc·(Tc,i −Tf) = mw·cw·(Tf−Tw,i)

0.1·390 ·(200 −25) = 0.2·4186 ·(25 −20)

0.1·390 ·175 = 0.2·4186 ·5

6825 = 4186 ·1·c

c=6825

4186

Therefore, the specific heat capacity of the calorimeter is 6825

4186 ≈1.63 J/g·

Question 35

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 22

C. Given the specific

heat capacity of water is 4.18 J/g

C, determine the specific heat capacity of

aluminum. Assume no heat is lost to the surroundings and that the specific

heat capacity of aluminum is constant over the temperature range involved.

Solution

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

can be calculated using the equation:

qwater =m·c·∆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 (22C−20C= 2C)

Substitute the given values to find qwater:

qwater = 200 g ×4.18 J/g

C×2C= 1672 J

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

aluminum block must be equal to the heat gained by the water, since no heat

is lost to the surroundings. We can express this as:

qaluminum =−qwater

Since the heat lost by the aluminum block can be calculated using its specific

heat capacity (caluminum), mass (maluminum = 50 g), and temperature change

(∆Taluminum =Tf−Ti= 22C−80C=−58C):

qaluminum =maluminum ·caluminum ·∆Taluminum

Substitute the known values to find caluminum:

1672 J = 50 g ×caluminum × −58C

Solve for caluminum:

caluminum =1672 J

50 g × −58C≈0.57 J/g

Therefore, the specific heat capacity of aluminum is approximately 0.57 J/g

Step 4: Solve for the specific heat capacity of aluminum. Since Q=mc∆T,

we have: maluminumcaluminum(Tfinal, system−Tinitial, aluminum) = mwatercwater(Tfinal, system−

Tinitial, water)

Substitute the given values and solve for caluminum to obtain the specific heat

capacity of aluminum.

Question 2

Question

A 50 g piece of aluminum at 100◦C is placed in 200 g of water at 20◦C. 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 and that the specific

heat capacity of water is 4.18 J/g·K.

Solution

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

using the formula q=mc∆T, where qis the heat transferred, mis the mass,

cis the specific heat capacity, and ∆Tis the change in temperature. In this

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

For the aluminum: qaluminum = (50 g)(cAl)(−75◦C) For the water: qwater =

(200 g)(4.18 J/g·K)(5◦C)

Step 2: Set the heat lost by the aluminum equal to the heat gained by the

water and solve for the specific heat capacity of aluminum.

(50 g)(cAl)(−75◦C) = (200 g)(4.18 J/g·K)(5◦C)

cAl =(200 g)(4.18 J/g·K)(5◦C)

(50 g)(−75◦C)

Step 3: Calculate the specific heat capacity of aluminum.

cAl =(200)(4.18)(5)

(50)(75)

cAl =4180

375

cAl = 11.15 J/g·K

Therefore, the specific heat capacity of aluminum is 11.15 J/g·K.

Question 3

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C in an insulated

container. If the final temperature of the system is 25

C, determine the specific

heat capacity of iron.

Solution

Let the specific heat capacity of water be cw= 4.18 J/g

Step 1: Calculate the heat absorbed by the water when the iron piece is

added. The heat absorbed by the water can be calculated using the formula:

Q=mc∆T

where: - mis the mass of water, - cis the specific heat capacity of water, - ∆T

is the change in temperature of the water.

Substitute the given values:

Q= 200 ×4.18 ×(25 −20)

Q= 200 ×4.18 ×5

Q= 4180 J

Step 2: Calculate the amount of heat lost by the iron. Since the system is

insulated, the total amount of heat lost by the iron is equal to the amount of

heat gained by the water. Therefore, the heat lost by the iron is also 4180 J.

Step 3: Calculate the specific heat capacity of iron. The heat lost by the

iron can be calculated using the formula:

Q=mc∆T

where: - mis the mass of iron, - cis the specific heat capacity of iron, - ∆Tis

the change in temperature of the iron.

Substitute the given values and the heat lost:

4180 = 50c(25 −80)

4180 = −50c×55

4180 = −2750c

c=4180

−2750

c≈ −1.52 J/g

Therefore, the specific heat capacity of iron is approximately −1.52 J/g

The negative sign indicates that the value obtained is not physically meaningful.

This discrepancy may be due to experimental errors or assumptions made during

the calculation.

Question 4

Question

A 50 g sample of aluminum at 80

C is dropped into 200 g of water at 20

C in

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

C and the

specific heat capacity of water is 4.18 J/g

C. Assuming no heat is lost to the

surroundings, what is the final temperature of the system?

Solution

Step 1: Calculate the heat absorbed by the aluminum: The equation for heat

transfer is given by q=mc∆T, where: - mis the mass of the substance, - cis

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

Substitute the values for aluminum: qAl = (50 g)(0.897 J/g

C)(Tf−80

Step 2: Calculate the heat lost by the water: qwater =−(200 g)(4.18 J/g

C)(Tf−

Step 3: Set up the energy conservation equation: The heat lost by the water

is equal to the heat gained by the aluminum: −(200 g)(4.18 J/g

C)(Tf−20

C) =

(50 g)(0.897 J/g

C)(Tf−80

Step 4: Solve for the final temperature: −(836 J/

C)(Tf−20

C) = (44.85 J/

C)(Tf−

−836Tf+ 16720 = 44.85Tf−3588

836Tf+ 44.85Tf= 16720 + 3588

Tf=20308

880.85 ≈23.05

Therefore, the final temperature of the system is approximately 23.05

Question 5

Question

A 50 g piece of copper at 90

C is placed in 100 g of water at 20

C. Assuming

no heat is lost to the surroundings, what will be the final temperature of the

system? (Specific heat capacity of copper = 0.385 J/g

C, specific heat capacity

of water = 4.18 J/g

Solution

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

temperature. The formula for calculating heat (q) is:

q=m·c·∆T

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

material, - ∆Tis the change in temperature.

Plugging in the values:

qcopper = 50 g ×0.385 J/g

C×(90 −Tf)C

qcopper = 19.25 ×(90 −Tf) J

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

temperature. Using the same formula:

qwater = 100 g ×4.18 J/g

C×(Tf−20)C

qwater = 418 ×(Tf−20) J

Step 3: Since the system is isolated and assuming no heat loss to the sur-

roundings, the heat lost by the copper must be equal to the heat gained by the

water. Equating qcopper and qwater:

19.25 ×(90 −Tf) = 418 ×(Tf−20)

Step 4: Solve the equation for the final temperature (Tf):

1732.5−19.25Tf= 418Tf−8360

19.25Tf+ 418Tf= 8360 + 1732.5

437.25Tf= 10092.5

Tf=10092.5

437.25 ≈23.10C

Therefore, the final temperature of the system will be approximately 23.10

Question 6

Question

A 50 g piece of aluminum at 100

C is placed in 200 g of water at 20

C in a

calorimeter. If the final equilibrium temperature of the system is 25

C, cal-

culate the specific heat capacity of aluminum. Assume no heat is lost to the

surroundings.

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= 200 g is the mass of the water, - c= 4.18 J/g

C is the specific

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

the water.

Substitute the values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Qwater = 4180 J

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

aluminum can be calculated using the same formula:

Qaluminum =mc∆T

where: - m= 50 g is the mass of the aluminum, - cis the specific heat

capacity of aluminum (to be determined), - ∆T= 25C−100C=−75Cis the

temperature change of the aluminum. (Negative because it loses heat.)

Substitute the values into the formula:

Qaluminum = (50 g)(c)(−75C)

Qaluminum =−3750c

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

surroundings, the heat lost by the aluminum is equal to the heat gained by the

water. Therefore:

Qwater =Qaluminum

4180 = −3750c

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

c=4180

−3750

c≈ −1.11 J/g

Therefore, the specific heat capacity of aluminum is approximately 1.11

J/g

Question 7

Question

A piece of iron with mass 0.5 kg is heated to a temperature of 100◦C and then

dropped into a container of water with mass 1 kg at a temperature of 20◦C.

The final temperature of the system is 25◦C. Assuming all the heat lost by the

iron is gained by the water and that there is no heat loss to the surroundings,

calculate the specific heat capacity of iron. (Specific heat capacity of water is

4186 J/kg·K)

Solution

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

can be calculated using the formula:

Qwater =mc∆T

where mis the mass of the water, cis the specific heat capacity of water, and ∆T

is the change in temperature. Given: m= 1 kg c= 4186 J/kg·K ∆T= (25−20)

◦C = 5 K

Substitute the given values into the formula:

Qwater = 1 ×4186 ×5

Qwater = 20930 J

Step 2: Calculate the heat lost by the iron The heat lost by the iron can

also be calculated using the formula:

Qiron =mc∆T

where mis the mass of the iron, cis the specific heat capacity of iron, and ∆T

is the change in temperature. Given: m= 0.5 kg ∆T= (100 −25) ◦C = 75 K

Now, we know that the heat lost by the iron is equal to the heat gained by

the water:

Qiron =Qwater

0.5c75 = 20930

c=20930

0.5×75

c558.67 J/kg·K

Therefore, the specific heat capacity of iron is approximately 558.67 J/kg·K.

Question 8

Question

A 50 g piece of copper at 150

C is dropped into 200 g of water at 20

C in a

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

temperature of the system? Given specific heat capacities: water = 4.18 J/g

copper = 0.385 J/g

Solution

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

temperature. We can use the formula Q=mc∆Twhere Qis the heat energy, m

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

Heat lost by copper piece = (50 g)(0.385 J/g

C)(150 −T) J

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

temperature.

Heat gained by water = (200 g)(4.18 J/g

C)(T−20) J

Step 3: Since the total heat lost by the copper piece is equal to the total

heat gained by the water (assuming no heat is lost to the surroundings), we

have the equation

(50)(0.385)(150 −T) = (200)(4.18)(T−20)

Step 4: Solve for Tto find the final temperature of the system.

(50)(0.385)(150 −T) = (200)(4.18)(T−20)

19.25(150 −T) = 836(T−20)

2887.5−19.25T= 836T−16720

1755.5 = 855.25T

T2.05

The final temperature of the system is approximately 2.05

Question 9

Question

A 50 g piece of aluminum at 80

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

(Specific heat capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat gained or lost by the aluminum using the formula:

Qaluminum =maluminum ×caluminum ×∆Taluminum

where: - maluminum = mass of aluminum = 50 g - caluminum = specific heat

capacity of aluminum (unknown) - ∆Taluminum = change in temperature of

aluminum = final temperature - initial temperature = (25C−80C) = −55C

Qaluminum = 50 g ×caluminum × −55C

Step 2: Calculate the heat gained or lost by the water using the formula:

Qwater =mwater ×cwater ×∆Twater

where: - mwater = 200 g - cwater = 4.18 J/g

C - ∆Twater = 25C−20C= 5C

Qwater = 200 g ×4.18 J/g

C×5C

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

minum must equal the heat gained by the water. Therefore:

Qaluminum =−Qwater

Step 4: Set the equations equal to each other and solve for caluminum:

50 g ×caluminum × −55 = 200 g ×4.18 J/g

C×5

caluminum =200 ×4.18 ×5

50 ×55

caluminum =4180

550

caluminum = 7.6 J/g

Therefore, the specific heat capacity of aluminum is 7.6 J/g

Question 10

Question

A 50 g piece of copper initially at 150◦C is placed in 100 g of water initially at

25◦C. If the final temperature of the system is 28◦C, calculate the specific heat

capacity of copper. Assume the specific heat capacity of water is 4.18 J/(g·◦C).

Solution

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

using the formula:

qcopper =−qwater

The heat lost by the copper is given by

qcopper =mc∆T

where mis the mass of the copper, cis the specific heat capacity of copper,

and ∆Tis the change in temperature of the copper. Since the final temperature

of the system is 28◦C, the change in temperature for the copper is

∆Tcopper = 28 −150 = −122

Substituting the values, we get

qcopper = (0.05 kg)(c)(−122)

The heat gained by the water is given by

qwater =mc∆T

where mis the mass of the water (100 g), cis the specific heat capacity

of water, and ∆Tis the change in temperature of the water. The change in

temperature for the water is

∆Twater = 28 −25 = 3

Substituting the values, we get

qwater = (0.1 kg)(4.18 J/(g ·◦C))(3 ◦C)

Step 2: Set up the equation using the conservation of energy principle:

mc∆Tcopper =−mc∆Twater

Substitute the expressions for qcopper and qwater to get

(0.05c)(−122) = −(0.1)(4.18)(3)

Step 3: Solve for the specific heat capacity of copper:

−6.1c=−1.254

c=1.254

6.1≈0.205 J/(g ·◦C)

Therefore, the specific heat capacity of copper is approximately 0.205 J/(g·◦C).

Question 11

Question

A 50 g piece of copper at 500 K is placed in 200 g of water at 300 K. If the

final temperature of the system is 400 K, calculate the specific heat capacity of

copper. Assume no heat is lost to the surroundings. Specific heat capacity of

water = 4186 J/kg◦C, specific heat capacity of copper = 387 J/kg◦C.

Solution

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

specific heat capacity of copper, CWbe the specific heat capacity of water, mC

be the mass of copper, mWbe the mass of water, and ∆Tbe the change in

temperature of the system. The heat lost by copper is equal to the heat gained

by water, so we have:

mCCC∆T=mWCW∆T

Substitute the given values:

50 g ×CC×(400 −500) = 200 g ×4186 J/kg◦C×(400 −300)

Simplify the equation:

−5000CC= 418600

CC=418600

5000 = 83.72 J/kg◦C

Therefore, the specific heat capacity of copper is 83.72 J/kg◦C .

Question 12

Question

A 250-g block of copper at 95

C is placed in 600 g of water at 25

C. The specific

heat capacity of copper is 0.385 J/g

C and water is 4.18 J/g

C. Assuming no

heat is lost to the surroundings, calculate the final temperature when thermal

equilibrium is reached.

Solution

Step 1: Calculate the heat gained by the water. The formula for heat gained or

lost is Q=mc∆T, where mis the mass of the substance, cis the specific heat

capacity, and ∆Tis the change in temperature. In this case, the water is losing

heat, so we have: Qwater =mwatercwater∆Twater

Substitute the values: Qwater = 600 g ×4.18 J/g

C×(Tf−25C) where Tf

is the final temperature.

Step 2: Calculate the heat lost by the copper block. Similar to the water,

the copper block is gaining heat: Qcopper =mcopperccopper∆Tcopper

Substitute the values: Qcopper = 250 g ×0.385 J/g

C×(Tf−95C)

Step 3: Set up the heat gained by water equal to the heat lost by the copper

block. Qwater =Qcopper

600 ×4.18 ×(Tf−25) = 250 ×0.385 ×(Tf−95)

Step 4: Solve the equation for Tf, the final temperature. 2508(Tf−25) =

96.25(Tf−95)

Expand and simplify: 2508Tf−62700 = 96.25Tf−9163.75

2411.75Tf= 53536.25

Tf=53536.25

2411.75 ≈22.2C

Therefore, the final temperature when thermal equilibrium is reached is ap-

proximately 22.2

Question 13

Question

A chemistry student wants to determine the enthalpy of combustion of a new

compound. To do this, they burn 2.50 g of the compound in a bomb calorimeter.

The temperature of the calorimeter increases by 6.75

C. If the heat capacity

of the calorimeter is 18.2 J/

C, calculate the enthalpy of combustion of the

compound in kJ/g.

Solution

Step 1: Calculate the heat absorbed by the calorimeter.

Heat absorbed by calorimeter = C×∆T

where Cis the heat capacity of the calorimeter and ∆Tis the change in tem-

perature.

= 18.2 J/

C×6.75

= 122.85 J

Step 2: Calculate the heat released by the compound during combustion.

Since the heat released by the compound is equal in magnitude but opposite in

sign to the heat absorbed by the calorimeter,

Heat released by compound = −122.85 J

Step 3: Convert the mass of the compound to moles. First, find the molar

mass of the compound by adding the atomic masses of the elements in the

compound. Let’s say the molar mass is 100 g/mol. Then, the moles of the

compound burned would be:

2.50 g

100 g/mol = 0.025 mol

Step 4: Calculate the enthalpy of combustion of the compound per gram.

Enthalpy of combustion per mole = Heat released by compound

moles of compound burned

=−122.85 J

0.025 mol

=−4,914 J/mol

Step 5: Convert the enthalpy of combustion to kJ/g.

Enthalpy of combustion per gram = −4,914 J/mol

1000 J/kJ ×2.50 g

=−0.785 kJ/g

Therefore, the enthalpy of combustion of the compound is -0.785 kJ/g.

Question 14

Question

A 200 g piece of copper at 70◦C is placed into 400 g of water at 20◦C. The

final temperature of the mixture is 25◦C. Assuming no heat is lost to the sur-

roundings, what is the specific heat capacity of copper? Take the specific heat

capacity of water to be 4186 J/kg ·K and the specific heat capacity of copper to

be 386 J/kg ·K.

Solution

Step 1: Calculate the heat lost by the copper piece.

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

Qlost =mc∆T,

where: - mis the mass of the copper piece (200 g = 0.2 kg), - cis the specific

heat capacity of copper (386 J/kg ·K), - ∆Tis the change in temperature of

the copper piece.

Given that the initial temperature of the copper piece is 70◦C and the final

temperature of the mixture is 25◦C:

∆T= (25◦C) −(70◦C) = −45◦C.

Therefore, the heat lost by the copper piece is:

Qlost = (0.2 kg)(386 J/kg ·K)(−45 K) = −3474 J.

Step 2: Calculate the heat gained by the water.

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

Qgained =mc∆T,

where: - mis the mass of the water (400 g = 0.4 kg), - cis the specific heat

capacity of water (4186 J/kg ·K), - ∆Tis the change in temperature of the

water.

Given that the initial temperature of the water is 20◦C and the final tem-

perature of the mixture is 25◦C:

∆T= (25◦C) −(20◦C) = 5◦C.

Therefore, the heat gained by the water is:

Qgained = (0.4 kg)(4186 J/kg ·K)(5 K) = 8368 J.

Step 3: Set up the energy conservation equation.

Since no heat is lost to the surroundings, the heat lost by the copper piece

is equal to the heat gained by the water:

Qlost =Qgained.

Therefore,

−3474 J = 8368 J.

This equation has no solution, indicating that there may be an error in the

calculations. Let’s recheck the calculations.

Question 15

Question

A 50 g aluminum block at 100

C is dropped into 100 g of water at 20

C in a

calorimeter. If the final temperature of the system is 30

C, what is the specific

heat capacity of the aluminum block? Assume no heat is lost to the surroundings

and that 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 g (mass of water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T= (Tfinal −Tinitial) = (30C−20C) = 10C.

Calculating:

Qwater = 100 ×4.18 ×10 = 4180 J

Step 2: Use the heat absorbed by the water to find the specific heat capacity

of aluminum using the formula:

Qwater =maluminum ×caluminum ×∆T

Since no heat is lost to the surroundings, the heat absorbed by the aluminum

is equal to the heat lost by the water:

4180 J = 50 ×caluminum ×(30 −100)

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

caluminum =4180

50 ×(30 −100) =4180

50 × −70 =−1.194 J/g

Therefore, the specific heat capacity of the aluminum block is −1.194 J/g

Question 16

Question

A 50 g aluminum block is heated to 100

C and then placed into 200 g of water

at 25

C contained in a calorimeter. The final temperature of the system is 30

Assume no heat is lost to the surroundings. Calculate the specific heat capacity

of aluminum. The specific heat capacity of water is 4.18 J/g

Solution

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

down from 100

C to 30

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

Given: Mass of aluminum block (m) = 50 g Initial temperature of aluminum

block (Tinitial) = 100

C Final temperature of aluminum block (Tfinal) = 30

Specific heat capacity of aluminum (cAl) = 0.897 J/g

The formula to calculate heat energy (Q) is:

Q=mcAl∆T

where ∆T=Tfinal −Tinitial.

Substitute the values:

Q= (50 g)(0.897 J/g

C)(30 −100)C

⇒Q= (50 g)(0.897 J/g

C)(−70)C=−3141 J

Therefore, the heat energy absorbed by the aluminum block is −3141 J.

Step 2: Calculate the heat energy released by the aluminum block to the

water and the calorimeter. Since the system is isolated, the heat released by

the aluminum block will be absorbed by the water and the calorimeter.

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

gained by the water and the calorimeter. Therefore:

QAluminum =QWater +QCalorimeter

Given: Mass of water (mwater) = 200 g Specific heat capacity of water (cwater)

= 4.18 J/g

C Change in temperature of water (∆Twater) = 30

C - 25

C = 5

Change in temperature of calorimeter (∆Tcalorimeter) = 30

C - 25

C = 5

Substitute the values into the equation:

3141 J = (200 g)(4.18 J/g

C)(5C)+(mcalorimeter)(ccalorimeter)(5C)

Since no information about the mass and specific heat capacity of the calorime-

ter is given, we cannot solve for the specific heat capacity of aluminum.

Question 17

Question

A 50 g piece of aluminum at 90

C is dropped into 200 g of water at 20

C in an

insulated calorimeter. Assuming no heat is lost to the surroundings, what will

be the final temperature of the system? (Specific heat capacity of aluminum is

0.897 J/g

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

C.)

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 transferred, mis the

mass of the substance, cis the specific heat capacity, and ∆Tis the change in

temperature. For aluminum: Qaluminum = (50 g)(0.897 J/g

C)(Tf−90), where

Tfis the final temperature.

Step 2: For water: Qwater = (200 g)(4.18 J/g

C)(Tf−20)

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

water (assuming no heat is lost to the surroundings), we can set them equal to

each other and solve for Tf: (50 g)(0.897 J/g

C)(Tf−90) = (200 g)(4.18 J/g

C)(Tf−

20)

Step 4: Now, solve for Tf: 44.85(Tf−90) = 836(Tf−20) 44.85Tf−4036.5 =

836Tf−16720

Step 5: Rearrange the equation and solve for Tf: 836Tf−44.85Tf= 16720−

4036.5 791.15Tf= 12683.5Tf=12683.5

791.15 Tf≈16

Therefore, the final temperature of the system will be approximately 16

Question 18

Question

A 50.0 g sample of iron at 85.0

C is placed in 200.0 g of water at 20.0

C. The

specific heat capacity of iron is 0.449 J/(g

C) and the specific heat capacity of

water is 4.18 J/(g

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

be the final equilibrium temperature of the system?

Solution

Step 1: Calculate the heat absorbed by the iron as it cools down to the final

temperature. The heat absorbed by the iron is given by the formula:

qiron =m·ciron ·∆Tiron

where: - m= mass of the iron sample = 50.0 g - ciron = specific heat capac-

ity of iron = 0.449 J/(g

C) - ∆Tiron = change in temperature of iron = final

temperature - initial temperature

∆Tiron =Tfinal −85.0

qiron = 50.0·0.449 ·(Tfinal −85.0)

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

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

qwater =m·cwater ·∆Twater

where: - m= mass of the water = 200.0 g - cwater = specific heat capacity of

water = 4.18 J/(g

C) - ∆Twater =Tfinal −20.0

qwater = 200.0·4.18 ·(Tfinal −20.0)

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

iron is equal to the heat gained by the water.

qiron =qwater

50.0·0.449 ·(Tfinal −85.0) = 200.0·4.18 ·(Tfinal −20.0)

Step 4: Solve for the final equilibrium temperature Tfinal.

50.0·0.449 ·(Tfinal −85.0) = 200.0·4.18 ·(Tfinal −20.0)

22.45 ·(Tfinal −85.0) = 836.0·(Tfinal −20.0)

22.45 ·Tfinal −22.45 ·85.0 = 836.0·Tfinal −836.0·20.0

22.45 ·Tfinal −1902.25 = 836.0·Tfinal −16720.0

813.55 = 813.55Tfinal

Tfinal =813.55

813.55 = 1.0

Therefore, the final equilibrium temperature of the system is 1.0

Question 19

Question

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

metal sample using a calorimeter. The student places a 200 g aluminum sample

at 100◦C into 400 g of water at 20◦C. After thermal equilibrium is reached, the

final temperature of the system is 25◦C. Calculate the specific heat capacity of

the metal.

Solution

Step 1: Calculate the heat transfer from the aluminum to the water. The heat

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

mal∆Tal=mwcw∆Tw

where mal= mass of aluminum (kg), ∆Tal= change in temperature of alu-

minum (K), mw= mass of water (kg), cw= specific heat capacity of water

(J/(kg·K)), ∆Tw= change in temperature of water (K).

Given: mal= 0.2 kg, ∆Tal= 100 −25 = 75 K, mw= 0.4 kg, cw= 4186

J/(kg·K), and ∆Tw= 25 −20 = 5 K.

Substitute the values and solve for l:

0.2×l×75 = 0.4×4186 ×5

l=0.4×4186 ×5

0.2×75

l= 4186 J/kg

Step 2: Calculate the specific heat capacity of the metal. The heat lost by

the metal is equal to the heat gained by the water:

malcal∆Tal=mwcw∆Treservoir

where cal= specific heat capacity of the metal (J/(kg·K)), ∆Treservoir = change

in temperature of the water after reaching equilibrium (K).

Given: mal= 0.2 kg, cal=l= 4186 J/(kg·K), ∆Tal= 75 K, mw= 0.4 kg,

cw= 4186 J/(kg·K), and ∆Treservoir = 25 −20 = 5 K.

Substitute the values and solve for cal:

0.2×4186 ×75 = 0.4×4186 ×5

cal=0.4×4186 ×5

0.2×75

cal= 4186 J/kg

Therefore, the specific heat capacity of the metal sample is 4186 J/(kg·K).

Question 20

Question

A 50 g piece of iron at 150

C is dropped into 200 g of water at 25

C in a 100

g aluminum cup. The final temperature of the system is 40

C. Assuming no

heat is lost to the surroundings, determine the specific heat capacity of iron.

The specific heat capacities of iron, water, and aluminum are 0.45 J/g

C, 4.18

J/g

C, and 0.897 J/g

C, respectively.

Solution

Step 1: Calculate the heat gained by the water and the aluminum cup. 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) - Tf= 40

C (final temperature) - Ti= 25

C (initial temper-

ature)

∆Twater =Tf−Ti= 40 −25 = 15

Qwater = 200 g ×4.18 J/g

C×15

Qwater = 12540 J

The heat gained by the aluminum cup can be calculated using the same

formula:

Qaluminum =maluminum ·caluminum ·∆Taluminum

Where: - maluminum = 100 g (mass of aluminum) - caluminum = 0.897 J/g

(specific heat capacity of aluminum) - ∆Taluminum = 40 −25 = 15

Qaluminum = 100 g ×0.897 J/g

C×15

Qaluminum = 1345.5 J

Step 2: Calculate the heat lost by the iron to the water and the aluminum

cup. Since no heat is lost to the surroundings, the heat lost by the iron is equal

to the heat gained by the water and the aluminum cup.

Qiron =Qwater +Qaluminum

Qiron = 12540 J + 1345.5 J

Qiron = 13885.5 J

Step 3: Use the formula for heat

Qiron =miron ·ciron ·∆Tiron

Since the mass of the iron is 50 g, we can solve for the specific heat capacity of

iron.

ciron =Qiron

miron ·∆Tiron

ciron =13885.5 J

50 g ×(40 −150)

ciron =13885.5

50 × −110

ciron =−2.523 J/g

Therefore, the specific heat capacity of iron is 2.523 J/g

C .

Question 21

Question

A 50 g piece of aluminum at 100

C is dropped into 100 g of water at 20

C in a

calorimeter. The final temperature of the system is 22

C. Assuming no heat is

lost to the surroundings, calculate the specific heat capacity of aluminum.

Solution

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

water.



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

QAluminum =m×cAluminum ×∆TAluminum

Where: m= 50 g (mass of aluminum), cAluminum = 0.903 J/g

C (specific

heat capacity of aluminum), and ∆TAluminum =Tf−Ti= 22 −100 =

−78

C (change in temperature of the aluminum).

Substituting the values, we get:

QAluminum = 50 ×0.903 ×(−78) = −3525.6 J



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

QWater =m×cWater ×∆TWater

Where: m= 100 g (mass of water), cWater = 4.18 J/g

C (specific heat

capacity of water), and ∆TWater =Tf−Ti= 22 −20 = 2

C (change in

temperature of the water).

Substituting the values, we get:

QWater = 100 ×4.18 ×2 = 836 J

Question 22

Question

A 100 g piece of aluminum at 80

C is placed in a calorimeter containing 200

g of water at 20

C. If the final temperature of the system is 25

C, what is the

heat capacity of the calorimeter? The specific heat capacities of aluminum and

water are 0.90 J/g

C and 4.18 J/g

C, respectively.

Solution

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

water. The heat lost by the aluminum can be calculated using the formula:

QAl =mAl ·cAl ·∆T

where: - mAl is the mass of aluminum - cAl is the specific heat capacity of

aluminum - ∆Tis the temperature change of aluminum

Substitute the given values into the formula:

QAl = 100 g ·0.90 J/g

C·(25 −80)

QAl = 100 ·0.90 ·(−55)

QAl =−4950 J

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

Qwater =mwater ·cwater ·∆T

where: - mwater is the mass of water - cwater is the specific heat capacity of water

- ∆Tis the temperature change of water

Substitute the given values into the formula:

Qwater = 200 g ·4.18 J/g

C·(25 −20)

Qwater = 200 ·4.18 ·5

Qwater = 4180 J

Step 2: Use the principle of conservation of energy to find the heat capacity

of the calorimeter. The heat lost by the aluminum is equal to the heat gained

by the water and calorimeter:

QAl =Qwater +Qcalorimeter

Substitute the calculated values:

−4950 = 4180 + Qcalorimeter

Qcalorimeter =−4950 −4180

Qcalorimeter =−9130 J

Step 3: Calculate the heat capacity of the calorimeter using the formula:

Qcalorimeter =Ccalorimeter ·∆T

where: - Ccalorimeter is the heat capacity of the calorimeter - ∆T=Tfinal −

Tinitial = 25 −20 = 5

Substitute the known values:

−9130 = Ccalorimeter ·5

Ccalorimeter =−9130

Ccalorimeter =−1826 J/

Therefore, the heat capacity of the calorimeter is 1826 J/

C .

Question 23

Question

A 50 g piece of iron at 80

C is dropped into a calorimeter containing 100 g of

water at 20

C. If the final temperature of the system is 30

C, what is the heat

capacity of the calorimeter? Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat gained by the water to warm up from 20

C to 30

Step 2: Calculate the heat lost by the iron to cool down from 80

C to 30

Step 3: Use the fact that the total heat lost by the iron is equal to the total

heat gained by the water and the calorimeter to find the heat capacity of the

calorimeter.

Step 1: The specific heat capacity of water is 4.18 J/g

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

qwater =m·c·∆T

where: - mis the mass of the water (100 g), - cis the specific heat capacity of

water (4.18 J/g

C), - ∆Tis the change in temperature (from 20

C to 30

C).

Substitute the values into the formula:

qwater = 100 g ×4.18 J/g

C×(30 −20)

qwater = 100 g ×4.18 J/g

C×10

qwater = 4180 J

Therefore, the heat gained by the water is 4180 J.

Step 2: The specific heat capacity of iron is 0.45 J/g

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

qiron =m·c·∆T

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

(0.45 J/g

C), - ∆Tis the change in temperature (from 80

C to 30

C).

Substitute the values into the formula:

qiron = 50 g ×0.45 J/g

C×(30 −80)

qiron = 50 g ×0.45 J/g

C× −50

qiron =−1125 J

Therefore, the heat lost by the iron is -1125 J. Note the negative sign indi-

cates heat loss.

Step 3: The total heat gained by the water and the calorimeter is equal

to the absolute value of the total heat lost by the iron. Therefore, |qwater +

qcalorimeter|=|qiron|.

Substitute the values of qwater and qiron into the equation:

|4180 J + qcalorimeter|=| − 1125 J|

4180 J + qcalorimeter = 1125 J

qcalorimeter = 1125 J −4180 J

qcalorimeter =−3055 J

Therefore, the heat capacity of the calorimeter is 3055 J.

Question 24

Question

A 50 g piece of copper at 150

C is placed in 200 g of water at 20

C. Assuming

no heat is lost to the surroundings, what will be the final temperature of the

system? (Specific heat capacity of copper = 0.386 J/g

C, specific heat capacity

of water = 4.18 J/g

Solution

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

the formula Q=mc∆Twhere Qis heat, mis mass, cis specific heat capacity,

and ∆Tis the change in temperature.

For copper: Qcopper = (50 g)(0.386 J/g

C)(Tf−150C)

For water: Qwater = (200 g)(4.18 J/g

C)(Tf−20C)

Since no heat is lost or gained from the surroundings, we have Qcopper =

−Qwater

Step 2: Set up the equation: (50 g)(0.386 J/g

C)(Tf−150C) = −(200 g)(4.18 J/g

C)(Tf−

20C)

Step 3: Solve for the final temperature Tf: 50(0.386)(Tf−150) = −200(4.18)(Tf−

20)

19.3Tf−2895 = −836Tf+ 16720

855Tf= 19615

Tf≈22.96C

Therefore, the final temperature of the system will be approximately 22.96C.

Question 25

Question

A 50 g piece of aluminum at 100

C is placed in 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 22

C. Assuming no heat

is lost 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 lost by the aluminum and the heat gained by the

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

Therefore, we have:

mAl ·cAl ·(Tf−Ti) = mw·cw·(Tf−Ti)

Substitute the given values:

50 g ·cAl ·(22 −100) = 200 g ·4.18 J/g

C·(22 −20)

Simplify and solve for cAl.

Step 2: Calculate the specific heat capacity of aluminum.

50 g ·cAl ·(−78) = 200 g ·4.18 J/g

C·2

cAl =200 g ·4.18 J/g

C·2

50 g ·78

cAl =1672 J

3900 g ·

cAl ≈0.43 J/g

Therefore, the specific heat capacity of aluminum is approximately 0.43 J/g

Question 26

Question

A student wants to determine the specific heat capacity of a metal using calorime-

try. They take a 150 g block of the metal at an initial temperature of 200

and place it in 200 g of water at an initial temperature of 20

C contained in a

calorimeter. The final temperature of the system is 23

C. What is the specific

heat capacity of the metal? Assume no heat is lost to the surroundings.

Solution

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

water can be calculated using the formula:

Qwater =m·c·∆T

where: - m= 200 g is the mass of water, - c= 4.18 J/g

C is the specific heat

capacity of water, - ∆T=Tf−Ti= 23C−20C= 3Cis the change in temper-

ature.

Substitute the values into the formula:

Qwater = 200 g ×4.18 J/g

C×3C= 2508 J

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

is equal to the heat absorbed by the water, as there is no heat lost to the

surroundings. We can use the formula:

Qmetal =−Qwater

Qmetal =−2508 J

Step 3: Calculate the specific heat capacity of the metal. The heat lost by

the metal can be calculated using the formula:

Qmetal =m·cmetal ·∆T

where: - m= 150 g is the mass of the metal and - ∆T=Tf−Ti= 200C−23C=

177Cis the change in temperature.

Substitute the known values and solve for cmetal:

−2508 J = 150 g ×cmetal ×177C

cmetal =−2508 J

150 g ×177C

cmetal ≈ −0.094 J/g

Therefore, the specific heat capacity of the metal is approximately -0.094

J/g

Question 27

Question

A piece of metal with mass 0.25 kg is heated to 100◦C and then placed in 0.5

kg of water at 20◦C. The final temperature of the system is 25◦C. If the specific

heat capacity of the metal is 450 J/kg·K, determine the specific heat capacity

of the metal.

Solution

Step 1: Calculate the heat lost by the metal as it cools down from 100◦C to

the final temperature. The heat lost by the metal can be calculated using the

formula:

Qmetal =mc∆T

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

- ∆Tis the change in temperature of the metal.

Substitute the values:

Qmetal = 0.25 kg ×450 J/kg·K×(100 −25) K

Qmetal = 0.25 ×450 ×75

Qmetal = 8437.5 J

Step 2: Calculate the heat gained by the water as it heats up from 20◦C 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, - cis the specific heat capacity of water, -

∆Tis the change in temperature of the water.

Substitute the values:

Qwater = 0.5 kg ×4186 J/kg·K×(25 −20) K

Qwater = 0.5×4186 ×5

Qwater = 10465 J

Step 3: Since energy is conserved, the heat lost by the metal is equal to the

heat gained by the water.

Qmetal =Qwater

8437.5 = 10465

Note: We reached an error in our calculation, please check the heat lost by

the metal and heat gained by the water and make sure they are equal.

Question 28

Question

A piece of metal of mass 0.5 kg at 150

C is placed in 1 kg of water at 20

contained in a 1 kg aluminum calorimeter cup at 20

C. The final temperature

of the system is 30

C. Given the specific heat capacities of the metal and water

are 0.5 J/g

C and 4.18 J/g

C respectively, calculate the specific heat capacity

of the metal.

Solution

Step 1: Calculate the heat gained by the water and the calorimeter cup.

The heat gained by the water and calorimeter cup can be calculated using the

formula:

Q=mc∆T

where Qis the heat gained, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature.

For water:

Qwater = (1 kg ×1000 g/kg ×4.18 J/g

C×(30 −20)

Qwater = 41800 J

For the calorimeter cup (assume specific heat capacity of aluminum is 0.9

J/g

C):

Qcup = (1 kg ×1000 g/kg ×0.9 J/g

C×(30 −20)

Qcup = 900 J

Step 2: Calculate the heat lost by the metal.

The heat lost by the metal is equal to the sum of the heat gained by the water

and calorimeter cup:

Qmetal =Qwater +Qcup

Qmetal = 41800 J + 900 J

Qmetal = 42700 J

Step 3: Calculate the specific heat capacity of the metal.

We know that the heat lost by the metal is given by:

Qmetal =mc∆T

Substitute the values we have:

42700 = 0.5×c×(30 −150)

42700 = −60c

c=−42700

c≈ −711.67 J/kg

Therefore, the specific heat capacity of the metal is approximately 711.67

J/kg

Question 29

Question

A piece of copper with a mass of 200 g is heated to a temperature of 100

and then dropped into a calorimeter containing 500 g of water at an initial

temperature of 20

C. If the final temperature of the system is 25

C, calculate

the specific heat capacity of copper. Assume no heat is lost to the surroundings

and 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.

Heat lost by copper = Heat gained by water

Step 2: The heat lost by the copper is given by the formula:

Qcopper =mc∆T

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

(unknown) - ∆Tis the change in temperature of the copper (final temperature

- initial temperature)

Qcopper = (200 g)(c)(25 −100)

Qcopper =−15000cJ

Step 3: The heat gained by the water is given by the formula:

Qwater =mc∆T

where: - mis the mass of water (500 g) - cis the specific heat capacity of water

(4.18 J/g

C) - ∆Tis the change in temperature of the water (final temperature

- initial temperature)

Qwater = (500 g)(4.18 J/g

C)(25 −20)

Qwater = 2090 J

Step 4: Set the heat lost by the copper equal to the heat gained by the water:

−15000c= 2090

Step 5: Solve for the specific heat capacity of copper:

c=2090

−15000 =−0.139 J/g

Therefore, the specific heat capacity of copper is -0.139 J/g

C. The negative

sign indicates that heat is being lost by the copper.

Question 30

Question

A 50 g piece of aluminum initially at 80

C is dropped into 200 g of water at

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

of the system? (Specific heat capacity of aluminum = 0.9 J/g

C, specific heat

capacity of water = 4.18 J/g

Solution

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

aluminum can be calculated using the formula:

qaluminum =mc∆T

where: - m= 50 g (mass of aluminum) - c= 0.9 J/g

C (specific heat capacity

of aluminum) - ∆T=Tfinal −Tinitial =Tfinal −80

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

water can be calculated using the formula:

qwater =mc∆T

where: - m= 200 g (mass of water) - c= 4.18 J/g

C (specific heat capacity of

water) - ∆T=Tfinal −Tinitial =Tfinal −20

Step 3: Set up the conservation of energy equation: Since the system is

isolated and no heat is lost to the surroundings, the heat lost by the aluminum

must be equal to the heat gained by the water, so we have:

qaluminum =qwater

Step 4: Substitute the calculated values and solve for the final temperature,

Tfinal.

maluminumcaluminum(Tfinal −80) = mwatercwater(Tfinal −20)

Step 5: Simplify the equation and solve for Tfinal.

50 ×0.9×(Tfinal −80) = 200 ×4.18 ×(Tfinal −20)

45(Tfinal −80) = 836(Tfinal −20)

45Tfinal −3600 = 836Tfinal −16720

791Tfinal = 13120

Tfinal =13120

791 ≈16.58

Therefore, the final temperature of the system is approximately 16.58

Question 31

Question

A 50 g aluminum block is heated to 100

C and then placed in 200 g of water

at 20

C inside a calorimeter. The final temperature of the system is 25

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

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

Solution

Step 1: Calculate the heat absorbed by the water during the cooling process.

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

qwater =mwater ×cwater ×∆T

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

C is the specific

heat capacity of water, and - ∆T= 25 −20 = 5

C is the temperature change

of the water. Substitute these values into the formula:

qwater = 200 ×4.18 ×5 = 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, so we

have:

qaluminum =−qwater =−4180 J

Step 3: Calculate the heat lost by the aluminum block using the formula:

qaluminum =maluminum ×caluminum ×∆T

where: - maluminum = 50 g is the mass of the aluminum block, - caluminum is

the specific heat capacity of aluminum (which we want to find), and - ∆T=

100 −25 = 75

C is the temperature change of the aluminum block.

Equating the heat lost by the aluminum block in step 2 and step 3, we have:

50 ×caluminum ×75 = −4180

caluminum =−4180

50 ×75

caluminum =−1.40 J/g

Therefore, the specific heat capacity of the aluminum block is 1.40 J/g

Question 32

Question

A piece of metal with mass 0.2 kg is heated to 100

C and then placed in 0.5 kg

of water at 25

C in an insulated container. The final temperature of the system

is 30

C. If the specific heat capacity of the metal is 450 J/kg

C and the specific

heat capacity of water is 4186 J/kg

C, calculate the specific heat capacity of the

container if none of the heat is lost to the surroundings.

Solution

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

formula:

Qmetal =Qwater

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

Qmetal =mc∆T

where: m= 0.2 kg (mass of the metal), c= 450 J/kg

C (specific heat capacity

of the metal), ∆T= 100 −30 = 70

C (final temperature - initial temperature).

Step 3: Substitute the values into the formula to find Qmetal:

Qmetal = 0.2×450 ×70

Step 4: Calculate the heat gained by the water using the formula:

Qwater =mc∆T

where: m= 0.5 kg (mass of the water), c= 4186 J/kg

C (specific heat capacity

of water), ∆T= 30 −25 = 5

C (final temperature - initial temperature).

Step 5: Substitute the values into the formula to find Qwater:

Qwater = 0.5×4186 ×5

Step 6: Set Qmetal =Qwater and solve for the specific heat capacity of the

container:

0.2×450 ×70 = 0.5×4186 ×5

Step 7: Solve for the specific heat capacity of the container.

Question 33

Question

A 50 g piece of aluminum initially at 100◦C is placed in a calorimeter containing

200 g of water at 20◦C. The final temperature of the system is 25◦C. Assuming

no heat is lost to the surroundings, what is the specific heat capacity of the

aluminum? (Specific heat capacity of water is 4.18 J/g◦C)

Solution

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

water using the formula:

q=m·c·∆T

where: q= heat energy in Joules (J), m= mass in grams (g), c= specific

heat capacity in J/g◦C, ∆T= change in temperature in ◦C.

The heat lost by aluminum is equal to the heat gained by water, since no

heat is lost to the surroundings.

For the aluminum:

qAl = (50 g) ·cAl ·(25 −100)◦C

For the water:

qH2O = (200 g) ·4.18 J/g◦C·(25 −20)◦C

Since qAl =qH2O:

(50 g) ·cAl ·(−75)◦C = (200 g) ·4.18 J/g◦C·5◦C

Simplify and solve for cAl in terms of the given values.

Question 34

Question

A 100 g piece of copper at 200

C is placed into a calorimeter containing 200 g of

water at 20

C. The final temperature of the system is 25

C. Assuming no heat

is lost to the surroundings, what is the specific heat capacity of the calorimeter?

Solution

Given: Mass of copper (mc) = 100 g = 0.1 kg

Initial temperature of copper (Tc,i) = 200

Specific heat capacity of copper (cc) = 390 J/kg ·

Mass of water in calorimeter (mw) = 200 g = 0.2 kg

Initial temperature of water (Tw,i) = 20

Final temperature of the system (Tf) = 25

Specific heat capacity of water (cw) = 4186 J/kg ·

We will assume that no heat is lost to the surroundings, so the total heat

lost by the copper is equal to the total heat gained by the water:

mc·cc·(Tc,i −Tf) = mw·cw·(Tf−Tw,i)

Step 1: Calculate the heat lost by the copper (Qc) and the heat

gained by the water (Qw).

Qc=mc·cc·(Tc,i −Tf)

Qw=mw·cw·(Tf−Tw,i)

Substitute the given values:

Qc= 0.1 kg ·390 J/kg ·

C·(200 −25)

Qw= 0.2 kg ·4186 J/kg ·

C·(25 −20)

Step 2: Equate the heat lost and gained equations. Solve for the

specific heat capacity of the calorimeter.

mc·cc·(Tc,i −Tf) = mw·cw·(Tf−Tw,i)

0.1·390 ·(200 −25) = 0.2·4186 ·(25 −20)

0.1·390 ·175 = 0.2·4186 ·5

6825 = 4186 ·1·c

c=6825

4186

Therefore, the specific heat capacity of the calorimeter is 6825

4186 ≈1.63 J/g·

Question 35

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 22

C. Given the specific

heat capacity of water is 4.18 J/g

C, determine the specific heat capacity of

aluminum. Assume no heat is lost to the surroundings and that the specific

heat capacity of aluminum is constant over the temperature range involved.

Solution

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

can be calculated using the equation:

qwater =m·c·∆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 (22C−20C= 2C)

Substitute the given values to find qwater:

qwater = 200 g ×4.18 J/g

C×2C= 1672 J

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

aluminum block must be equal to the heat gained by the water, since no heat

is lost to the surroundings. We can express this as:

qaluminum =−qwater

Since the heat lost by the aluminum block can be calculated using its specific

heat capacity (caluminum), mass (maluminum = 50 g), and temperature change

(∆Taluminum =Tf−Ti= 22C−80C=−58C):

qaluminum =maluminum ·caluminum ·∆Taluminum

Substitute the known values to find caluminum:

1672 J = 50 g ×caluminum × −58C

Solve for caluminum:

caluminum =1672 J

50 g × −58C≈0.57 J/g

Therefore, the specific heat capacity of aluminum is approximately 0.57 J/g

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set

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