CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 2 | CHEM 107 - Essentials of General and Organic Chemistry

CHEM 107 - ESSENTIALS OF

GENERAL AND ORGANIC

CHEMISTRY - Calorimetry

Question Bank - Set 2

Liberty University

Question 1

Question

A 50.0 g piece of iron at 120.0

C is placed into 200 g of water at 20.0

C in a

calorimeter. The final temperature of the system is 25.0

C. Assuming no heat

loss to the surroundings, calculate the specific heat capacity of iron.

Given: Specific heat capacity of water = 4.18 J/(g

Solution

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

can be calculated using the formula:

qwater =m×c×∆T

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

C) (specific heat capacity of

water), ∆T= 25.0C−20.0C= 5.0C(change in temperature).

Substitute the values into the formula:

qwater = 200 g ×4.18 J/(g

C) ×5.0C

qwater = 4180 J

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

also be calculated using the formula:

qiron =m×c×∆T

where: m= 50.0 g (mass of iron), c(specific heat capacity of iron), ∆T=

25.0C−120.0C=−95.0C(change in temperature).

Substitute the values into the formula and equate it to the negative of the

heat gained by the water:

50.0 g ×c× −95.0C=−4180 J

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

c=−4180 J

50.0 g × −95.0C

c≈0.878 J/(g

Therefore, the specific heat capacity of iron is approximately 0.878 J/(g

C).

Question 2

Question

A 50 g piece of aluminum at 80

C is dropped into 200 g of water at 20

in a calorimeter. The final temperature of the aluminum and water is 25

Assuming no heat is lost to the surroundings and the specific heat capacity of

water is 4.18 J/(g

C), 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 is equal to the heat gained by the water.

Let the specific heat capacity of aluminum be cAl. The equation to calculate

the heat lost by aluminum is:

QAl =mAl ·cAl ·∆TAl

Where: - mAl = 50 g (mass of aluminum) - cAl =? (specific heat capacity

of aluminum) - Ti,Al = 80

C (initial temperature of aluminum) - Tf,Al = 25

(final temperature of aluminum) - ∆TAl =Tf,Al −Ti,Al

The equation to calculate the heat gained by water is:

Qwater =mwater ·cwater ·∆Twater

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

C) (specific heat

capacity of water) - Ti,water = 20

C (initial temperature of water) - Tf,water =

C (final temperature of water) - ∆Twater =Tf,water −Ti,water

Since no heat is lost to the surroundings, we have:

QAl =−Qwater

Thus:

mAl ·cAl ·∆TAl =mwater ·cwater ·∆Twater

Step 2: Solve for the specific heat capacity of aluminum: Substitute the

known values:

50 g ·cAl ·(25

C−80

C) = 200 g ·4.18 J/(g

C) ·(25

C−20

50 g ·cAl ·(−55

C) = 200 g ·4.18 J

−2750 g

C·cAl = 836 J

Finally, solve for cAl:

cAl =836 J

−2750 g

cAl ≈ −0.3036 J/(g

Therefore, the specific heat capacity of aluminum is approximately -0.3036

J/(g

C). Since this value is nonsensical (specific heat capacity should be posi-

tive), there may have been an error in the calculation or assumption.

Question 3

Question

A 50.0 g piece of aluminum at 75.0

C is dropped into 100.0 g of water at 20.0

in a calorimeter. The final temperature of the system is 23.5

C. Assuming 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 absorbed by the water when it warms up. The heat

absorbed by the water can be calculated using the formula:

qwater =m·c·∆T

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

C is the specific heat

capacity of water, - ∆T=Tfinal −Tinitial = 23.5C−20.0C= 3.5Cis the change

in temperature.

Plugging in the values, we get:

qwater = 100.0 g ×4.18 J/g

C×3.5C= 1463 J

Step 2: Calculate the heat released by the aluminum when it cools down.

The heat released by the aluminum can be calculated using the same formula:

qaluminum =m·c·∆T

where: - m= 50.0 g is the mass of aluminum, - cis the specific heat capacity of

aluminum, - ∆T=Tfinal −Tinitial = 23.5C−75.0C=−51.5C(negative because

the aluminum is cooling down).

Plugging in the values, we get:

qaluminum = 50.0 g ×c× −51.5C=−2575cJ

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

gained by water (assuming no heat loss to the surroundings), we can set up the

equation:

qaluminum =−qwater

−2575c=−1463 J

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

c=−1463 J

2575 J ≈0.57 J/g

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

Question 4

Question

A piece of iron weighing 150 g at 85

C is placed into 200 g of water at 25

The final temperature of the mixture is measured at 30

C. Assuming no heat is

lost to the surroundings, calculate the specific heat capacity of iron. (Specific

heat capacity of water = 4.18 J/g

Solution

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

The heat lost by the iron (Qiron) is given by the formula:

Qiron =mc∆T

where: - mis the mass of the iron (150 g) - cis the specific heat capacity

of iron (unknown) - ∆Tis the change in temperature experienced by the iron

(30C−85C=−55C)

Therefore,

Qiron = 150 g ×c×(−55C)

The heat gained by the water (Qwater) is given by the formula:

Qwater =mc∆T

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

water (4.18 J/g

C) - ∆Tis the change in temperature experienced by the water

(30C−25C= 5C)

Therefore,

Qwater = 200 g ×4.18 J/g

C×5C

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

water (assuming no heat is lost to the surroundings), we can set up the equation:

150c(−55) = 200 ×4.18 ×5

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

150c(−55) = 200 ×4.18 ×5

−8250c= 4180

c=4180

−8250

c=−0.506 J/g

Therefore, the specific heat capacity of iron is approximately −0.506 J/g

Question 5

Question

A 50.0 g sample of gold at 100.0

C is added to 100.0 g of water at 20.0

C in a

calorimeter. The final temperature of the system is 23.0

C. Assuming no heat

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

specific heat capacity of water is 4.18 J/g

Solution

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

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

surroundings. We use the formula:

qlost =qgain

Step 2: Calculate the heat lost by gold. The formula to calculate the heat

lost or gained by a substance is:

q=m×c×∆T

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

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

Substitute the given values for gold:

qgold = 50.0 g ×cgold ×(23.0−100.0) ◦C

Step 3: Calculate the heat gained by water. Substitute the given values for

water:

qwater = 100.0 g ×4.18 J/g

C×(23.0−20.0) ◦C

Step 4: Equate the heat lost by gold to the heat gained by water and find

the specific heat capacity of gold. We have:

50.0 g ×cgold ×(23.0−100.0) = 100.0 g ×4.18 J/g

C×(23.0−20.0)

Now solve for cgold.

Question 6

Question

A 50 g piece of aluminum at 80

C is dropped into 200 g of water at 20

C. If the

final temperature of the mixture is 24

C, what is the specific heat capacity of

the aluminum?

Given: Specific heat capacity of water: cw= 4.18 J/g

Solution

Step 1: Calculate the energy lost by the aluminum piece as it cools down to the

final temperature.

∆QAl =mAl ·cAl ·∆TAl

where mAl = 50 g (mass of aluminum), cAl is the specific heat capacity of

aluminum (in J/g

C), and ∆TAl = 80 −24 = 56

C. Substitute the known

values:

∆QAl = 50 g ·cAl ·56

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

temperature.

∆Qw=mw·cw·∆Tw

where mw= 200 g (mass of water), cw= 4.18 J/g

C, and ∆Tw= 24 −20 = 4

Substitute the known values:

∆Qw= 200 g ·4.18 J/g

C·4

Step 3: Since energy is conserved, we have:

∆QAl =−∆Qw

50 g ·cAl ·56

C = −200 g ·4.18 J/g

C·4

cAl =−200 g ·4.18 J/g

C·4

50 g ·56

Question 7

Question

A 50 g piece of aluminum at 120

C is placed in 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.

(Specific heat capacity of water is 4.18 J/g

Solution

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

the formula q=mc∆T.

Heat lost by aluminum = Heat gained by water

mAlcAl∆TAl =mH2OcH2O∆TH2O

Step 2: Substitute the known values into the equation.

50 g ×cAl ×(22 −120) = 100 g ×4.18 J/g

C×(22 −20)

−7000cAl = 836.0

Step 3: Solve for cAl.

cAl =836.0

−7000

cAl =−0.119 J/g

Therefore, the specific heat capacity of aluminum is 0.119 J/g

Question 8

Question

A 50 g piece of iron at 250

C is dropped into 200 g of water at 20

C inside

a perfectly insulated container. Assuming no heat is lost to the surroundings,

what will be the final temperature of the system? (Specific heat capacity of iron

= 0.45 J/g

C, specific heat capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the iron as it cools down from 250

C to the

final temperature. The heat lost 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), and ∆Tis the change in temperature. Since the initial temperature of

the iron is 250

C and the final temperature is T

C (unknown), the change in

temperature is 250 −T. Therefore,

Qiron = 50 g ·0.45 J/g

C·(250 −T)

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

C to

the final temperature. The heat gained can be calculated using the formula:

Qwater =m·c·∆T

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

water (4.18 J/g

C), and ∆Tis the change in temperature. Since the initial

temperature of the water is 20

C and the final temperature is T

C (unknown),

the change in temperature is T−20. Therefore,

Qwater = 200 g ·4.18 J/g

C·(T−20)

Step 3: Since energy is conserved in this system, the heat lost by the iron is

equal to the heat gained by the water. Therefore,

50 g ·0.45 J/g

C·(250 −T) = 200 g ·4.18 J/g

C·(T−20)

Step 4: Simplify the equation and solve for T to find the final temperature

of the system.

22.5(250 −T) = 836(T−20)

5625 −22.5T= 836T−16720

0 = 858.5T−22345

T≈22345

858.5≈26.04C

Therefore, the final temperature of the system will be approximately 26.04

Question 9

Question

A 50 g aluminum block at 80

C is placed in 200 g of water at 20

C in a calorime-

ter. Assuming all the heat lost by the aluminum block is gained by the water

and the calorimeter, what is the final temperature of the system? (Specific

heat capacity of aluminum = 0.902 J/g

C, specific heat capacity of water =

4.18 J/g

C, specific heat capacity of the calorimeter = 1.0 J/g

Solution

Step 1: Calculate the heat lost by the aluminum block: Given: Mass of the

aluminum block, mAl = 50 g Initial temperature of the aluminum block, TAl,i =

80CSpecific heat capacity of aluminum, cAl = 0.902 J/g

Using the formula for heat energy:

QAl =mcAl∆TAl

where ∆TAl =Tfinal −TAl,i is the change in temperature of the aluminum block.

Step 2: Calculate the heat gained by the water and the calorimeter: Given:

Mass of the water, mwater = 200 g Initial temperature of the water, Twater,i =

20CSpecific heat capacity of water, cwater = 4.18 J/g

C Specific heat capacity

of the calorimeter, ccalorimeter = 1.0 J/g

Using the same formula for heat energy:

Qwater+calorimeter = (mwater +mcalorimeter)c∆Twater+calorimeter

where mcalorimeter is the mass of the calorimeter and ∆Twater+calorimeter =Tfinal−

Twater,i is the change in temperature of the water and the calorimeter.

Step 3: Set the heat lost equal to the heat gained:

QAl =Qwater+calorimeter

Step 4: Solve for the final temperature, Tfinal:

mAlcAl(Tfinal−TAl,i)=(mwater+mcalorimeter)(cwater+ccalorimeter)(Tfinal−Twater,i)

Now, solve the equation above to find the final temperature of the system.

Question 10

Question

A 50 g piece of copper initially at 200

C is dropped into 200 g of water at 20

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

C. Assuming no

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

(Specific heat capacity of water = 4.18 J/g

Solution

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

temperature. The heat lost by the copper can be calculated using the formula

Q=mc∆T, where: - mis the mass of the copper (50 g), - cis the specific

heat capacity of copper (to be determined), - ∆Tis the change in temperature

(200

C - 25

C).

Substitute the values into the formula: Qcopper = (50 g)(c)(200 −25) =

50c(175) = 8750cJ

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

temperature. The heat gained by the water can be calculated using the formula

Q=mc∆T, where: - mis the mass of the water (200 g), - cis the specific

heat capacity of water (4.18 J/g

C), - ∆Tis the change in temperature (25

C -

C).

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

C)(25 −

20) = 200(4.18)(5) = 4180 J

Step 3: Since heat is conserved, the heat lost by the copper must be equal to

the heat gained by the water. Set Qcopper =Qwater and solve for c: 8750c= 4180

Step 4: Solve for c.c=4180

8750 ≈0.4777 J/g

Therefore, the specific heat capacity of copper is approximately 0.4777 J/g

Question 11

Question

A 50-g piece of aluminum at 120

C is dropped into 200 g of water at 20

C. If

the final temperature of the system is 25

C, calculate the specific heat capacity

of aluminum (cAl) given that the specific heat capacity of water is 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:

mAlcAl∆TAl =mwatercwater∆Twater

where: mAl = mass of aluminum = 50 g, cwater = specific heat capacity of water

= 4.18 J/g

C, mwater = mass of water = 200 g, ∆TAl = change in temperature

of aluminum = 120

C - 25

C = 95

C, ∆Twater = change in temperature of water

= 25

C - 20

C = 5

Substitute the values into the equation:

50 ×cAl ×95 = 200 ×4.18 ×5

Step 2: Solve for cAl.

50 ×cAl ×95 = 200 ×4.18 ×5

50 ×cAl =200 ×4.18 ×5

cAl =200 ×4.18 ×5

95 ×50

cAl =4180

cAl = 44.0 J/g

Therefore, the specific heat capacity of aluminum is 44.0 J/g

Question 12

Question

A 50 g piece of copper at 200

C is placed in 200 g of water at 20

C in a calorime-

ter. The final temperature of the system is 25

C. Assuming no heat is lost to

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

capacity of water = 4.18 J/g

Solution

Given: Mass of copper (mc) = 50 g, Initial temperature of copper (Tc1) = 200

Mass of water (mw) = 200 g, Initial temperature of water (Tw1) = 20

C, Final

temperature of system (Tf) = 25

C, Specific heat capacity of water (cw) = 4.18

J/g

Let’s denote the specific heat capacity of copper as cc.

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

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

mc·cc·∆Tc=−mw·cw·∆Tw

Where ∆Tcand ∆Tware the changes in temperature for copper and water

respectively.

Step 2: Calculate ∆Tcand ∆TwStep 2.1: Change in temperature for

copper:

∆Tc=Tf−Tc1= 25C−200C=−175C

Step 2.2: Change in temperature for water:

∆Tw=Tf−Tw1= 25C−20C= 5C

Step 3: Substitute into the heat equation Substitute the values into

the heat equation from Step 1:

50 ·cc·(−175) = −200 ·4.18 ·5

Step 4: Solve for cc

cc=−200 ·4.18 ·5

50 ·(−175)

cc=−4180

−8750

cc= 0.4771 J/g

Therefore, the specific heat capacity of copper is 0.4771 J/g

Question 13

Question

A 50.0 g piece of copper at 95.0

C is placed in 100.0 g of water at 20.0

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

of the system? (Specific heat capacity of copper = 0.387 J/g

C, specific heat

capacity of water = 4.18 J/g

Solution

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

final temperature. 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: mcopper = 50.0 g, ccopper = 0.387 J/g

C, Tinitial = 95.0C,Tfinal =

Tsystem qcopper =mcopper ·ccopper ·(Tfinal −Tinitial)

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

temperature. Given: mwater = 100.0 g, cwater = 4.18 J/g

C, Tinitial, water =

20.0C,Tfinal, water =Tsystem qwater =mwater ·cwater ·(Tfinal −Tinitial, water)

Step 3: Since heat lost by copper piece is equal to heat gained by water

when no heat is lost to the surroundings, we have: qcopper =−qwater mcopper ·

ccopper ·(Tfinal −Tinitial) = −mwater ·cwater ·(Tfinal −Tinitial, water)

Step 4: Solve the equation from step 3 for the final temperature Tfinal.

50.0·0.387 ·(Tfinal −95.0) = −100.0·4.18 ·(Tfinal −20.0)

19.35Tfinal −1866.5 = −418Tfinal + 8360

437.35Tfinal = 10226.5

Tfinal23.4C

Therefore, the final temperature of the system is approximately 23.4

Question 14

Question

A 50 g piece of copper at 200

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 and the heat gained by the water

using the formula:

qlost =−qgain

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

qcopper =mc∆T

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

∆Tis the change in temperature.

Substitute the given values:

qcopper = (50 g)(0.385 J/g

C)(Tf−200C)

Step 3: Calculate the heat gained by the water using the formula:

qwater =mc∆T

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

∆Tis the change in temperature.

Substitute the given values:

qwater = (100 g)(4.18 J/g

C)(Tf−20C)

Step 4: Set the two heat equations equal to each other:

(50 g)(0.385 J/g

C)(Tf−200C) = (100 g)(4.18 J/g

C)(Tf−20C)

Step 5: Solve for Tfby simplifying and rearranging the equation.

19.25 ·(Tf−200) = 418 ·(Tf−20)

19.25Tf−3850 = 418Tf−8360

3985 = 398.75Tf

Tf= 10C

Therefore, the final temperature of the system will be 10

Question 15

Question

A sample of gold at 200.0◦C is placed in 100.0 g of water at 20.0◦C. If the final

temperature of the system is 25.0◦C, determine the mass of the gold sample.

Assume the specific heat capacity of gold is 0.129 J/g◦C and the specific heat

capacity of water is 4.18 J/g◦C.

Solution

Step 1: Use the heat gained by the water is equal to the heat lost by the gold

sample: Let mbe the mass of the gold sample and cgold be the specific heat

capacity of gold. We can set up the equation:

m·cgold ·(Tfinal −Tinitial, gold) = mwater ·cwater ·(Tfinal, water −Tinitial, water)

where: - Tfinal is the final temperature of the system, - Tinitial, gold is the initial

temperature of the gold sample, - Tinitial, water is the initial temperature of the

water, - mwater is the mass of the water, and - Tfinal, water is the final temperature

of the water.

Step 2: Substitute the known values into the equation: We know Tfinal =

25.0◦C, Tinitial, gold = 200.0◦C, Tinitial, water = 20.0◦C, mwater = 100.0 g, cgold =

0.129 J/g◦C, and cwater = 4.18 J/g◦C.

m·0.129 ·(25.0−200.0) = 100.0·4.18 ·(25.0−20.0)

Step 3: Solve for the mass of the gold sample:

m·0.129 ·(−175.0) = 100.0·4.18 ·5.0

−22.575m= 209.0

m=209.0

−22.575

m≈ −9.27 g

Since we cannot have a negative mass, the mass of the gold sample must be

9.27 g.

Question 16

Question

A 50 g aluminum block at 80

C is added to 200 g of water at 20

C in a calorime-

ter. If the final temperature of the system is 25

C, calculate the specific heat

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

Solution

Step 1: Calculate the heat lost by aluminum (Q loss). The heat lost by the

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

The formula to calculate heat is given by Q=mc∆T, where: - mis the mass

of the substance (in kg), - cis the specific heat capacity of the substance (in

J/(kg

·°

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

C).

Given: - Mass of aluminum block, mAl = 50 g = 0.05 kg, - Initial tempera-

ture of aluminum block, TAl, initial = 80 C, - Final temperature of the system,

Tfinal = 25 C.

To get ∆Tfor aluminum, we use the formula: ∆T=Tfinal −TAl, initial.

∆T= 25 C −80 C = −55 C

Therefore, the heat lost by aluminum, Qloss =mAl ·cAl ·∆TAl.

Step 2: Calculate the heat gained by water and calorimeter (Q gain). The

heat gained by the water and calorimeter is equal to the heat lost by the alu-

minum block. This is given by Qloss =Qgain.

Next, we calculate Qgain using the formula Qgain =mwater ·cwater ·∆Twater +

ccal ·∆Tcal, where: - mwater = 200 g = 0.2 kg is the mass of water, - cwater =

4186 J/(kg

·°

C) is the specific heat capacity of water, - ∆Twater =Tfinal −

Twater, initial, - ccal is the specific heat capacity of the calorimeter, - ∆Tcal =

Tfinal −Tcal, initial.

Step 3: Setting Qloss =Qgain and calculating cAl.

mAl ·cAl ·∆TAl =mwater ·cwater ·∆Twater +ccal ·∆Tcal

Substitute the known values and solve for cAl to find the specific heat ca-

pacity of aluminum.

Question 17

Question

A piece of copper of mass 200 g at a temperature of 100◦C is placed in 400 g

of water at 20◦C contained in a calorimeter of mass 100 g. The temperature of

water and the calorimeter rises to 28◦C. If the specific heat capacity of copper

is 0.387 J/g◦C, determine the specific heat capacity of the calorimeter. Assume

no heat is lost to the surroundings.

Solution

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

calorimeter. Let Cpbe the specific heat capacity of the calorimeter. The heat

lost by copper is given by

Qcopper =m·c·∆T

where m= mass of copper = 200 g, c= specific heat capacity of copper = 0.387

J/g◦C, ∆T= change in temperature of copper = (temperature of water and

calorimeter - initial temperature of copper) = (28◦C - 100◦C).

The heat gained by the water and calorimeter is given by

Qwater+calorimeter = (mwater +mcalorimeter)·cwater+calorimeter ·∆T

where mwater = mass of water = 400 g, mcalorimeter = mass of calorimeter =

100 g, cwater+calorimeter = specific heat capacity of water and calorimeter, ∆T=

change in temperature of water and calorimeter = (temperature of water and

calorimeter - initial temperature of water and calorimeter) = (28◦C - 20◦C).

Since heat is conserved, we have

Qcopper =Qwater+calorimeter

m·c·∆T= (mwater +mcalorimeter)·cwater+calorimeter ·∆T

Step 2: Solve for cwater+calorimeter, the specific heat capacity of water and

calorimeter. Substitute the given values into the equation:

200 g ·0.387 J/g◦C·(28 −100) = (400 g + 100 g) ·cwater+calorimeter ·(28 −20)

cwater+calorimeter =200 ·0.387 ·(−72)

500 ·8

Step 3: Calculate the specific heat capacity of the calorimeter. The specific

heat capacity of the calorimeter, Cp, is equal to the specific heat capacity of the

water and calorimeter, cwater+calorimeter.

Cp=cwater+calorimeter =200 ·0.387 ·(−72)

500 ·8

Question 18

Question

A 50.0 g piece of aluminum at 95.0

C is placed into 100.0 g of water at 20.0

The final temperature of the system is 22.0

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: Determine the heat gained by the water and the heat lost by the alu-

minum.

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

qwater =mwater ×cwater ×∆T

where: - mwater = 100.0 g (mass of water) - cwater = 4.18 J/g

C (specific heat

capacity of water) - ∆T=Tfinal −Tinitial = 22.0C −20.0C = 2.0C(change in

temperature)

Plugging in the values, we get:

qwater = 100.0 g ×4.18 J/g

C×2.0C= 836 J

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

qaluminum =maluminum ×caluminum ×∆T

where: - maluminum = 50.0 g (mass of aluminum) - caluminum (specific heat ca-

pacity of aluminum) - ∆T=Tfinal −Tinitial = 22.0C −95.0C = −73.0C(change

in temperature)

We can rewrite the equation as:

qaluminum =maluminum ×caluminum × −73.0C

qaluminum =−50.0 g ×caluminum ×73.0 J

Step 2: Since the system is isolated, the heat lost by aluminum is equal to

the heat gained by water:

qwater =qaluminum

836 J = −50.0 g ×caluminum ×73.0 J

Step 3: Solve for the specific heat capacity of aluminum (caluminum):

caluminum =836

−50.0×73.0

caluminum =−0.228 J/g

Therefore, the specific heat capacity of aluminum is −0.228 J/g

Question 19

Question

A 50 g piece of copper at 150

C is placed in 200 g of water at 25

C. Assuming

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

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

C, specific heat of water = 4.184

J/g

C, and the specific heat of fusion of copper = 205 J/g)

Solution

Step 1: Calculate the heat lost by the copper piece to reach equilibrium with

the water. The heat lost by the copper can be calculated as:

qcopper =mcopper ·ccopper ·∆T

where: mcopper = 50 g (mass of copper), ccopper = 0.386 J/g

C (specific heat of

copper), and ∆T=Tfinal −150

Step 2: Calculate the heat gained by the water to reach equilibrium with

the copper. The heat gained by the water can be calculated as:

qwater =mwater ·cwater ·∆T

where: mwater = 200 g (mass of water), cwater = 4.184 J/g

C (specific heat of

water), and ∆T=Tfinal −25

Step 3: Since the total heat loss by the copper is equal to the total heat gain

by the water when they reach equilibrium, we can set up the equation:

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

Step 4: Solve for the final temperature. Solving the equation from Step 3

for Tfinal, we have:

50 ·0.386 ·(Tfinal −150) = 200 ·4.184 ·(Tf inal −25)

Step 5: Simplify the equation and solve for Tfinal to find the final temperature

of the system.

Question 20

Question

A 50 g ice cube at -10

C is placed in a calorimeter containing 200 g of water

at 20

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

temperature of the system when the ice has completely melted? (Specific heat

capacity of water = 4.18 J/g

C, specific heat capacity of ice = 2.09 J/g

C, heat

of fusion for ice = 334 J/g)

Solution

Step 1: Calculate the heat absorbed by the ice cube to melt into water.

Q1=m·Lf

= 50 g ×334 J/g

= 16700 J

Step 2: Calculate the heat gained by the ice cube to reach the final temper-

ature.

Q2=m·cice ·∆T

= 50 g ×2.09 J/g

C×(0 −(−10))

= 1045 J

Step 3: Calculate the heat gained by the water to reach the final temperature.

Q3=m·cwater ·∆T

= 200 g ×4.18 J/g

C×(Tf−20)

= 836 J ×(Tf−20)

Step 4: Since the total heat gained equals the total heat lost, we have:

Q1+Q2=−Q3

Substitute the calculated values:

16700 J + 1045 J = −836 J ×(Tf−20)

Step 5: Solve for Tf.

17745 = −836 J ×(Tf−20)

Tf−20 = −17745

836

Tf=−17745

836 + 20

Therefore, the final temperature of the system when the ice has completely

melted is approximately −2.38

Question 21

Question

A 50 g piece of aluminum at 80

C is placed in 200 g of water at 20

C. After

reaching thermal equilibrium, the final temperature of the system is 25

C. Cal-

culate the specific heat capacity of aluminum. (Specific heat capacities: water

= 4.18 J/g

C, aluminum = 0.897 J/g

Solution

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

Let’s assume no heat loss to the surroundings. The heat lost by aluminum is

equal to the heat gained by water:

m1c1(Tf−T1) = m2c2(Tf−T2)

where: - m1= 50 g (mass of aluminum) - c1= 0.897 J/g

C (specific heat capac-

ity of aluminum) - Tf= 25C(final temperature) - T1= 80C(initial temperature

of aluminum) - m2= 200 g (mass of water) - c2= 4.18 J/g

C (specific heat ca-

pacity of water) - T2= 20C(initial temperature of water)

Substitute the values into the equation:

(50)(0.897)(25 −80) = (200)(4.18)(25 −20)

Step 2: Solve the equation for c1.

−22425 = 4180(5)

Step 3: Rearrange the formula to solve for c1.

c1=4180(5)

50(55) = 3.82 J/g

Therefore, the specific heat capacity of aluminum is 3.82 J/g

Question 22

Question

A 50.0 g sample of aluminum at 87.0

C is dropped into 300.0 g of water at

20.0

C. If the final temperature of the system is 21.9

C, what is the specific

heat capacity of aluminum? Assume the specific heat capacity of water is 4.18

J/g

Solution

Step 1: Calculate the heat released by the aluminum when it cools down.

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

QAluminum =m·c·∆T

where: - m= 50.0 g is the mass of aluminum, - cis the specific heat capacity

of aluminum (to be determined), - ∆T= 87.0−21.9 = 65.1

C is the change in

temperature.

Substitute the values into the formula:

QAluminum = 50.0 g ×c×65.1

Step 2: Calculate the heat absorbed by the water as it heats up.

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

QWater =m·c·∆T

where: - m= 300.0 g is the mass of water, - c= 4.18 J/g

C is the specific heat

capacity of water, - ∆T= 21.9−20.0=1.9

C is the change in temperature.

Substitute the values into the formula:

QWater = 300.0 g ×4.18 J/g

C×1.9

Step 3: Set up the heat exchange equation.

According to the principle of conservation of energy, the heat released by

the aluminum is equal to the heat absorbed by the water. This can be written

as:

QAluminum =QWater

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

Set QAluminum =QWater and solve for cto find the specific heat capacity of

aluminum.

Question 23

Question

A 50 g piece of iron is heated to 100

C and then dropped into a calorimeter

containing 200 g of water at 20

C. If the final temperature of the system is

C, calculate the specific heat capacity of iron. Assume no heat is lost to the

surroundings.

Solution

Step 1: Calculate the heat gained by the water. The heat gained 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 (4.18

J/g

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

Given that: - m= 200 g, - c= 4.18 J/g

C, - ∆T= 30 −20 = 10

we can substitute these values into the formula:

Q= 200 ×4.18 ×10

Q= 8360 J

Step 2: Calculate the heat lost by the iron. The heat lost by the iron is the

same as the heat gained by the water (by the law of conservation of energy).

So, we have:

Qiron =−Qwater

Qiron =−8360 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 (50 g), - cis the specific heat capacity of iron, - ∆T

is the change in temperature of iron (final temperature - initial temperature).

Given that: - m= 50 g, - ∆T= 100 −30 = 70

we can substitute these values into the formula:

−8360 = 50c×70

c=−8360

50 ×70

c=−3.77 J/g

Therefore, the specific heat capacity of iron is approximately 3.77 J/g

Question 24

Question

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

metal using a calorimeter. The student places 200 g of the metal, initially at

150

C, into 300 g of water at 20

C in the calorimeter, resulting in a final equi-

librium temperature of 25

C. If the specific heat capacity of water is 4.18 J/g

calculate the specific heat capacity of the metal.

Solution

Step 1: First, calculate the heat absorbed or released by the metal. The heat

absorbed by the metal can be calculated using the formula:

Qmetal =mmetal ×cmetal ×∆T

where: - mmetal = 200 g (mass of the metal), - cmetal is the specific heat capacity

of the metal to be determined, - ∆T=Tf−Tiis the change in temperature of

the metal. Given Ti= 150Cand Tf= 25C:

∆T= 25C−150C=−125C

Qmetal = 200 g ×cmetal ×(−125C)

Step 2: Next, calculate the heat absorbed or released by the water. The

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

Qwater =mwater ×cwater ×∆T

where: - mwater = 300 g (mass of the water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T=Tf−Tiis the change in temperature of the water.

Given Ti= 20Cand Tf= 25C:

∆T= 25C−20C= 5C

Qwater = 300 g ×4.18 J/g

C×5C

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

the heat lost by the metal must equal the heat gained by the water:

Qmetal =−Qwater

200 g ×cmetal ×(−125C) = −300 g ×4.18 J/g

C×5C

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

cmetal =−300 g ×4.18 J/g

C×5C

200 g ×(−125C)

Question 25

Question

A 50.0 g iron bar at 120.0

C is placed in 200.0 g of water at 20.0

C. If the final

temperature of the system is 25.0

C, calculate the specific heat capacity of iron.

Solution

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

can be calculated using the formula:

Q=mc∆T

where: - Qis the heat gained, - mis the mass of the water, - cis the specific

heat capacity of water (4.18 J/g

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

water.

Plugging in the values:

Q= (200.0 g)(4.18 J/g

C)(25.0−20.0)

Q= (200.0 g)(4.18 J/g

C)(5.0)

Q= 4180 J

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

be calculated using the formula:

Q=mc∆T

where: - Qis the heat lost, - mis the mass of the iron (50.0 g), - cis the specific

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

Plugging in the values:

4180 J = (50.0 g)c(120.0−25.0)

4180 J = (50.0 g)c(95.0)

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

c=4180 J

(50.0 g)(95.0

c=4180 J

4750.0 g

c≈0.88 J/g

Therefore, the specific heat capacity of iron is approximately 0.88 J/g

Question 26

Question

A 50 g piece of copper is heated to 100

C and then placed in a calorimeter

containing 200 g of water at 20

C. If the final temperature of the system is

C, determine the specific heat capacity of the calorimeter assuming no heat

is lost to the surroundings. The specific heat capacity of copper is 0.386 J/g

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

Solution

Step 1: Calculate the heat lost by the copper piece to cool down from 100

C to

C. The formula 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.

Using this formula for copper: Qcopper = (50 g)(0.386 J/g

C)(25C−100C)

Qcopper =−1025 J

Step 2: Calculate the heat gained by the water to warm up from 20

C to

C. Using the same formula but for water: Qwater = (200 g)(4.18 J/g

C)(25C−

20C)Qwater = 418 J

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

is equal to the heat gained by the water. Therefore, Qcopper =Qwater −1025 J =

418 J −1025 J + 418 J = 0 −1025 J + 418 J = 0

Step 4: Calculate the heat absorbed by the calorimeter. Since the calorime-

ter’s specific heat capacity (ccalorimeter) is unknown, we denote it with ccalorimeter.

Using the same formula for the calorimeter: Qcalorimeter = (50 g+200 g)ccalorimeter(25C−

20C)−1025 J = 250ccalorimeter5C−1025 J = 1250ccalorimeter ccalorimeter =−1025 J

1250 J/g

ccalorimeter ≈ −0.82 J/g

Therefore, the specific heat capacity of the calorimeter is approximately

−0.82 J/g

Question 27

Question

A 50.0 g piece of aluminum, initially at 100.0

C, is dropped into a calorimeter

containing water. The water has a mass of 200.0 g and is at 20.0

C. If the

final temperature of the system is 25.0

C, calculate the specific heat capacity

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

Solution

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

water.

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

Qlost =m·c·∆T

where: - m= 50.0 g (mass of aluminum) - c=? (specific heat capacity of

aluminum) - ∆T=Tf−Ti= 25.0

C−100.0

C = −75.0

Substitute in the values:

Qlost = 50.0 g ·c·(−75.0

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

Qgain =m·cwater ·∆T

where: - m= 200.0 g (mass of water) - cwater = 4.18 J/g

C (specific heat

capacity of water) - ∆T= 25.0

C−20.0

C=5.0

Substitute in the values:

Qgain = 200.0 g ·4.18 J/g

C·5.0

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

According to the principle of conservation of energy, the heat lost by the

aluminum is equal to the heat gained by the water.

Qlost =Qgain

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

Equating the two equations and solving for c:

50.0 g ·c·(−75.0

C) = 200.0 g ·4.18 J/g

C·5.0

c=200.0 g ·4.18 J/g

C·5.0

50.0 g ·(−75.0

c≈4180 J

−3750 g ≈ −1.114 J/g

So, the specific heat capacity of aluminum is approximately 1.114 J/g

C .

Question 28

Question

A 50 gram ice cube at -10

C is placed in a calorimeter containing 200 grams

of water at 25

C. If the final temperature of the system is 10

C, calculate the

specific heat capacity of the calorimeter.

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

C and the latent heat

of fusion of ice is 334 J/g)

Solution

Step 1: Calculate the energy needed to heat the ice to 0

Q1=m1c1∆T

= (50 g)(2.09 J/g

C)(0

C−(−10

C))

= 1045 J

Step 2: Calculate the energy needed to melt the ice at 0

Q2=m1Lf

= (50 g)(334 J/g)

= 16700 J

Step 3: Calculate the energy needed to heat the water from 25

C to 0

Q3=m2c2∆T

= (200 g)(4.18 J/g

C)(0

C−25

=−20900 J

Step 4: Calculate the energy needed to heat the water from 0

C to 10

Q4=m2c2∆T

= (200 g)(4.18 J/g

C)(10

C−0

= 8360 J

Step 5: Since energy is conserved, the total energy gained by the system is

equal to the total energy lost by the system:

Qgained =Qlost

Q1+Q2+Q3+Q4= 0

1045 J + 16700 J + (−20900 J) + 8360 J = 0

1205 J = 0 J

Step 6: Since there is no heat exchange with the surroundings, the total

energy is absorbed by the system. The energy absorbed by the calorimeter is:

Qcalorimeter =mcalorimeterccalorimeter∆T

=Qgained

ccalorimeter =Qgained

mcalorimeter∆T

=1205 J

200 g ×20

= 0.3025 J/g

Therefore, the specific heat capacity of the calorimeter is 0.3025 J/g

Question 29

Question

A 50 g block of copper at 200

C is dropped into an insulated vessel containing

200 g of water at 20

C. Assuming no heat is lost to the surroundings and that

the specific heat capacity of copper is 0.385 J/g

C, the specific heat capacity of

water is 4.18 J/g

C, and neglecting the heat capacity of the vessel, what will be

the final temperature of the system?

Solution

Step 1: First, we can calculate the heat lost by the copper block as it cools

down to the final temperature of the system using the formula:

qcopper =mcopper ·ccopper ·∆T

where mcopper is the mass of the copper block, ccopper is the specific heat

capacity of copper, and ∆Tis the temperature change of the copper block.

Plugging in the values, we have:

qcopper = 50 g ·0.385 J/g

C·(Tf−200C)

Step 2: Next, let’s calculate the heat gained by the water as it warms up to

the final temperature of the system using the formula:

qwater =mwater ·cwater ·∆T

where mwater is the mass of the water, cwater is the specific heat capacity of

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

Plugging in the values, we have:

qwater = 200 g ·4.18 J/g

C·(Tf−20C)

Step 3: Since no heat is lost in the system, we can set qcopper =qwater and

solve for the final temperature, Tf.

50 g ·0.385 J/g

C·(Tf−200C) = 200 g ·4.18 J/g

C·(Tf−20C)

Solving for Tf, we find:

50 ·0.385 ·(Tf−200) = 200 ·4.18 ·(Tf−20)

19.25 ·Tf−7700 = 836 ·Tf−16720

816 ·Tf= 9020

Tf=9020

816 = 11.029

Therefore, the final temperature of the system will be 11.029

Question 30

Question

A 100 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

mixture? (Specific heat capacities: water = 4.18 J/g

C, copper = 0.387 J/g

Solution

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

water can be calculated using the formula:

qwater =mwatercwater∆T

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

C (specific heat

capacity of water) - ∆T=Tfinal −Tinitial =Tfinal −20 (change in temperature

of water)

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

can be calculated using the formula:

qcopper =mcopperccopper∆T

where: - mcopper = 100 g (mass of copper) - ccopper = 0.387 J/g

C (specific heat

capacity of copper) - ∆T= 150 −Tfinal (change in temperature of copper)

Step 3: Since heat is conserved in this isolated system, the heat lost by the

copper must equal the heat gained by the water:

qcopper =qwater

mcopperccopper∆T=mwatercwater∆T

Step 4: Substitute the expressions for qcopper and qwater into the equation

from step 3 and solve for Tfinal.

Step 5: After solving the equation, you will find the final temperature of the

mixture to be the point at which the heat lost by the copper equals the heat

gained by the water.

Question 31

Question

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

calorimetry. The student places a 0.2 kg metal block, initially at a temperature

of 100◦C, into 0.5 kg of water at 20◦C contained in a calorimeter. The final

temperature of the system is 30◦C. If the specific heat capacity of water is 4186

J/kg·K, calculate the specific heat capacity of the metal block.

Solution

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

can be calculated using the formula

Q=mc∆T

where: - mis the mass of the water (0.5 kg), - cis the specific heat capacity of

water (4186 J/kg·K), - ∆Tis the change in temperature of the water (30◦C -

20◦C = 10◦C).

Substitute the given values into the formula:

Q= (0.5 kg)(4186 J/kg ·K)(10 K)

Q= 20930 J

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

block can be calculated using the same formula as above. Since the system is

isolated and no heat is lost to the surroundings, the heat lost by the metal block

is equal to the heat gained by the water. Therefore, the heat lost by the metal

block is also 20930 J.

Step 3: Use the equation for heat transfer: Since the heat lost by the metal

block is equal to the heat gained by the water, we can write:

Qmetal =mc∆T

where: - mis the mass of the metal block (0.2 kg), - cis the specific heat capacity

of the metal block (which we want to find), - ∆Tis the change in temperature

of the metal block (final temperature - initial temperature).

Since the temperature of the metal decreases from 100◦C to 30◦C, we have

∆T= 70 ◦C. Substitute the known values of m, ∆T, and Qinto the equation:

20930 J = (0.2 kg)c(70 K)

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

c=20930 J

(0.2 kg)(70 K)

c≈299 J/kg ·K

Therefore, the specific heat capacity of the metal block is approximately 299

J/kg·K.

Question 32

Question

A piece of unknown metal weighing 95 g is heated to 99

C and then placed

in a calorimeter containing 120 g of water at 22

C. The final temperature of

the system is 25

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

the specific heat capacity of the metal. The specific heat capacity of water is

4.184 J/g

Solution

Step 1: Calculate the heat absorbed by the metal.

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

qmetal =mmetal ×cmetal ×∆Tmetal

where: - mmetal = 95 g (mass of the metal), - cmetal is the specific heat

capacity of the metal (to be found), - ∆Tmetal =Tfinal −Tinitial = 25C−99C=

−74C(temperature change of the metal).

Plugging in the values gives:

qmetal = 95 g ×cmetal ×(−74C)

Step 2: Calculate the heat released by the water.

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

qwater =mwater ×cwater ×∆Twater

where: - mwater = 120 g (mass of the water), - cwater = 4.184 J/g

C (spe-

cific heat capacity of water), - ∆Twater =Tfinal −Tinitial = 25C−22C= 3C

(temperature change of the water).

Plugging in the values gives:

qwater = 120 g ×4.184 J/g

C×3C

Step 3: Set up the heat exchange equation.

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

equal to the heat released by the water. Thus,

qmetal =qwater

Step 4: Solve for cmetal.

Equating qmetal and qwater, we have:

95 g ×cmetal ×(−74C) = 120 g ×4.184 J/g

C×3C

Now, solve for cmetal.

Question 33

Question

A piece of gold of mass 50 g at 80

C is placed in 200 g of water at 20

C. Assuming

no heat is lost to the surroundings, what is the final temperature of the system?

The specific heat capacity of gold is 0.13 J/g·

C and that of water is 4.18 J/g·

Solution

Step 1: Calculate the heat lost by the gold piece and the heat gained by the

water.

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

Qgold =mgold ·cgold ·∆T

where: - mgold = 50 g (mass of gold), - cgold = 0.13 J/g ·

C (specific heat

capacity of gold), - ∆T=Tfinal −Tinitial =Tf−Ti=Tf−80 (temperature

change of gold).

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

Qwater =mwater ·cwater ·∆T

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

C (specific heat

capacity of water), - ∆T=Tfinal −Tinitial =Tf−Ti=Tf−20 (temperature

change of water).

Step 2: Set up the heat lost by the gold piece equal to the heat gained by

the water and solve for the final temperature.

Qgold =Qwater

mgold ·cgold ·(Tf−80) = mwater ·cwater ·(Tf−20)

Substitute the given values:

50 ·0.13 ·(Tf−80) = 200 ·4.18 ·(Tf−20)

Now, solve this equation to find the final temperature Tf.

Question 34

Question

A 50 g piece of iron at 80

C is placed in 200 g of water at 20

C. Assuming all

the heat is transferred to the water and that no heat is lost to the surroundings,

what is the final temperature of the system? (Specific heat capacity of iron =

0.449 J/g ·

C, specific heat capacity of water = 4.18 J/g ·

Solution

Step 1: Calculate the heat lost by the iron: The formula for heat transfer is

given by Q=mc∆T, where: - Qis the heat transfer - mis the mass of the

material - cis the specific heat capacity of the material - ∆Tis the change in

temperature

Given that the iron piece is cooling down: Qiron =−mciron∆Tiron

Substitute the values given: Qiron =−(50 g)(0.449 J/g ·

C)(Tf−80

Step 2: Calculate the heat gained by the water: Since all the heat lost by

the iron is gained by the water: Qwater =mcwater∆Twater

Substitute the values given: Qwater = (200 g)(4.18 J/g ·

C)(Tf−20

Step 3: Set the heat lost equal to the heat gained: −(50)(0.449)(Tf−80) =

(200)(4.18)(Tf−20)

Step 4: Solve for the final temperature Tf:−22.45Tf+359.2 = 836Tf−16720

858.45Tf= 17079.2

Tf≈19.9

Therefore, the final temperature of the system is approximately 19.9

Question 35

Question

A sample of metal with an initial temperature of 100

C is placed in a calorimeter

containing 200 g of water at 20

C. The final temperature of the system is 25

If the specific heat capacity of the metal is 0.5 J/g

C, calculate the mass of the

metal sample.

Solution

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

by the metal can be calculated using the formula:

Qmetal =mc∆T

where: m = mass of the metal (g) c = specific heat capacity of the metal (J/g

∆T= change in temperature of the metal

Given: c = 0.5 J/g

C, initial temperature = 100

C, final temperature = 25

∆T= 25C−100C=−75C

Substitute the values into the formula:

Qmetal =m·0.5·(−75)

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

be calculated using the formula:

Qwater =mc∆T

where: m = mass of the water (g) = 200 g c = specific heat capacity of water

(J/g

C) = 4.18 J/g

C ∆T= 25C−20C= 5C

Substitute the values into the formula:

Qwater = 200 ·4.18 ·5

Step 3: Since the total heat gained by the metal is equal to the total heat

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

Qmetal =Qwater

Question 7

Question

A 50 g piece of aluminum at 120

C is placed in 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.

(Specific heat capacity of water is 4.18 J/g

Solution

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

the formula q=mc∆T.

Heat lost by aluminum = Heat gained by water

mAlcAl∆TAl =mH2OcH2O∆TH2O

Step 2: Substitute the known values into the equation.

50 g ×cAl ×(22 −120) = 100 g ×4.18 J/g

C×(22 −20)

−7000cAl = 836.0

Step 3: Solve for cAl.

cAl =836.0

−7000

cAl =−0.119 J/g

Therefore, the specific heat capacity of aluminum is 0.119 J/g

Question 8

Question

A 50 g piece of iron at 250

C is dropped into 200 g of water at 20

C inside

a perfectly insulated container. Assuming no heat is lost to the surroundings,

what will be the final temperature of the system? (Specific heat capacity of iron

= 0.45 J/g

C, specific heat capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the iron as it cools down from 250

C to the

final temperature. The heat lost 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), and ∆Tis the change in temperature. Since the initial temperature of

the iron is 250

C and the final temperature is T

C (unknown), the change in

temperature is 250 −T. Therefore,

Qiron = 50 g ·0.45 J/g

C·(250 −T)

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

C to

the final temperature. The heat gained can be calculated using the formula:

Qwater =m·c·∆T

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

water (4.18 J/g

C), and ∆Tis the change in temperature. Since the initial

temperature of the water is 20

C and the final temperature is T

C (unknown),

the change in temperature is T−20. Therefore,

Qwater = 200 g ·4.18 J/g

C·(T−20)

Step 3: Since energy is conserved in this system, the heat lost by the iron is

equal to the heat gained by the water. Therefore,

50 g ·0.45 J/g

C·(250 −T) = 200 g ·4.18 J/g

C·(T−20)

Step 4: Simplify the equation and solve for T to find the final temperature

of the system.

22.5(250 −T) = 836(T−20)

5625 −22.5T= 836T−16720

0 = 858.5T−22345

T≈22345

858.5≈26.04C

Therefore, the final temperature of the system will be approximately 26.04

Question 9

Question

A 50 g aluminum block at 80

C is placed in 200 g of water at 20

C in a calorime-

ter. Assuming all the heat lost by the aluminum block is gained by the water

and the calorimeter, what is the final temperature of the system? (Specific

heat capacity of aluminum = 0.902 J/g

C, specific heat capacity of water =

4.18 J/g

C, specific heat capacity of the calorimeter = 1.0 J/g

Solution

Step 1: Calculate the heat lost by the aluminum block: Given: Mass of the

aluminum block, mAl = 50 g Initial temperature of the aluminum block, TAl,i =

80CSpecific heat capacity of aluminum, cAl = 0.902 J/g

Using the formula for heat energy:

QAl =mcAl∆TAl

where ∆TAl =Tfinal −TAl,i is the change in temperature of the aluminum block.

Step 2: Calculate the heat gained by the water and the calorimeter: Given:

Mass of the water, mwater = 200 g Initial temperature of the water, Twater,i =

20CSpecific heat capacity of water, cwater = 4.18 J/g

C Specific heat capacity

of the calorimeter, ccalorimeter = 1.0 J/g

Using the same formula for heat energy:

Qwater+calorimeter = (mwater +mcalorimeter)c∆Twater+calorimeter

where mcalorimeter is the mass of the calorimeter and ∆Twater+calorimeter =Tfinal−

Twater,i is the change in temperature of the water and the calorimeter.

Step 3: Set the heat lost equal to the heat gained:

QAl =Qwater+calorimeter

Step 4: Solve for the final temperature, Tfinal:

mAlcAl(Tfinal−TAl,i)=(mwater+mcalorimeter)(cwater+ccalorimeter)(Tfinal−Twater,i)

Now, solve the equation above to find the final temperature of the system.

Question 10

Question

A 50 g piece of copper initially at 200

C is dropped into 200 g of water at 20

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

C. Assuming no

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

(Specific heat capacity of water = 4.18 J/g

Solution

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

temperature. The heat lost by the copper can be calculated using the formula

Q=mc∆T, where: - mis the mass of the copper (50 g), - cis the specific

heat capacity of copper (to be determined), - ∆Tis the change in temperature

(200

C - 25

C).

Substitute the values into the formula: Qcopper = (50 g)(c)(200 −25) =

50c(175) = 8750cJ

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

temperature. The heat gained by the water can be calculated using the formula

Q=mc∆T, where: - mis the mass of the water (200 g), - cis the specific

heat capacity of water (4.18 J/g

C), - ∆Tis the change in temperature (25

C -

C).

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

C)(25 −

20) = 200(4.18)(5) = 4180 J

Step 3: Since heat is conserved, the heat lost by the copper must be equal to

the heat gained by the water. Set Qcopper =Qwater and solve for c: 8750c= 4180

Step 4: Solve for c.c=4180

8750 ≈0.4777 J/g

Therefore, the specific heat capacity of copper is approximately 0.4777 J/g

Question 11

Question

A 50-g piece of aluminum at 120

C is dropped into 200 g of water at 20

C. If

the final temperature of the system is 25

C, calculate the specific heat capacity

of aluminum (cAl) given that the specific heat capacity of water is 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:

mAlcAl∆TAl =mwatercwater∆Twater

where: mAl = mass of aluminum = 50 g, cwater = specific heat capacity of water

= 4.18 J/g

C, mwater = mass of water = 200 g, ∆TAl = change in temperature

of aluminum = 120

C - 25

C = 95

C, ∆Twater = change in temperature of water

= 25

C - 20

C = 5

Substitute the values into the equation:

50 ×cAl ×95 = 200 ×4.18 ×5

Step 2: Solve for cAl.

50 ×cAl ×95 = 200 ×4.18 ×5

50 ×cAl =200 ×4.18 ×5

cAl =200 ×4.18 ×5

95 ×50

cAl =4180

cAl = 44.0 J/g

Therefore, the specific heat capacity of aluminum is 44.0 J/g

Question 12

Question

A 50 g piece of copper at 200

C is placed in 200 g of water at 20

C in a calorime-

ter. The final temperature of the system is 25

C. Assuming no heat is lost to

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

capacity of water = 4.18 J/g

Solution

Given: Mass of copper (mc) = 50 g, Initial temperature of copper (Tc1) = 200

Mass of water (mw) = 200 g, Initial temperature of water (Tw1) = 20

C, Final

temperature of system (Tf) = 25

C, Specific heat capacity of water (cw) = 4.18

J/g

Let’s denote the specific heat capacity of copper as cc.

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

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

mc·cc·∆Tc=−mw·cw·∆Tw

Where ∆Tcand ∆Tware the changes in temperature for copper and water

respectively.

Step 2: Calculate ∆Tcand ∆TwStep 2.1: Change in temperature for

copper:

∆Tc=Tf−Tc1= 25C−200C=−175C

Step 2.2: Change in temperature for water:

∆Tw=Tf−Tw1= 25C−20C= 5C

Step 3: Substitute into the heat equation Substitute the values into

the heat equation from Step 1:

50 ·cc·(−175) = −200 ·4.18 ·5

Step 4: Solve for cc

cc=−200 ·4.18 ·5

50 ·(−175)

cc=−4180

−8750

cc= 0.4771 J/g

Therefore, the specific heat capacity of copper is 0.4771 J/g

Question 13

Question

A 50.0 g piece of copper at 95.0

C is placed in 100.0 g of water at 20.0

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

of the system? (Specific heat capacity of copper = 0.387 J/g

C, specific heat

capacity of water = 4.18 J/g

Solution

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

final temperature. 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: mcopper = 50.0 g, ccopper = 0.387 J/g

C, Tinitial = 95.0C,Tfinal =

Tsystem qcopper =mcopper ·ccopper ·(Tfinal −Tinitial)

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

temperature. Given: mwater = 100.0 g, cwater = 4.18 J/g

C, Tinitial, water =

20.0C,Tfinal, water =Tsystem qwater =mwater ·cwater ·(Tfinal −Tinitial, water)

Step 3: Since heat lost by copper piece is equal to heat gained by water

when no heat is lost to the surroundings, we have: qcopper =−qwater mcopper ·

ccopper ·(Tfinal −Tinitial) = −mwater ·cwater ·(Tfinal −Tinitial, water)

Step 4: Solve the equation from step 3 for the final temperature Tfinal.

50.0·0.387 ·(Tfinal −95.0) = −100.0·4.18 ·(Tfinal −20.0)

19.35Tfinal −1866.5 = −418Tfinal + 8360

437.35Tfinal = 10226.5

Tfinal23.4C

Therefore, the final temperature of the system is approximately 23.4

Question 14

Question

A 50 g piece of copper at 200

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 and the heat gained by the water

using the formula:

qlost =−qgain

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

qcopper =mc∆T

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

∆Tis the change in temperature.

Substitute the given values:

qcopper = (50 g)(0.385 J/g

C)(Tf−200C)

Step 3: Calculate the heat gained by the water using the formula:

qwater =mc∆T

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

∆Tis the change in temperature.

Substitute the given values:

qwater = (100 g)(4.18 J/g

C)(Tf−20C)

Step 4: Set the two heat equations equal to each other:

(50 g)(0.385 J/g

C)(Tf−200C) = (100 g)(4.18 J/g

C)(Tf−20C)

Step 5: Solve for Tfby simplifying and rearranging the equation.

19.25 ·(Tf−200) = 418 ·(Tf−20)

19.25Tf−3850 = 418Tf−8360

3985 = 398.75Tf

Tf= 10C

Therefore, the final temperature of the system will be 10

Question 15

Question

A sample of gold at 200.0◦C is placed in 100.0 g of water at 20.0◦C. If the final

temperature of the system is 25.0◦C, determine the mass of the gold sample.

Assume the specific heat capacity of gold is 0.129 J/g◦C and the specific heat

capacity of water is 4.18 J/g◦C.

Solution

Step 1: Use the heat gained by the water is equal to the heat lost by the gold

sample: Let mbe the mass of the gold sample and cgold be the specific heat

capacity of gold. We can set up the equation:

m·cgold ·(Tfinal −Tinitial, gold) = mwater ·cwater ·(Tfinal, water −Tinitial, water)

where: - Tfinal is the final temperature of the system, - Tinitial, gold is the initial

temperature of the gold sample, - Tinitial, water is the initial temperature of the

water, - mwater is the mass of the water, and - Tfinal, water is the final temperature

of the water.

Step 2: Substitute the known values into the equation: We know Tfinal =

25.0◦C, Tinitial, gold = 200.0◦C, Tinitial, water = 20.0◦C, mwater = 100.0 g, cgold =

0.129 J/g◦C, and cwater = 4.18 J/g◦C.

m·0.129 ·(25.0−200.0) = 100.0·4.18 ·(25.0−20.0)

Step 3: Solve for the mass of the gold sample:

m·0.129 ·(−175.0) = 100.0·4.18 ·5.0

−22.575m= 209.0

m=209.0

−22.575

m≈ −9.27 g

Since we cannot have a negative mass, the mass of the gold sample must be

9.27 g.

Question 16

Question

A 50 g aluminum block at 80

C is added to 200 g of water at 20

C in a calorime-

ter. If the final temperature of the system is 25

C, calculate the specific heat

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

Solution

Step 1: Calculate the heat lost by aluminum (Q loss). The heat lost by the

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

The formula to calculate heat is given by Q=mc∆T, where: - mis the mass

of the substance (in kg), - cis the specific heat capacity of the substance (in

J/(kg

·°

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

C).

Given: - Mass of aluminum block, mAl = 50 g = 0.05 kg, - Initial tempera-

ture of aluminum block, TAl, initial = 80 C, - Final temperature of the system,

Tfinal = 25 C.

To get ∆Tfor aluminum, we use the formula: ∆T=Tfinal −TAl, initial.

∆T= 25 C −80 C = −55 C

Therefore, the heat lost by aluminum, Qloss =mAl ·cAl ·∆TAl.

Step 2: Calculate the heat gained by water and calorimeter (Q gain). The

heat gained by the water and calorimeter is equal to the heat lost by the alu-

minum block. This is given by Qloss =Qgain.

Next, we calculate Qgain using the formula Qgain =mwater ·cwater ·∆Twater +

ccal ·∆Tcal, where: - mwater = 200 g = 0.2 kg is the mass of water, - cwater =

4186 J/(kg

·°

C) is the specific heat capacity of water, - ∆Twater =Tfinal −

Twater, initial, - ccal is the specific heat capacity of the calorimeter, - ∆Tcal =

Tfinal −Tcal, initial.

Step 3: Setting Qloss =Qgain and calculating cAl.

mAl ·cAl ·∆TAl =mwater ·cwater ·∆Twater +ccal ·∆Tcal

Substitute the known values and solve for cAl to find the specific heat ca-

pacity of aluminum.

Question 17

Question

A piece of copper of mass 200 g at a temperature of 100◦C is placed in 400 g

of water at 20◦C contained in a calorimeter of mass 100 g. The temperature of

water and the calorimeter rises to 28◦C. If the specific heat capacity of copper

is 0.387 J/g◦C, determine the specific heat capacity of the calorimeter. Assume

no heat is lost to the surroundings.

Solution

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

calorimeter. Let Cpbe the specific heat capacity of the calorimeter. The heat

lost by copper is given by

Qcopper =m·c·∆T

where m= mass of copper = 200 g, c= specific heat capacity of copper = 0.387

J/g◦C, ∆T= change in temperature of copper = (temperature of water and

calorimeter - initial temperature of copper) = (28◦C - 100◦C).

The heat gained by the water and calorimeter is given by

Qwater+calorimeter = (mwater +mcalorimeter)·cwater+calorimeter ·∆T

where mwater = mass of water = 400 g, mcalorimeter = mass of calorimeter =

100 g, cwater+calorimeter = specific heat capacity of water and calorimeter, ∆T=

change in temperature of water and calorimeter = (temperature of water and

calorimeter - initial temperature of water and calorimeter) = (28◦C - 20◦C).

Since heat is conserved, we have

Qcopper =Qwater+calorimeter

m·c·∆T= (mwater +mcalorimeter)·cwater+calorimeter ·∆T

Step 2: Solve for cwater+calorimeter, the specific heat capacity of water and

calorimeter. Substitute the given values into the equation:

200 g ·0.387 J/g◦C·(28 −100) = (400 g + 100 g) ·cwater+calorimeter ·(28 −20)

cwater+calorimeter =200 ·0.387 ·(−72)

500 ·8

Step 3: Calculate the specific heat capacity of the calorimeter. The specific

heat capacity of the calorimeter, Cp, is equal to the specific heat capacity of the

water and calorimeter, cwater+calorimeter.

Cp=cwater+calorimeter =200 ·0.387 ·(−72)

500 ·8

Question 18

Question

A 50.0 g piece of aluminum at 95.0

C is placed into 100.0 g of water at 20.0

The final temperature of the system is 22.0

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: Determine the heat gained by the water and the heat lost by the alu-

minum.

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

qwater =mwater ×cwater ×∆T

where: - mwater = 100.0 g (mass of water) - cwater = 4.18 J/g

C (specific heat

capacity of water) - ∆T=Tfinal −Tinitial = 22.0C −20.0C = 2.0C(change in

temperature)

Plugging in the values, we get:

qwater = 100.0 g ×4.18 J/g

C×2.0C= 836 J

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

qaluminum =maluminum ×caluminum ×∆T

where: - maluminum = 50.0 g (mass of aluminum) - caluminum (specific heat ca-

pacity of aluminum) - ∆T=Tfinal −Tinitial = 22.0C −95.0C = −73.0C(change

in temperature)

We can rewrite the equation as:

qaluminum =maluminum ×caluminum × −73.0C

qaluminum =−50.0 g ×caluminum ×73.0 J

Step 2: Since the system is isolated, the heat lost by aluminum is equal to

the heat gained by water:

qwater =qaluminum

836 J = −50.0 g ×caluminum ×73.0 J

Step 3: Solve for the specific heat capacity of aluminum (caluminum):

caluminum =836

−50.0×73.0

caluminum =−0.228 J/g

Therefore, the specific heat capacity of aluminum is −0.228 J/g

Question 19

Question

A 50 g piece of copper at 150

C is placed in 200 g of water at 25

C. Assuming

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

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

C, specific heat of water = 4.184

J/g

C, and the specific heat of fusion of copper = 205 J/g)

Solution

Step 1: Calculate the heat lost by the copper piece to reach equilibrium with

the water. The heat lost by the copper can be calculated as:

qcopper =mcopper ·ccopper ·∆T

where: mcopper = 50 g (mass of copper), ccopper = 0.386 J/g

C (specific heat of

copper), and ∆T=Tfinal −150

Step 2: Calculate the heat gained by the water to reach equilibrium with

the copper. The heat gained by the water can be calculated as:

qwater =mwater ·cwater ·∆T

where: mwater = 200 g (mass of water), cwater = 4.184 J/g

C (specific heat of

water), and ∆T=Tfinal −25

Step 3: Since the total heat loss by the copper is equal to the total heat gain

by the water when they reach equilibrium, we can set up the equation:

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

Step 4: Solve for the final temperature. Solving the equation from Step 3

for Tfinal, we have:

50 ·0.386 ·(Tfinal −150) = 200 ·4.184 ·(Tf inal −25)

Step 5: Simplify the equation and solve for Tfinal to find the final temperature

of the system.

Question 20

Question

A 50 g ice cube at -10

C is placed in a calorimeter containing 200 g of water

at 20

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

temperature of the system when the ice has completely melted? (Specific heat

capacity of water = 4.18 J/g

C, specific heat capacity of ice = 2.09 J/g

C, heat

of fusion for ice = 334 J/g)

Solution

Step 1: Calculate the heat absorbed by the ice cube to melt into water.

Q1=m·Lf

= 50 g ×334 J/g

= 16700 J

Step 2: Calculate the heat gained by the ice cube to reach the final temper-

ature.

Q2=m·cice ·∆T

= 50 g ×2.09 J/g

C×(0 −(−10))

= 1045 J

Step 3: Calculate the heat gained by the water to reach the final temperature.

Q3=m·cwater ·∆T

= 200 g ×4.18 J/g

C×(Tf−20)

= 836 J ×(Tf−20)

Step 4: Since the total heat gained equals the total heat lost, we have:

Q1+Q2=−Q3

Substitute the calculated values:

16700 J + 1045 J = −836 J ×(Tf−20)

Step 5: Solve for Tf.

17745 = −836 J ×(Tf−20)

Tf−20 = −17745

836

Tf=−17745

836 + 20

Therefore, the final temperature of the system when the ice has completely

melted is approximately −2.38

Question 21

Question

A 50 g piece of aluminum at 80

C is placed in 200 g of water at 20

C. After

reaching thermal equilibrium, the final temperature of the system is 25

C. Cal-

culate the specific heat capacity of aluminum. (Specific heat capacities: water

= 4.18 J/g

C, aluminum = 0.897 J/g

Solution

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

Let’s assume no heat loss to the surroundings. The heat lost by aluminum is

equal to the heat gained by water:

m1c1(Tf−T1) = m2c2(Tf−T2)

where: - m1= 50 g (mass of aluminum) - c1= 0.897 J/g

C (specific heat capac-

ity of aluminum) - Tf= 25C(final temperature) - T1= 80C(initial temperature

of aluminum) - m2= 200 g (mass of water) - c2= 4.18 J/g

C (specific heat ca-

pacity of water) - T2= 20C(initial temperature of water)

Substitute the values into the equation:

(50)(0.897)(25 −80) = (200)(4.18)(25 −20)

Step 2: Solve the equation for c1.

−22425 = 4180(5)

Step 3: Rearrange the formula to solve for c1.

c1=4180(5)

50(55) = 3.82 J/g

Therefore, the specific heat capacity of aluminum is 3.82 J/g

Question 22

Question

A 50.0 g sample of aluminum at 87.0

C is dropped into 300.0 g of water at

20.0

C. If the final temperature of the system is 21.9

C, what is the specific

heat capacity of aluminum? Assume the specific heat capacity of water is 4.18

J/g

Solution

Step 1: Calculate the heat released by the aluminum when it cools down.

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

QAluminum =m·c·∆T

where: - m= 50.0 g is the mass of aluminum, - cis the specific heat capacity

of aluminum (to be determined), - ∆T= 87.0−21.9 = 65.1

C is the change in

temperature.

Substitute the values into the formula:

QAluminum = 50.0 g ×c×65.1

Step 2: Calculate the heat absorbed by the water as it heats up.

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

QWater =m·c·∆T

where: - m= 300.0 g is the mass of water, - c= 4.18 J/g

C is the specific heat

capacity of water, - ∆T= 21.9−20.0=1.9

C is the change in temperature.

Substitute the values into the formula:

QWater = 300.0 g ×4.18 J/g

C×1.9

Step 3: Set up the heat exchange equation.

According to the principle of conservation of energy, the heat released by

the aluminum is equal to the heat absorbed by the water. This can be written

as:

QAluminum =QWater

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

Set QAluminum =QWater and solve for cto find the specific heat capacity of

aluminum.

Question 23

Question

A 50 g piece of iron is heated to 100

C and then dropped into a calorimeter

containing 200 g of water at 20

C. If the final temperature of the system is

C, calculate the specific heat capacity of iron. Assume no heat is lost to the

surroundings.

Solution

Step 1: Calculate the heat gained by the water. The heat gained 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 (4.18

J/g

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

Given that: - m= 200 g, - c= 4.18 J/g

C, - ∆T= 30 −20 = 10

we can substitute these values into the formula:

Q= 200 ×4.18 ×10

Q= 8360 J

Step 2: Calculate the heat lost by the iron. The heat lost by the iron is the

same as the heat gained by the water (by the law of conservation of energy).

So, we have:

Qiron =−Qwater

Qiron =−8360 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 (50 g), - cis the specific heat capacity of iron, - ∆T

is the change in temperature of iron (final temperature - initial temperature).

Given that: - m= 50 g, - ∆T= 100 −30 = 70

we can substitute these values into the formula:

−8360 = 50c×70

c=−8360

50 ×70

c=−3.77 J/g

Therefore, the specific heat capacity of iron is approximately 3.77 J/g

Question 24

Question

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

metal using a calorimeter. The student places 200 g of the metal, initially at

150

C, into 300 g of water at 20

C in the calorimeter, resulting in a final equi-

librium temperature of 25

C. If the specific heat capacity of water is 4.18 J/g

calculate the specific heat capacity of the metal.

Solution

Step 1: First, calculate the heat absorbed or released by the metal. The heat

absorbed by the metal can be calculated using the formula:

Qmetal =mmetal ×cmetal ×∆T

where: - mmetal = 200 g (mass of the metal), - cmetal is the specific heat capacity

of the metal to be determined, - ∆T=Tf−Tiis the change in temperature of

the metal. Given Ti= 150Cand Tf= 25C:

∆T= 25C−150C=−125C

Qmetal = 200 g ×cmetal ×(−125C)

Step 2: Next, calculate the heat absorbed or released by the water. The

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

Qwater =mwater ×cwater ×∆T

where: - mwater = 300 g (mass of the water), - cwater = 4.18 J/g

C (specific heat

capacity of water), - ∆T=Tf−Tiis the change in temperature of the water.

Given Ti= 20Cand Tf= 25C:

∆T= 25C−20C= 5C

Qwater = 300 g ×4.18 J/g

C×5C

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

the heat lost by the metal must equal the heat gained by the water:

Qmetal =−Qwater

200 g ×cmetal ×(−125C) = −300 g ×4.18 J/g

C×5C

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

cmetal =−300 g ×4.18 J/g

C×5C

200 g ×(−125C)

Question 25

Question

A 50.0 g iron bar at 120.0

C is placed in 200.0 g of water at 20.0

C. If the final

temperature of the system is 25.0

C, calculate the specific heat capacity of iron.

Solution

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

can be calculated using the formula:

Q=mc∆T

where: - Qis the heat gained, - mis the mass of the water, - cis the specific

heat capacity of water (4.18 J/g

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

water.

Plugging in the values:

Q= (200.0 g)(4.18 J/g

C)(25.0−20.0)

Q= (200.0 g)(4.18 J/g

C)(5.0)

Q= 4180 J

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

be calculated using the formula:

Q=mc∆T

where: - Qis the heat lost, - mis the mass of the iron (50.0 g), - cis the specific

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

Plugging in the values:

4180 J = (50.0 g)c(120.0−25.0)

4180 J = (50.0 g)c(95.0)

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

c=4180 J

(50.0 g)(95.0

c=4180 J

4750.0 g

c≈0.88 J/g

Therefore, the specific heat capacity of iron is approximately 0.88 J/g

Question 26

Question

A 50 g piece of copper is heated to 100

C and then placed in a calorimeter

containing 200 g of water at 20

C. If the final temperature of the system is

C, determine the specific heat capacity of the calorimeter assuming no heat

is lost to the surroundings. The specific heat capacity of copper is 0.386 J/g

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

Solution

Step 1: Calculate the heat lost by the copper piece to cool down from 100

C to

C. The formula 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.

Using this formula for copper: Qcopper = (50 g)(0.386 J/g

C)(25C−100C)

Qcopper =−1025 J

Step 2: Calculate the heat gained by the water to warm up from 20

C to

C. Using the same formula but for water: Qwater = (200 g)(4.18 J/g

C)(25C−

20C)Qwater = 418 J

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

is equal to the heat gained by the water. Therefore, Qcopper =Qwater −1025 J =

418 J −1025 J + 418 J = 0 −1025 J + 418 J = 0

Step 4: Calculate the heat absorbed by the calorimeter. Since the calorime-

ter’s specific heat capacity (ccalorimeter) is unknown, we denote it with ccalorimeter.

Using the same formula for the calorimeter: Qcalorimeter = (50 g+200 g)ccalorimeter(25C−

20C)−1025 J = 250ccalorimeter5C−1025 J = 1250ccalorimeter ccalorimeter =−1025 J

1250 J/g

ccalorimeter ≈ −0.82 J/g

Therefore, the specific heat capacity of the calorimeter is approximately

−0.82 J/g

Question 27

Question

A 50.0 g piece of aluminum, initially at 100.0

C, is dropped into a calorimeter

containing water. The water has a mass of 200.0 g and is at 20.0

C. If the

final temperature of the system is 25.0

C, calculate the specific heat capacity

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

Solution

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

water.

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

Qlost =m·c·∆T

where: - m= 50.0 g (mass of aluminum) - c=? (specific heat capacity of

aluminum) - ∆T=Tf−Ti= 25.0

C−100.0

C = −75.0

Substitute in the values:

Qlost = 50.0 g ·c·(−75.0

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

Qgain =m·cwater ·∆T

where: - m= 200.0 g (mass of water) - cwater = 4.18 J/g

C (specific heat

capacity of water) - ∆T= 25.0

C−20.0

C=5.0

Substitute in the values:

Qgain = 200.0 g ·4.18 J/g

C·5.0

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

According to the principle of conservation of energy, the heat lost by the

aluminum is equal to the heat gained by the water.

Qlost =Qgain

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

Equating the two equations and solving for c:

50.0 g ·c·(−75.0

C) = 200.0 g ·4.18 J/g

C·5.0

c=200.0 g ·4.18 J/g

C·5.0

50.0 g ·(−75.0

c≈4180 J

−3750 g ≈ −1.114 J/g

So, the specific heat capacity of aluminum is approximately 1.114 J/g

C .

Question 28

Question

A 50 gram ice cube at -10

C is placed in a calorimeter containing 200 grams

of water at 25

C. If the final temperature of the system is 10

C, calculate the

specific heat capacity of the calorimeter.

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

C and the latent heat

of fusion of ice is 334 J/g)

Solution

Step 1: Calculate the energy needed to heat the ice to 0

Q1=m1c1∆T

= (50 g)(2.09 J/g

C)(0

C−(−10

C))

= 1045 J

Step 2: Calculate the energy needed to melt the ice at 0

Q2=m1Lf

= (50 g)(334 J/g)

= 16700 J

Step 3: Calculate the energy needed to heat the water from 25

C to 0

Q3=m2c2∆T

= (200 g)(4.18 J/g

C)(0

C−25

=−20900 J

Step 4: Calculate the energy needed to heat the water from 0

C to 10

Q4=m2c2∆T

= (200 g)(4.18 J/g

C)(10

C−0

= 8360 J

Step 5: Since energy is conserved, the total energy gained by the system is

equal to the total energy lost by the system:

Qgained =Qlost

Q1+Q2+Q3+Q4= 0

1045 J + 16700 J + (−20900 J) + 8360 J = 0

1205 J = 0 J

Step 6: Since there is no heat exchange with the surroundings, the total

energy is absorbed by the system. The energy absorbed by the calorimeter is:

Qcalorimeter =mcalorimeterccalorimeter∆T

=Qgained

ccalorimeter =Qgained

mcalorimeter∆T

=1205 J

200 g ×20

= 0.3025 J/g

Therefore, the specific heat capacity of the calorimeter is 0.3025 J/g

Question 29

Question

A 50 g block of copper at 200

C is dropped into an insulated vessel containing

200 g of water at 20

C. Assuming no heat is lost to the surroundings and that

the specific heat capacity of copper is 0.385 J/g

C, the specific heat capacity of

water is 4.18 J/g

C, and neglecting the heat capacity of the vessel, what will be

the final temperature of the system?

Solution

Step 1: First, we can calculate the heat lost by the copper block as it cools

down to the final temperature of the system using the formula:

qcopper =mcopper ·ccopper ·∆T

where mcopper is the mass of the copper block, ccopper is the specific heat

capacity of copper, and ∆Tis the temperature change of the copper block.

Plugging in the values, we have:

qcopper = 50 g ·0.385 J/g

C·(Tf−200C)

Step 2: Next, let’s calculate the heat gained by the water as it warms up to

the final temperature of the system using the formula:

qwater =mwater ·cwater ·∆T

where mwater is the mass of the water, cwater is the specific heat capacity of

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

Plugging in the values, we have:

qwater = 200 g ·4.18 J/g

C·(Tf−20C)

Step 3: Since no heat is lost in the system, we can set qcopper =qwater and

solve for the final temperature, Tf.

50 g ·0.385 J/g

C·(Tf−200C) = 200 g ·4.18 J/g

C·(Tf−20C)

Solving for Tf, we find:

50 ·0.385 ·(Tf−200) = 200 ·4.18 ·(Tf−20)

19.25 ·Tf−7700 = 836 ·Tf−16720

816 ·Tf= 9020

Tf=9020

816 = 11.029

Therefore, the final temperature of the system will be 11.029

Question 30

Question

A 100 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

mixture? (Specific heat capacities: water = 4.18 J/g

C, copper = 0.387 J/g

Solution

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

water can be calculated using the formula:

qwater =mwatercwater∆T

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

C (specific heat

capacity of water) - ∆T=Tfinal −Tinitial =Tfinal −20 (change in temperature

of water)

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

can be calculated using the formula:

qcopper =mcopperccopper∆T

where: - mcopper = 100 g (mass of copper) - ccopper = 0.387 J/g

C (specific heat

capacity of copper) - ∆T= 150 −Tfinal (change in temperature of copper)

Step 3: Since heat is conserved in this isolated system, the heat lost by the

copper must equal the heat gained by the water:

qcopper =qwater

mcopperccopper∆T=mwatercwater∆T

Step 4: Substitute the expressions for qcopper and qwater into the equation

from step 3 and solve for Tfinal.

Step 5: After solving the equation, you will find the final temperature of the

mixture to be the point at which the heat lost by the copper equals the heat

gained by the water.

Question 31

Question

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

calorimetry. The student places a 0.2 kg metal block, initially at a temperature

of 100◦C, into 0.5 kg of water at 20◦C contained in a calorimeter. The final

temperature of the system is 30◦C. If the specific heat capacity of water is 4186

J/kg·K, calculate the specific heat capacity of the metal block.

Solution

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

can be calculated using the formula

Q=mc∆T

where: - mis the mass of the water (0.5 kg), - cis the specific heat capacity of

water (4186 J/kg·K), - ∆Tis the change in temperature of the water (30◦C -

20◦C = 10◦C).

Substitute the given values into the formula:

Q= (0.5 kg)(4186 J/kg ·K)(10 K)

Q= 20930 J

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

block can be calculated using the same formula as above. Since the system is

isolated and no heat is lost to the surroundings, the heat lost by the metal block

is equal to the heat gained by the water. Therefore, the heat lost by the metal

block is also 20930 J.

Step 3: Use the equation for heat transfer: Since the heat lost by the metal

block is equal to the heat gained by the water, we can write:

Qmetal =mc∆T

where: - mis the mass of the metal block (0.2 kg), - cis the specific heat capacity

of the metal block (which we want to find), - ∆Tis the change in temperature

of the metal block (final temperature - initial temperature).

Since the temperature of the metal decreases from 100◦C to 30◦C, we have

∆T= 70 ◦C. Substitute the known values of m, ∆T, and Qinto the equation:

20930 J = (0.2 kg)c(70 K)

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

c=20930 J

(0.2 kg)(70 K)

c≈299 J/kg ·K

Therefore, the specific heat capacity of the metal block is approximately 299

J/kg·K.

Question 32

Question

A piece of unknown metal weighing 95 g is heated to 99

C and then placed

in a calorimeter containing 120 g of water at 22

C. The final temperature of

the system is 25

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

the specific heat capacity of the metal. The specific heat capacity of water is

4.184 J/g

Solution

Step 1: Calculate the heat absorbed by the metal.

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

qmetal =mmetal ×cmetal ×∆Tmetal

where: - mmetal = 95 g (mass of the metal), - cmetal is the specific heat

capacity of the metal (to be found), - ∆Tmetal =Tfinal −Tinitial = 25C−99C=

−74C(temperature change of the metal).

Plugging in the values gives:

qmetal = 95 g ×cmetal ×(−74C)

Step 2: Calculate the heat released by the water.

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

qwater =mwater ×cwater ×∆Twater

where: - mwater = 120 g (mass of the water), - cwater = 4.184 J/g

C (spe-

cific heat capacity of water), - ∆Twater =Tfinal −Tinitial = 25C−22C= 3C

(temperature change of the water).

Plugging in the values gives:

qwater = 120 g ×4.184 J/g

C×3C

Step 3: Set up the heat exchange equation.

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

equal to the heat released by the water. Thus,

qmetal =qwater

Step 4: Solve for cmetal.

Equating qmetal and qwater, we have:

95 g ×cmetal ×(−74C) = 120 g ×4.184 J/g

C×3C

Now, solve for cmetal.

Question 33

Question

A piece of gold of mass 50 g at 80

C is placed in 200 g of water at 20

C. Assuming

no heat is lost to the surroundings, what is the final temperature of the system?

The specific heat capacity of gold is 0.13 J/g·

C and that of water is 4.18 J/g·

Solution

Step 1: Calculate the heat lost by the gold piece and the heat gained by the

water.

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

Qgold =mgold ·cgold ·∆T

where: - mgold = 50 g (mass of gold), - cgold = 0.13 J/g ·

C (specific heat

capacity of gold), - ∆T=Tfinal −Tinitial =Tf−Ti=Tf−80 (temperature

change of gold).

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

Qwater =mwater ·cwater ·∆T

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

C (specific heat

capacity of water), - ∆T=Tfinal −Tinitial =Tf−Ti=Tf−20 (temperature

change of water).

Step 2: Set up the heat lost by the gold piece equal to the heat gained by

the water and solve for the final temperature.

Qgold =Qwater

mgold ·cgold ·(Tf−80) = mwater ·cwater ·(Tf−20)

Substitute the given values:

50 ·0.13 ·(Tf−80) = 200 ·4.18 ·(Tf−20)

Now, solve this equation to find the final temperature Tf.

Question 34

Question

A 50 g piece of iron at 80

C is placed in 200 g of water at 20

C. Assuming all

the heat is transferred to the water and that no heat is lost to the surroundings,

what is the final temperature of the system? (Specific heat capacity of iron =

0.449 J/g ·

C, specific heat capacity of water = 4.18 J/g ·

Solution

Step 1: Calculate the heat lost by the iron: The formula for heat transfer is

given by Q=mc∆T, where: - Qis the heat transfer - mis the mass of the

material - cis the specific heat capacity of the material - ∆Tis the change in

temperature

Given that the iron piece is cooling down: Qiron =−mciron∆Tiron

Substitute the values given: Qiron =−(50 g)(0.449 J/g ·

C)(Tf−80

Step 2: Calculate the heat gained by the water: Since all the heat lost by

the iron is gained by the water: Qwater =mcwater∆Twater

Substitute the values given: Qwater = (200 g)(4.18 J/g ·

C)(Tf−20

Step 3: Set the heat lost equal to the heat gained: −(50)(0.449)(Tf−80) =

(200)(4.18)(Tf−20)

Step 4: Solve for the final temperature Tf:−22.45Tf+359.2 = 836Tf−16720

858.45Tf= 17079.2

Tf≈19.9

Therefore, the final temperature of the system is approximately 19.9

Question 35

Question

A sample of metal with an initial temperature of 100

C is placed in a calorimeter

containing 200 g of water at 20

C. The final temperature of the system is 25

If the specific heat capacity of the metal is 0.5 J/g

C, calculate the mass of the

metal sample.

Solution

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

by the metal can be calculated using the formula:

Qmetal =mc∆T

where: m = mass of the metal (g) c = specific heat capacity of the metal (J/g

∆T= change in temperature of the metal

Given: c = 0.5 J/g

C, initial temperature = 100

C, final temperature = 25

∆T= 25C−100C=−75C

Substitute the values into the formula:

Qmetal =m·0.5·(−75)

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

be calculated using the formula:

Qwater =mc∆T

where: m = mass of the water (g) = 200 g c = specific heat capacity of water

(J/g

C) = 4.18 J/g

C ∆T= 25C−20C= 5C

Substitute the values into the formula:

Qwater = 200 ·4.18 ·5

Step 3: Since the total heat gained by the metal is equal to the total heat

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

Qmetal =Qwater

m·0.5·(−75) = 200 ·4.18 ·5

Step 4: Solve for the mass of the metal, m.

m=200 ·4.18 ·5

0.5·(−75)

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 2

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