CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 3 | CHEM 107 - Essentials of General and Organic Chemistry

CHEM 107 - ESSENTIALS OF

GENERAL AND ORGANIC

CHEMISTRY - Calorimetry

Question Bank - Set 3

Liberty University

Question 1

Question

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

a calorimeter. The student places the metal sample, initially at a temperature

of 100◦C, into a calorimeter containing 200 g of water at 20◦C. After thermal

equilibrium is reached, the final temperature of the system is 25◦C. If the metal

sample has a mass of 50 g, what is the specific heat capacity of the metal?

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: - m= 200 g is the mass of the water, - cis the specific heat capacity of

water (4.18 J/g◦C), - ∆T=Tfinal −Tinitial = 25 −20 = 5 ◦C.

Substitute these values into the formula to find Qwater:

Qwater = 200 ×4.18 ×5 = 4180J.

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

be calculated using the formula:

Qmetal =mc∆T,

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

of the metal (to be determined), - ∆T=Tfinal −Tinitial = 25 −100 = −75 ◦C

(negative because the metal is losing heat).

Substitute these values into the formula to find Qmetal.

Qmetal = 50 ×c×(−75) = −3750cJ.

Step 3: Since energy is conserved, the heat gained by the water is equal to

the heat lost by the metal.

Qwater =Qmetal

4180 = −3750c

c=4180

−3750 =−1.115 J/g◦C

Therefore, the specific heat capacity of the metal is 1.115 J/g◦C.

Question 2

Question

A 100 g piece of aluminum at 100

C is dropped 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, calculate the specific heat capacity of the

calorimeter (assume it is the same as water, 4.18 J/g

C).

Solution

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

pacity of aluminum is 0.903 J/g

C. The formula for heat transfer is: q=mcT ,

where - qis the heat transferred, - mis the mass of the substance, - cis the

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

The heat lost by the aluminum piece is given by:

qAl =mAlcAlTAl

Plugging in the values:

qAl = (100 g)(0.903 J/g

C)(100 −25)C

qAl = 64725 J

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

formula for heat transfer is: q=mcT , where now mis the total mass of water

and the calorimeter.

The specific heat capacity of water is 4.18 J/g

C. The total mass of water

and calorimeter is 200 g+mcal, where mcal is the mass of the calorimeter (water

equivalent).

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

qw+cal = (200 g + mcal)(4.18 J/g

C)(25 −20)C

qw+cal = (200 g + mcal)(20.9 J)

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

must be equal to the heat gained by the water and calorimeter. Therefore,

64725 J = (200 g + mcal)(20.9 J)

Step 4: Solve for mcal to determine the mass of the calorimeter.

mcal =64725 J

20.9 J −200 g

mcal = 3085.65 g −200 g = 2885.65 g

Step 5: Calculate the specific heat capacity of the calorimeter. Since the

specific heat capacity of the calorimeter is assumed to be the same as water

(4.18 J/g

C), the specific heat capacity of the calorimeter is:

ccal = 4.18 J/g

Question 3

Question

A 50 g piece of iron at 100◦C is placed in 100 g of water at 20◦C. If the final

temperature of the system is 25◦C, what is the specific heat capacity of the

iron?

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

= 0.449 J/g◦C)

Solution

Let the specific heat capacity of iron be denoted by cFe.

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

water. The heat lost by the iron is equal to the heat gained by the water:

mFecFe∆TFe =−mwcw∆Tw

where: mFe = 50 g (mass of iron), cw= 4.18 J/g◦C (specific heat capacity of

water), cFe =? (specific heat capacity of iron), ∆TFe = (25 −100) ◦C, ∆Tw=

(25 −20) ◦C.

Substitute the given values into the equation:

50 ×cFe ×(25 −100) = −100 ×4.18 ×(25 −20)

Step 2: Solve for cFe.

50 ×cFe ×(−75) = −100 ×4.18 ×5

cFe =−100 ×4.18 ×5

50 ×(−75)

cFe = 0.558 J/g◦C

Therefore, the specific heat capacity of iron is 0.558 J/g◦C.

Question 4

Question

A 50 g piece of aluminum at 150◦C is placed in a container of water at 25◦C.

If the final temperature of the system is 30◦C, calculate the mass of water in

the container. (Specific heat capacity of aluminum = 0.90 J/g◦C, specific heat

capacity of water = 4.18 J/g◦C)

Solution

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

2: Set the two heat quantities equal and then solve for the mass of water.

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

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

qAl =mc∆T

where: m= mass of aluminum = 50 g, c= specific heat capacity of aluminum

= 0.90 J/g◦C, ∆T= change in temperature of aluminum = (30 −150)◦C =

−120◦C.

Substitute the values into the formula:

qAl = 50 g ×0.90 J/g◦C× −120◦C

qAl =−5400 J

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

qH2O =mc∆T

where: m= mass of water, c= specific heat capacity of water = 4.18 J/g◦C,

∆T= change in temperature of water = (30 −25)◦C=5◦C.

Step 2: Set the two heat quantities equal and then solve for the mass of

water.

Since energy is conserved:

qAl =qH2O

−5400 J = m×4.18 J/g◦C×5◦C

Solving for m:

−5400 J = 20.9m

m=−5400 J

20.9 J/g ≈ −258.3 g

Since mass cannot be negative, the mass of water in the container is approx-

imately 258.3 g .

Question 5

Question

A student is performing a calorimetry experiment by mixing 100 g of water at

C with 50 g of water at 60

C. Assuming no heat is lost to the surroundings,

calculate the final temperature of the system.

Solution

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

using the formula:

q=mc∆T

For the hot water:

qhot =mhotcwater(Tfinal −Thot)

qhot = 50 g ×4.18 J/g

C×(Tfinal −60C)

For the cold water:

qcold =mcoldcwater(Tfinal −Tcold)

qcold = 100 g ×4.18 J/g

C×(Tfinal −20C)

Since energy is conserved in the system when the two waters combine:

qhot =−qcold

Step 2: Substitute the expressions for qhot and qcold into the conservation of

energy equation.

50 ·4.18 ·(Tfinal −60) = −100 ·4.18 ·(Tfinal −20)

Step 3: Solve the equation for the final temperature, Tfinal.

50 ·4.18 ·Tfinal −50 ·4.18 ·60 = −100 ·4.18 ·Tfinal + 100 ·4.18 ·20

184.6·Tfinal −12516 = −418 ·Tfinal + 8360

602 ·Tfinal = 20876

Tfinal =20876

602

Tfinal ≈34.7C

Therefore, the final temperature of the system is approximately 34.7

Question 6

Question

A 100 g piece of copper at 100◦C is placed in 200 g of water at 20◦C in a

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

what will be the final temperature of the system? The specific heat capacity of

copper is 0.385 J/g◦C and that of water is 4.18 J/g◦C.

Solution

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

copper can be calculated using the formula q=mc∆T, where mis the mass of

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

temperature. Given: mCu = 100 g, cCu = 0.385 J/g◦C, Tinitial = 100◦C, and

Tfinal =Teq (final equilibrium temperature).

The heat lost by the copper is:

qCu =mCu ·cCu ·(Tfinal −Tinitial)

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

water can also be calculated using the formula q=mc∆T, with the mass and

specific heat capacity of water. Given: mH2O = 200 g, cH2O = 4.18 J/g◦C,

Tinitial = 20◦C, and Tfinal =Teq.

The heat gained by the water is:

qH2O =mH2O ·cH2O ·(Tfinal −Tinitial)

Step 3: Since the total heat lost by the copper equals the total heat gained

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

qCu =qH2O

mCu ·cCu ·(Tfinal −Tinitial) = mH2O ·cH2O ·(Tfinal −Tinitial)

Step 4: Solve for the final temperature Tfinal: Substitute the given values

into the equation and solve for Tfinal:

100 ·0.385 ·(Tfinal −100) = 200 ·4.18 ·(Tfinal −20)

Solving for Tfinal will give us the final equilibrium temperature of the system.

Question 7

Question

A piece of copper of mass 150 g at 200

C is placed in a calorimeter containing

500 g of water at 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.

Solution

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

formula of heat transfer:

Q=mc∆T

where: - Qis the heat transfer - mis the mass - cis the specific heat capacity

- ∆Tis the change in temperature

For copper:

Qcopper =mc∆T

Qcopper = (0.15 kg)(387 J/kg ·

C)(200 −25)

Qcopper = 6007.5 J

For water:

Qwater =mc∆T

Qwater = (0.5 kg)(4186 J/kg ·

C)(25 −20)

Qwater = 10465 J

Step 2: Since no heat is lost to the surroundings, the heat lost by the copper

is equal to the heat gained by the water:

Qcopper =Qwater

6007.5 = 10465

ccopper =Qwater

mcopper∆Tcopper

ccopper =10465

0.15 kg ·175

ccopper =10465

26.25

ccopper ≈398.86 J/kg ·

Therefore, the specific heat capacity of copper is approximately 398.86 J/kg

Question 8

Question

A 50 g aluminum block at 100

C is dropped into 200 g of water at 20

C. The final

temperature of the mixture is 25

C. Assuming no heat is lost to the surroundings

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

C, calculate the specific heat

capacity of aluminum.

Solution

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

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

specific heat capacity of aluminum. We have:

maCa∆Ta=mwCw∆Tw

Where: ma= 0.05 kg (mass of aluminum), Cw= 4186 J/kg

C (specific heat

capacity of water), mw= 0.2 kg (mass of water), ∆Ta=Tf−Ta,i = 25 −100 =

−75

C, and ∆Tw=Tf−Tw,i = 25 −20 = 5

Step 2: Substitute the known values into the equation and solve for Ca.

0.05Ca(−75) = 0.2×4186 ×5

−3.75Ca= 4186

Ca=4186

−3.75

Ca≈ −1116.27 J/kg

Therefore, the specific heat capacity of aluminum is approximately 1116.27

J/kg

Question 9

Question

A 50 g block of copper at 80

C is placed in 200 g of water at 20

C in an insulated

container. The final temperature of the mixture is 25

C. Assuming no heat is

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

specific heat capacity of water is 4.18 J/g

C and the specific heat capacity of

copper is 0.385 J/g

Solution

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

using the formula:

qcopper =−qwater

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

qcopper =mcopper ×ccopper ×∆T

where: - mcopper = mass of copper = 50 g - ccopper = specific heat capacity

of copper = 0.385 J/g

C-∆T= change in temperature of copper = final

temperature - initial temperature - Initial temperature of copper = 80

C - Final

temperature of the mixture = 25

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

qcopper = 50 g ×0.385 J/g

C×(25 −80)

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

qwater =mwater ×cwater ×∆T

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

water = 4.18 J/g

C-∆T= change in temperature of water = final temperature

- initial temperature - Initial temperature of water = 20

C - Final temperature

of the mixture = 25

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

qwater = 200 g ×4.18 J/g

C×(25 −20)

Step 6: Use the fact that qcopper =−qwater to solve for the specific heat

capacity of copper:

mcopper ×ccopper ×(Tf−Ti) = −mwater ×cwater ×(Tf−Ti)

Step 7: Plug in the values and solve for ccopper:

50 g ×ccopper ×(25 −80) = −200 g ×4.18 J/g

C×(25 −20)

Question 10

Question

A 50.0 g piece of copper at 150.0

C is placed in 100.0 g of water at 20.0

Assuming no heat is lost to the surroundings, what is the final equilibrium

temperature of the system?

Given: Specific heat capacity of copper = 0.385 J/g

C Specific heat capacity

of water = 4.184 J/g

Solution

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

Q=mc∆T, where Qis the heat transfer, mis the mass, cis the specific

heat capacity, and ∆Tis the change in temperature. For copper: Qcopper =

(50.0 g)(0.385 J/g

C)(Tfinal −150.0)

Step 2: For water: Qwater = (100.0 g)(4.184 J/g

C)(Tfinal −20.0)

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

no heat loss to the surroundings), we have: Qcopper =Qwater

Step 4: Set the two equations equal to each other: (50.0)(0.385)(Tfinal −

150.0) = (100.0)(4.184)(Tfinal −20.0)

Step 5: Simplify and solve for Tfinal: 19.25(Tfinal−150.0) = 418.4(Tfinal−20.0)

Step 6: Expand and solve for Tfinal: 19.25Tfinal −2887.5 = 418.4Tfinal −8368

Step 7: Rearrange the equation: 418.4Tfinal −19.25Tfinal = 8368 −2887.5

Step 8: Solve for Tfinal: 399.15Tfinal = 5480.5

Step 9: Calculate the final equilibrium temperature: Tfinal =5480.5

399.15 ≈13.72C

Therefore, the final equilibrium temperature of the system is approximately

13.72

Question 11

Question

A 50 g piece of copper metal is heated to 100◦C and then dropped into 200

g of water at 20◦C in a calorimeter. If the final temperature of the system

is 25◦C, calculate the specific heat capacity of the copper metal. The specific

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

calorimeter is negligible.

Solution

Step 1: Calculate the heat transfer from the copper piece to the water using the

formula:

Qcopper =−Qwater

Where Qcopper is the heat lost by the copper piece and Qwater is the heat

gained by the water. The formula for heat transfer is:

Q=mc∆T

For the copper piece, the heat lost is:

Qcopper =mcopperccopper∆T

Substitute the values mcopper = 50 g, cwater = 0.387 J/(g·◦C) (specific heat

capacity of copper), ∆T= 100 −25 = 75:

Qcopper = 50 ×ccopper ×75

For the water, the heat gained is:

Qwater =mwatercwater∆T

Substitute the values mwater = 200 g, cwater = 4.18 J/(g·◦C) (specific heat

capacity of water), ∆T= 25 −20 = 5:

Qwater = 200 ×4.18 ×5

Set Qcopper equal to −Qwater and solve for ccopper.

50 ×ccopper ×75 = −200 ×4.18 ×5

ccopper =−200 ×4.18 ×5

50 ×75

ccopper =−0.557 J/(g ·◦C)

So, the specific heat capacity of copper is 0.557 J/(g·◦C).

Question 12

Question

A piece of copper weighing 200 g at a temperature of 100◦Cis placed in a

calorimeter containing 500 g of water at 20◦C. The final temperature of the

system is 25◦C. Assuming no heat loss to the surroundings, what is the specific

heat capacity of copper? (Specific heat capacities: copper = 0.385 J/g◦C, water

= 4.18 J/g◦C)

Solution

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

water can be calculated using the formula:

Q=mc∆T

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

water (4.18 J/g◦C), - ∆Tis the temperature change of the water (25◦C−20◦C=

5◦C).

Substitute these values into the formula:

Q= (500 g)(4.18 J/g◦C)(5◦C) = 10450 J

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

is equal to the heat absorbed by the water (by the principle of conservation of

energy), so Qcopper =−Qwater.

Substitute the value of Qwater into the formula:

Qcopper =−10450 J

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

copper can be calculated using the formula:

Qcopper =mc∆T

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

copper (0.385 J/g◦C), - ∆Tis the temperature change of the copper (25◦C−

100◦C=−75◦C).

Substitute the known values into the formula and solve for c:

−10450 = (200 g)(0.385 J/g◦C)(−75◦C)

−10450 = −5782.5 Jg×(−75◦C)

−10450 = 433875 J

c=−10450

433875

c≈ −0.024 J/g◦C

Therefore, the specific heat capacity of copper is approximately 0.024 J/g◦C.

Question 13

Question

A 50 g piece of copper 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, determine the specific heat capacity of copper. (Specific

heat capacity of water = 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 =mwater ·cwater ·∆Twater

where: - mwater = 200 g = 0.2 kg (mass of water), - cwater = 4186 J/kg·K

(specific heat capacity of water), and - ∆Twater =Tf−Ti= 22 −20 = 2 K

(change in temperature of water). Therefore,

Qwater = 0.2 kg ×4186 J/kg ·K×2 K = 1674.4 J

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

can be calculated using the formula:

Qcopper =−mcopper ·ccopper ·∆Tcopper

where: - mcopper = 50 g = 0.05 kg (mass of copper), - ccopper (specific heat

capacity of copper), and - ∆Tcopper =Tf−Ti= 22 −100 = −78 K (change in

temperature of copper). We need to find ccopper.

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

surroundings, the total heat lost by the copper must equal the total heat gained

by the water:

Qcopper =Qwater

−0.05 kg ×ccopper × −78 K = 1674.4 J

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

0.05 kg ×ccopper ×78 K = 1674.4 J

ccopper =1674.4 J

0.05 kg ×78 K = 429.69 J/kg ·K

Therefore, the specific heat capacity of copper is approximately 429.69 J/kg·K.

Question 14

Question

A piece of metal weighing 150 g at a temperature of 180

C is placed in 300 g of

water at 20

C. The final temperature of the system is 30

C. Assuming no heat

loss to the surroundings, calculate the specific heat capacity of the metal. The

specific heat capacity of water is 4.18 J/g

Solution

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

qwater =m·c·∆T

where: - m= 300 g (mass of water) - c= 4.18 J/g

C (specific heat capacity

of water) - ∆T= 30 −20 = 10

C (change in temperature)

Calculating:

qwater = 300 g ×4.18 J/g

C×10

qwater = 12540 J

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

qmetal =m·cm·∆T

where: - m= 150 g (mass of metal) - cmis the specific heat capacity of the

metal (unknown) - ∆T= 30 −180 = −150

C (change in temperature)

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

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

qwater =−qmetal

Substitute the known values and solve for the specific heat capacity of the

metal:

12540 J = 150 g ×cm× −150

cm=12540 J

150 g × −150

cm=−0.56 J/g

Therefore, the specific heat capacity of the metal is −0.56 J/g

Question 15

Question

A 200 g aluminum block at 100◦C is dropped into 400 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, calculate the specific heat capacity of aluminum. The

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

Solution

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

can be calculated using the formula:

Q=mc∆T

where: - m= 400 g is the mass of water, - c= 4.18 J/(g·◦C) is the specific heat

capacity of water, - ∆T= 25 −20 = 5 ◦C is the change in temperature.

Substitute the values into the formula:

Qwater = 400 g ×4.18 J/(g·◦C) ×5◦C = 400 ×4.18 ×5 = 8360 J

Step 2: Find the heat lost by the aluminum block: The heat lost by the

aluminum block will be equal to the heat absorbed by the water, since no heat

is lost to the surroundings. Therefore,

Qaluminum =−Qwater =−8360 J

Step 3: Find the specific heat capacity of aluminum: The heat lost by the

aluminum block can be calculated using the formula:

Q=mc∆T

where: - m= 200 g is the mass of aluminum, - cis the specific heat capacity

of aluminum (to be determined), - ∆T= 25 −100 = −75 ◦C is the change in

temperature.

Substitute the values into the formula and solve for c:

−8360 J = 200 g ×c× −75 ◦C

8360 = 15000c

c=8360

15000 = 0.557 J/(g·◦C)

Therefore, the specific heat capacity of aluminum is 0.557 J/(g·◦C).

Question 16

Question

A piece of aluminum (specific heat, c= 0.897 J/g

C) at 100

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? (specific heat of water, c= 4.18 J/g

Solution

Step 1: First, calculate the heat gained or lost by the aluminum as it cools down

to the final temperature. Given: - Mass of aluminum, mAl = 200 g - Initial

temperature of aluminum, TAl,i = 100

C - Final temperature of the system,

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

qAl =−mAlcAl(TAl,i −Tfinal)

qAl =−200 ×0.897 ×(100 −Tfinal)

Step 2: Next, calculate the heat gained or lost by the water as it warms up to

the final temperature. Given: - Mass of water, mw= 200 g - Initial temperature

of water, Tw,i = 20

C - Final temperature of the system, Tfinal The heat gained

by the water can be calculated using the formula:

qw=mwcw(Tfinal −Tw,i)

qw= 200 ×4.18 ×(Tfinal −20)

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

minum is equal to the heat gained by the water. Set qAl =qw.

−200 ×0.897 ×(100 −Tfinal) = 200 ×4.18 ×(Tfinal −20)

Step 4: Solve the equation for Tfinal to find the final temperature of the

system. This involves algebraic manipulation and solving for Tfinal in

Question 17

Question

A 50 g piece of iron 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, calculate the specific heat capacity of iron. (Specific heat capacity

of water = 4.18 J/g

C).

Solution

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

lost by the iron is equal to the heat gained by the water. Using the formula

Q=mc∆T, where - Qis the heat energy absorbed or released, - mis the mass

of the substance, - cis the specific heat capacity of the substance, - ∆Tis the

change in temperature.

Let’s denote the specific heat capacity of iron as cFe. The heat lost by the

iron is QFe =mFe ·cFe ·∆TFe, where - mFe = 50 g, - ∆TFe =Tfinal −Tinitial =

25 −80 = −55

The heat gained by the water is Qwater =mwater ·cwater ·∆Twater, where -

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

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

Since the heat lost by the iron is equal to the heat gained by the water, we

have: QFe =Qwater

Step 2: Set up the equation and solve for cFe.mFe ·cFe ·∆TFe =mwater ·

cwater ·∆Twater

Substitute the given values: 50 ·cFe ·(−55) = 200 ·4.18 ·5

Solving for cFe:cFe =200·4.18·5

50·55 cFe = 0.381 J/g

Therefore, the specific heat capacity of iron is 0.381 J/g

Question 18

Question

A 50.0 g piece of iron at 150.0

C is placed in a calorimeter containing 200.0 g of

water at 20.0

C. The final temperature of the iron and water mixture is 25.0

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

of the iron?

Solution

Step 1: First, calculate the heat lost by the iron and the heat gained by the

water using the formula:

Q=mc∆T

where Qis the heat energy, mis the mass, cis the specific heat capacity,

and ∆Tis the change in temperature.

For the iron:

Qiron =mironciron∆Tiron

Qiron = (0.050 kg)(ciron)(25 −150)

Qiron = (0.050 kg)(ciron)(−125)

For the water:

Qwater =mwatercwater∆Twater

Qwater = (0.200 kg)(4190 J/kg ·

C)(25 −20)

Qwater = (0.200 kg)(4190 J/kg ·

C)(5)

Step 2: Since the system is isolated:

Qiron =−Qwater

(0.050 kg)(ciron)(−125) = (0.200 kg)(4190 J/kg ·

C)(5)

Step 3: Solve for ciron:

ciron =(0.200 kg)(4190 J/kg ·

C)(5)

0.050 kg(−125)

ciron =((0.200)(4190)(5))

(0.050)(−125)

ciron =4190

−25 =−167.6 J/kg ·

Therefore, the specific heat capacity of iron is −167.6 J/kg·

C. The negative

sign indicates that heat is lost by the iron.

Question 19

Question

A 50.0 g piece of iron at 95.0

C is placed in a calorimeter containing 100.0 g of

water at 20.0

C. The final temperature of the system is 25.0

C. Assuming no

heat is transferred to the surroundings, calculate the specific heat capacity of

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

Solution

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

can be calculated using the formula:

Qlost =m·c·∆T

Where: - m= 50.0 g (mass of iron) - cis the specific heat capacity of iron

(unknown) - ∆T= 95.0−25.0 = 70.0

C (change in temperature of the iron)

Plugging in the values, we get:

Qlost = 50.0 g ×c×70.0

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

water can be calculated using the formula:

Qgain =m·c·∆T

Where: - m= 100.0 g (mass of water) - c= 4.18 J/g◦C (specific heat capacity

of water) - ∆T= 25.0−20.0=5.0

C (change in temperature of the water)

Plugging in the values, we get:

Qgain = 100.0 g ×4.18 J/g◦C×5.0

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

According to the law of conservation of energy, the heat lost by the iron must

be equal to the heat gained by the water. Therefore:

Qlost =Qgain

Step 4: Solve for the specific heat capacity of iron. Equating Qlost and Qgain,

we have:

50.0 g ·c·70.0

C = 100.0 g ·4.18 J/g◦C·5.0

c=100.0 g ·4.18 J/g◦C·5.0

50.0 g ·70.0

Question 20

Question

A piece of metal of mass 0.2 kg is heated to 100◦C and then dropped into 0.5

kg of water at 20◦C. The final temperature of the mixture is 30◦C. Given that

the specific heat capacity of the metal is 500 J/kg◦C, calculate the specific heat

capacity of the water.

Solution

Step 1: First, we need to calculate the heat absorbed by the metal as it cools

down to the final temperature. The heat absorbed by the metal is given by the

formula:

Qmetal =mcmetal∆T

where: - m= 0.2 kg is the mass of the metal, - cmetal = 500 J/kg◦C is the

specific heat capacity of the metal, and - ∆T=Tf−Ti= 30◦C−100◦C = −70◦C

is the change in temperature of the metal.

Therefore,

Qmetal = 0.2×500 ×(−70)

Qmetal =−7000 J

Step 2: Next, calculate the heat lost by the metal is gained by the water

in order for the system to reach thermal equilibrium. Let’s assume the specific

heat capacity of water is cwater (in J/kg◦C). The heat lost by the metal is equal

to the heat gained by the water:

mcmetal∆T=mcwater∆T

Substitute the known values:

(−7000) = (0.5)(cwater)(10)

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

cwater =−7000

0.5×10

cwater =−1400 J/kg◦C

Therefore, the specific heat capacity of the water is 1400 J/kg◦C.

Question 21

Question

A piece of metal with mass 200 g is heated to 100

C and then transferred

to a calorimeter of mass 150 g containing 400 g of water at 20

C. The final

temperature of the system is 25

C. If the specific heat capacity of the metal is

0.2 J/g

C, calculate the specific heat capacity of the calorimeter. Assume no

heat is lost to the surroundings.

Solution

Step 1: Calculate the heat transfer from the metal to the water in the calorime-

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

calorimeter:

mmetal ·cmetal ·∆T= (mwater +mcalorimeter)·cwater ·∆T

Substitute the given values:

200 g ·0.2 J/g

C·(100 −25)C= (400 + 150) g ·cwater ·(25 −20)C

3500 J = 550 g ·cwater ·5C

cwater =3500

2750 ≈1.27 J/g

Step 2: Calculate the specific heat capacity of the calorimeter. To find

the specific heat capacity of the calorimeter, we can rewrite the heat transfer

equation in terms of the calorimeter:

mcalorimeter ·ccalorimeter ·∆T= (mwater +mmetal)·cwater ·∆T

Substitute the known values:

150 g ·ccalorimeter ·(25 −20)C= 550 g ·1.27 J/g

C·5C

30 g ·ccalorimeter = 3183.5

ccalorimeter =3183.5

30 ≈106.12 J/g

Therefore, the specific heat capacity of the calorimeter is approximately

106.12 J/g

Question 22

Question

A student carries out an experiment to determine the specific heat capacity of

a metal using a calorimeter. The student places 200 g of the metal at an initial

temperature of 120

C into 400 g of water at an initial temperature of 20

C. The

final temperature of the mixture is 30

C. If the specific heat capacity of water

is 4.18 J/g ·

C, determine the specific heat capacity of the metal.

Solution

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

can be calculated using the equation:

Qwater =mwater ·cwater ·∆T

where: - mwater = 400 g is the mass of the water, - cwater = 4.18 J/g ·

C is the

specific heat capacity of water, and - T=Tfinal −Tinitial = 30

C−20

C = 10

is the temperature change.

Therefore:

Qwater = 400 g ×4.18 J/g ·

C×10

C = 16720 J

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

is equal in magnitude but opposite in sign to the heat gained by the water.

Therefore, the heat lost by the metal can be calculated as:

Qmetal =−Qwater =−16720 J

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

the metal can be calculated using the equation:

Qmetal =mmetal ·cmetal ·∆T

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

capacity of the metal, and - T=Tfinal −Tinitial = 30

C−120

C = −90

C is

the temperature change.

Substitute in the values:

−16720 J = 200 g ×cmetal × −90

Solve for cmetal:

cmetal =−16720 J

200 g × −90

C= 9.29 J/g ·

Therefore, the specific heat capacity of the metal is 9.29 J/g ·

Question 23

Question

A 50 g aluminum sample 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, determine the

specific heat capacity of the material of the calorimeter. Assume no heat is

lost to the surroundings. (Specific heat capacity of aluminum is 0.9 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 sample.

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

qaluminum =m·c·∆T

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

capacity of aluminum (0.9 J/g◦C) - ∆Tis the change in temperature of the

aluminum sample (80◦C−25◦C = 55◦C)

Substitute the values into the formula:

qaluminum = 50 g ·0.9 J/g◦C·55◦C

qaluminum = 2475 J

Step 2: Calculate the heat gained by the water in the calorimeter.

The heat gained by the water in the calorimeter can be calculated using the

formula:

qwater =m·c·∆T

where: - mis the mass of the water in the calorimeter (200 g) - cis the specific

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

water (25◦C−20◦C=5◦C)

Substitute the values into the formula:

qwater = 200 g ·4.18 J/g◦C·5◦C

qwater = 4180 J

Step 3: Determine the specific heat capacity of the calorimeter material.

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

sample is equal to the heat gained by the water:

qaluminum =qwater

2475 J = 4180 J + m·ccalorimeter ·∆T

m·ccalorimeter = 2475 J −4180 J

m·ccalorimeter =−1705 J

Since the mass of the calorimeter material is not given, we can solve for the

specific heat capacity of the calorimeter material:

ccalorimeter =−1705 J

m·∆T

Therefore, the specific heat capacity of the material of the calorimeter is

ccalorimeter =−1705 J

m·∆T.

Question 24

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? The specific heat capacity of copper is 0.385 J/g

C and that of water

is 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 is Q=m·c·∆T, where: - Qis

the heat energy, - mis the mass of the substance, - cis the specific heat capacity

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

Given that the initial temperature of the copper piece is 200

C and the final

temperature is T(in

C), the change in temperature is ∆T= 200 −T

Therefore, the heat lost by the copper piece is:

Qcopper = 50 g ·0.385 J/g

C·(200 −T)

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

temperature. Following the same formula as above and with the same reasoning,

the heat gained by the water is:

Qwater = 100 g ·4.18 J/g

C·(T−20)

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

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

the water. Therefore,

Qcopper =Qwater

Step 4: Set the two expressions for heat equal to each other and solve for T.

50 ·0.385 ·(200 −T) = 100 ·4.18 ·(T−20)

Step 5: Solve the equation for T to find the final temperature of the system.

Question 25

Question

A 50 g gold coin at a temperature of 85

C is dropped into 200 g of water at

C. If the final temperature of the mixture is 25

C, calculate the specific heat

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

Solution

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

temperature of the mixture. The heat lost by the gold coin can be calculated

using the formula:

Qlost, gold =mgold ·cgold ·(Tf−Ti)

where: - mgold = 50 g = 0.05 kg is the mass of the gold coin, - cgold is the specific

heat capacity of gold (to be determined), - Tf= 25◦C = 25 + 273 = 298 K is

the final temperature of the mixture, - Ti= 85◦C = 85 + 273 = 358 K is the

initial temperature of the gold coin.

Substitute the given values into the formula:

Qlost, gold = 0.05 kg ·cgold ·(298 K −358 K)

Qlost, gold =−15 J ·cgold

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

temperature of the mixture. The heat gained by the water can be calculated

using the formula:

Qgained, water =mwater ·cwater ·(Tf−Ti)

where: - mwater = 200 g = 0.2 kg is the mass of the water, - cwater = 4186 J/kg·K

is the specific heat capacity of water, - Tf= 298 K is the final temperature of

the mixture, - Ti= 15◦C = 15 + 273 = 288 K is the initial temperature of the

water.

Substitute the given values into the formula:

Qgained, water = 0.2 kg ·4186 J/kg ·K·(298 K −288 K)

Qgained, water = 4186 J

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

coin must equal the heat gained by the water. Set the expressions for heat lost

and heat gained equal to each other:

−15 J ·cgold = 1846 J

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

cgold =4186 J

0.05 kg ·(358 K −298 K)

cgold =4186 J

0.05 kg ·60 K

cgold =4186 J

3 J/kg ·K

cgold = 1395.33 J/kg ·K

Therefore, the specific heat capacity of gold is approximately 1395.33 J/kg·K.

Question 26

Question

A 100 g piece of aluminum at 80

C is placed in 200 g of water at 20

C in an

insulated container. The final equilibrium temperature of the system is 30

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◦C)

Solution

Step 1: First, we need to 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×c×∆T

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

aluminum (to be calculated) - ∆Tis the change in temperature of aluminum

Given that the initial temperature of aluminum (Tinitial, Al) is 80

C and the

final temperature of the system is 30

C, the change in temperature of aluminum

is:

∆TAluminum =Tfinal −Tinitial, Al = 30C−80C=−50C

(We use a negative sign because the temperature of aluminum is decreasing)

Step 2: Now, we calculate the heat gained by the water. This can be calcu-

lated using the formula:

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 water

Given that the initial temperature of water (Tinitial, Water) is 20

C and the

final temperature of the system is 30

C, the change in temperature of water is:

∆TWater =Tfinal −Tinitial, Water = 30C−20C= 10C

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

we can assume that the heat lost by the aluminum is equal to the heat gained

by the water:

QAluminum =QWater

mAl ×cAl ×∆TAl =mWater ×cWater ×∆TWater

Substitute the given values:

100 ×cAl ×(−50) = 200 ×4.18 ×10

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

cAl =200 ×4.18 ×10

100 ×(−50)

cAl =8360

−5000

cAl =−1.672 J/g◦C

Therefore, the specific heat capacity of aluminum is −1.672 J/g◦C.

Question 27

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C. Assuming no

heat is lost to the surroundings, calculate the final temperature of the system.

The specific heat capacity of iron is 0.45 J/g

C and the specific heat capacity

of water is 4.18 J/g

Solution

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

or lost is given by: Q=mc∆T, where Qis the heat, mis the mass, cis the

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

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

C, Tinitial, water = 20

C, Tfinal, water =

x(unknown).

Qwater =mwater ×cwater ×∆Twater

Qwater = 200 ×4.18 ×(x−20)

Step 2: Calculate the heat lost by the iron. Given: miron = 50 g, ciron = 0.45

J/g

C, Tinitial, iron = 80

C, Tfinal, iron =x(unknown).

Qiron =miron ×ciron ×∆Tiron

Qiron = 50 ×0.45 ×(x−80)

Step 3: Since the total heat gained by the water should be equal to the total

heat lost by the iron (neglecting any heat loss to the surroundings), we can set

Qwater equal to Qiron and solve for the final temperature, x.

200 ×4.18 ×(x−20) = 50 ×0.45 ×(x−80)

Question 28

Question

A 50 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 30

C, what is the heat

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

Solution

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

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

QAl =mAl ·cAl ·∆TAl

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

of aluminum (0.903 J/g

C), - ∆TAl is the change in temperature of aluminum

(final temperature - initial temperature).

Substitute the given values:

QAl = 50 g ×0.903 J/g

C×(30 −80)

QAl = 50 ×0.903 ×(−50)

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

water can be calculated using the formula:

QH2O =mH2O ·cH2O ·∆TH2O

where: - mH2O is the mass of water (200 g), - cH2O is the specific heat capacity

of water (4.18 J/g

C), - ∆TH2O is the change in temperature of water.

Substitute the given values:

QH2O = 200 g ×4.18 J/g

C×(30 −20)

Step 3: Recall that in an isolated system, heat lost equals heat gained. Set

up an equation: QAl =QH2O and solve for the heat capacity of the calorimeter,

denoted as Ccal.

50 ×0.903 ×(−50) = 200 ×4.18 ×(30 −20) + Ccal ×(30 −20)

This equation can be solved to find the heat capacity of the calorimeter.

Question 29

Question

A student wants to determine the specific heat capacity of an unknown metal.

To do this, the student heats a 150 g metal sample to 100

C and then drops it

into 200 g of water initially at 20

C in a calorimeter. The final temperature of

the system is 25

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

specific heat capacity of the metal.

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

C is the specific

heat capacity of water, - ∆T= (Tf−Ti) = (25C−20C)=5Cis the change in

temperature.

Substitute the values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Qwater = 4180 J

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

Step 2: Calculate the heat lost by the metal.

The heat lost by the metal is equal to the heat gained by the water, as no

heat is lost to the surroundings (assuming an insulated calorimeter). Therefore,

the heat lost by the metal can be calculated as:

Qmetal = 4180 J

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

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

Qmetal =mc∆T

where: - m= 150 g is the mass of the metal, - cis the specific heat capacity

of the metal, - ∆T= 100C−25C= 75Cis the change in temperature.

Substitute the values into the formula and solve for c:

4180 J = (150 g)c(75C)

c=4180 J

(150 g)(75C)

c≈37.07 J/g

Therefore, the specific heat capacity of the metal is approximately 37.07

J/g

Question 30

Question

A 50 g piece of aluminum at 100

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 25

C. Assuming no heat is

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

calculate the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by aluminum. The formula for calculating heat

is:

Q=mc∆T

where: - Q is the heat absorbed or lost, - m is the mass of the material, - c is

the specific heat capacity of the material, - T is the change in temperature.

Substitute the given values:

QAl = (50 g)(cAl)(25 −100)

QAl =−50cAl

Step 2: Calculate the heat gained by water.

QH2O= (200 g)(4.18 J/g

C)(25 −20)

QH2O= 4180 J

Step 3: Since the system is isolated, the heat lost by aluminum is equal to

the heat gained by water.

−50cAl = 4180

cAl =−4180

cAl =−83.6 J/g

Therefore, the specific heat capacity of aluminum is 83.6 J/g

Question 31

Question

A 50.0 g piece of aluminum at 90.0◦C is added to 100.0 g of water at 20.0◦C

in a calorimeter. The final temperature of the mixture is 27.0◦C. Assume all

the heat lost by the aluminum is gained by the water and the calorimeter itself.

The specific heat capacity of aluminum is 0.900 J/g◦C, and the specific heat

capacity of water is 4.184 J/g◦C. What is the heat capacity (in J◦C) of the

calorimeter?

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: m= mass of water in grams c= specific heat capacity of water in J/g◦C

∆T= change in temperature of the water

Given that m= 100.0 g, c= 4.184 J/g◦C, and ∆T= 27.0−20.0=7.0◦C,

we can calculate:

qwater = 100.0 g ×4.184 J/g◦C×7.0◦C

qwater = 2928.8 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= mass of aluminum in grams c= specific heat capacity of aluminum

in J/g◦C ∆T= change in temperature of the aluminum

Given that m= 50.0 g, c= 0.900 J/g◦C, and ∆T= 27.0−90.0 = −63.0◦C

(negative because the aluminum is losing heat), we can calculate:

qaluminum = 50.0 g ×0.900 J/g◦C× −63.0◦C

qaluminum =−2835.0 J

Step 3: Calculate the heat capacity of the calorimeter. Since the heat lost

by the aluminum is equal to the heat gained by the water and the calorimeter,

we can write:

qaluminum =qwater +qcalorimeter

Substitute the calculated values for qaluminum and qwater:

−2835.0 = 2928.8 + qcalorimeter

qcalorimeter =−2835.0−2928.8

qcalorimeter =−5763.8 J

Therefore, the heat capacity of the calorimeter is 5763.8 J◦C .

Question 32

Question

A 100 g piece of copper at 200

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 22.5

C, find the specific

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

specific heat capacity of water is 4.18 J/g

Solution

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

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

(mcopper)(ccopper)(∆Tcopper) = (mwater)(cwater)(∆Twater)

where: - mcopper is the mass of copper (100 g) - ccopper is the specific heat

capacity of copper (to be found) - ∆Tcopper is the change in temperature of

copper (final temp - initial temp) - mwater is the mass of water (200 g) - cwater

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

C) - ∆Twater is the change in

temperature of water (final temp - initial temp)

Step 2: Calculate the change in temperature for copper and water.

∆Tcopper = final temp −initial temp = 22.5−200 = −177.5C

∆Twater = final temp −initial temp = 22.5−20 = 2.5C

Step 3: Substitute the known values into the heat equation and solve for

ccopper.

(100)(ccopper)(−177.5) = (200)(4.18)(2.5)

−17750ccopper = 2095

ccopper =2095

−17750 =−0.118 J/g

Therefore, the specific heat capacity of copper is −0.118 J/g

C .

Question 33

Question

A 50 g block of copper at 95

C is dropped into 200 g of water at 25

C in a

calorimeter. If the final temperature of the copper-water mixture is 30

C, what

is the specific heat capacity of the copper block? The specific heat capacity of

water is 4.18 J/g

Solution

Step 1: Calculate the heat absorbed by the water. Given: Mass of water,

mw= 200 g Initial temperature of water, Twi = 25

C Final temperature of

water and copper mixture, Tf= 30

C Specific heat capacity of water, cw= 4.18

J/g

Using the formula for heat absorbed or released:

Q=mc∆T

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

the change in temperature.

The heat absorbed by the water can be calculated as:

Qw=mwcw∆T

Qw= (200 g)(4.18 J/g

C)(30 −25)

Qw= 4180 J

Step 2: Calculate the heat lost by the copper block. Given: Mass of copper

block, mc= 50 g Initial temperature of copper block, Tci = 95

C Final temper-

ature of copper-water mixture, Tf= 30

C Specific heat capacity of copper, cc

(to be determined)

The heat lost by the copper block can also be calculated using the formula

for heat:

Qc=mc∆T

Step 3: Set up the conservation of energy equation. According to the prin-

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

heat gained by the water:

Qc=Qw

Step 4: Solve for the specific heat capacity of the copper block. Substitute

the known values and solve for cc:

mccc∆T=mwcw∆T

(50 g)cc(30 −95)

C = 4180 J

cc(−65) = 4180

cc=4180

−65

cc≈ −64.31 J/g

Therefore, the specific heat capacity of the copper block is approximately

−64.31 J/g

C. Note the negative sign indicates that the direction of heat transfer

for the copper block is opposite to that of the water.

Question 34

Question

A 50 g piece of metal at 150

C is placed into a calorimeter containing 200 g of

water at 20

C. The final temperature of the system is 25

C. If the specific heat

capacity of the metal is 0.5 J/g

C and the specific heat capacity of water is 4.18

J/g

C, calculate the initial temperature of the metal before it was added to the

calorimeter.

Solution

Step 1: Calculate the heat gained by the water Let the initial temperature of

the metal be Tm

C. The heat lost by the metal = heat gained by the water

mm·cm·(Tf−Tm) = mw·cw·(Tf−Tw)

50 ·0.5·(25 −Tm) = 200 ·4.18 ·(25 −20)

25(25 −Tm) = 836

625 −25Tm= 836

25Tm=−211

Tm=−8.44

Therefore, the initial temperature of the metal was -8.44

C before it was

added to the calorimeter.

Question 35

Question

A 50 g cube of iron at 80

C is placed in 200 g of water at 20

C in 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 = 450 J/kg·

K, specific heat capacity of water = 4186 J/kg ·K)

Solution

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

can be expressed as Q=mc∆T, where mis the mass of the water, cis the

specific heat capacity of water, and ∆Tis the change in temperature. Given:

m= 200 g = 0.2 kg, c= 4186 J/kg ·K, and ∆T=Tf−20 (since the water is

originally at 20

C). Therefore, Q= (0.2)(4186)(Tf−20).

Step 2: Calculate the heat lost by the iron: Similarly, the heat lost by the

iron can be calculated as Q=mc∆T, where mis the mass of the iron, cis

the specific heat capacity of iron, and ∆Tis the change in temperature. Given:

Question 4

Question

A 50 g piece of aluminum at 150◦C is placed in a container of water at 25◦C.

If the final temperature of the system is 30◦C, calculate the mass of water in

the container. (Specific heat capacity of aluminum = 0.90 J/g◦C, specific heat

capacity of water = 4.18 J/g◦C)

Solution

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

2: Set the two heat quantities equal and then solve for the mass of water.

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

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

qAl =mc∆T

where: m= mass of aluminum = 50 g, c= specific heat capacity of aluminum

= 0.90 J/g◦C, ∆T= change in temperature of aluminum = (30 −150)◦C =

−120◦C.

Substitute the values into the formula:

qAl = 50 g ×0.90 J/g◦C× −120◦C

qAl =−5400 J

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

qH2O =mc∆T

where: m= mass of water, c= specific heat capacity of water = 4.18 J/g◦C,

∆T= change in temperature of water = (30 −25)◦C=5◦C.

Step 2: Set the two heat quantities equal and then solve for the mass of

water.

Since energy is conserved:

qAl =qH2O

−5400 J = m×4.18 J/g◦C×5◦C

Solving for m:

−5400 J = 20.9m

m=−5400 J

20.9 J/g ≈ −258.3 g

Since mass cannot be negative, the mass of water in the container is approx-

imately 258.3 g .

Question 5

Question

A student is performing a calorimetry experiment by mixing 100 g of water at

C with 50 g of water at 60

C. Assuming no heat is lost to the surroundings,

calculate the final temperature of the system.

Solution

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

using the formula:

q=mc∆T

For the hot water:

qhot =mhotcwater(Tfinal −Thot)

qhot = 50 g ×4.18 J/g

C×(Tfinal −60C)

For the cold water:

qcold =mcoldcwater(Tfinal −Tcold)

qcold = 100 g ×4.18 J/g

C×(Tfinal −20C)

Since energy is conserved in the system when the two waters combine:

qhot =−qcold

Step 2: Substitute the expressions for qhot and qcold into the conservation of

energy equation.

50 ·4.18 ·(Tfinal −60) = −100 ·4.18 ·(Tfinal −20)

Step 3: Solve the equation for the final temperature, Tfinal.

50 ·4.18 ·Tfinal −50 ·4.18 ·60 = −100 ·4.18 ·Tfinal + 100 ·4.18 ·20

184.6·Tfinal −12516 = −418 ·Tfinal + 8360

602 ·Tfinal = 20876

Tfinal =20876

602

Tfinal ≈34.7C

Therefore, the final temperature of the system is approximately 34.7

Question 6

Question

A 100 g piece of copper at 100◦C is placed in 200 g of water at 20◦C in a

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

what will be the final temperature of the system? The specific heat capacity of

copper is 0.385 J/g◦C and that of water is 4.18 J/g◦C.

Solution

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

copper can be calculated using the formula q=mc∆T, where mis the mass of

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

temperature. Given: mCu = 100 g, cCu = 0.385 J/g◦C, Tinitial = 100◦C, and

Tfinal =Teq (final equilibrium temperature).

The heat lost by the copper is:

qCu =mCu ·cCu ·(Tfinal −Tinitial)

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

water can also be calculated using the formula q=mc∆T, with the mass and

specific heat capacity of water. Given: mH2O = 200 g, cH2O = 4.18 J/g◦C,

Tinitial = 20◦C, and Tfinal =Teq.

The heat gained by the water is:

qH2O =mH2O ·cH2O ·(Tfinal −Tinitial)

Step 3: Since the total heat lost by the copper equals the total heat gained

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

qCu =qH2O

mCu ·cCu ·(Tfinal −Tinitial) = mH2O ·cH2O ·(Tfinal −Tinitial)

Step 4: Solve for the final temperature Tfinal: Substitute the given values

into the equation and solve for Tfinal:

100 ·0.385 ·(Tfinal −100) = 200 ·4.18 ·(Tfinal −20)

Solving for Tfinal will give us the final equilibrium temperature of the system.

Question 7

Question

A piece of copper of mass 150 g at 200

C is placed in a calorimeter containing

500 g of water at 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.

Solution

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

formula of heat transfer:

Q=mc∆T

where: - Qis the heat transfer - mis the mass - cis the specific heat capacity

- ∆Tis the change in temperature

For copper:

Qcopper =mc∆T

Qcopper = (0.15 kg)(387 J/kg ·

C)(200 −25)

Qcopper = 6007.5 J

For water:

Qwater =mc∆T

Qwater = (0.5 kg)(4186 J/kg ·

C)(25 −20)

Qwater = 10465 J

Step 2: Since no heat is lost to the surroundings, the heat lost by the copper

is equal to the heat gained by the water:

Qcopper =Qwater

6007.5 = 10465

ccopper =Qwater

mcopper∆Tcopper

ccopper =10465

0.15 kg ·175

ccopper =10465

26.25

ccopper ≈398.86 J/kg ·

Therefore, the specific heat capacity of copper is approximately 398.86 J/kg

Question 8

Question

A 50 g aluminum block at 100

C is dropped into 200 g of water at 20

C. The final

temperature of the mixture is 25

C. Assuming no heat is lost to the surroundings

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

C, calculate the specific heat

capacity of aluminum.

Solution

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

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

specific heat capacity of aluminum. We have:

maCa∆Ta=mwCw∆Tw

Where: ma= 0.05 kg (mass of aluminum), Cw= 4186 J/kg

C (specific heat

capacity of water), mw= 0.2 kg (mass of water), ∆Ta=Tf−Ta,i = 25 −100 =

−75

C, and ∆Tw=Tf−Tw,i = 25 −20 = 5

Step 2: Substitute the known values into the equation and solve for Ca.

0.05Ca(−75) = 0.2×4186 ×5

−3.75Ca= 4186

Ca=4186

−3.75

Ca≈ −1116.27 J/kg

Therefore, the specific heat capacity of aluminum is approximately 1116.27

J/kg

Question 9

Question

A 50 g block of copper at 80

C is placed in 200 g of water at 20

C in an insulated

container. The final temperature of the mixture is 25

C. Assuming no heat is

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

specific heat capacity of water is 4.18 J/g

C and the specific heat capacity of

copper is 0.385 J/g

Solution

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

using the formula:

qcopper =−qwater

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

qcopper =mcopper ×ccopper ×∆T

where: - mcopper = mass of copper = 50 g - ccopper = specific heat capacity

of copper = 0.385 J/g

C-∆T= change in temperature of copper = final

temperature - initial temperature - Initial temperature of copper = 80

C - Final

temperature of the mixture = 25

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

qcopper = 50 g ×0.385 J/g

C×(25 −80)

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

qwater =mwater ×cwater ×∆T

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

water = 4.18 J/g

C-∆T= change in temperature of water = final temperature

- initial temperature - Initial temperature of water = 20

C - Final temperature

of the mixture = 25

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

qwater = 200 g ×4.18 J/g

C×(25 −20)

Step 6: Use the fact that qcopper =−qwater to solve for the specific heat

capacity of copper:

mcopper ×ccopper ×(Tf−Ti) = −mwater ×cwater ×(Tf−Ti)

Step 7: Plug in the values and solve for ccopper:

50 g ×ccopper ×(25 −80) = −200 g ×4.18 J/g

C×(25 −20)

Question 10

Question

A 50.0 g piece of copper at 150.0

C is placed in 100.0 g of water at 20.0

Assuming no heat is lost to the surroundings, what is the final equilibrium

temperature of the system?

Given: Specific heat capacity of copper = 0.385 J/g

C Specific heat capacity

of water = 4.184 J/g

Solution

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

Q=mc∆T, where Qis the heat transfer, mis the mass, cis the specific

heat capacity, and ∆Tis the change in temperature. For copper: Qcopper =

(50.0 g)(0.385 J/g

C)(Tfinal −150.0)

Step 2: For water: Qwater = (100.0 g)(4.184 J/g

C)(Tfinal −20.0)

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

no heat loss to the surroundings), we have: Qcopper =Qwater

Step 4: Set the two equations equal to each other: (50.0)(0.385)(Tfinal −

150.0) = (100.0)(4.184)(Tfinal −20.0)

Step 5: Simplify and solve for Tfinal: 19.25(Tfinal−150.0) = 418.4(Tfinal−20.0)

Step 6: Expand and solve for Tfinal: 19.25Tfinal −2887.5 = 418.4Tfinal −8368

Step 7: Rearrange the equation: 418.4Tfinal −19.25Tfinal = 8368 −2887.5

Step 8: Solve for Tfinal: 399.15Tfinal = 5480.5

Step 9: Calculate the final equilibrium temperature: Tfinal =5480.5

399.15 ≈13.72C

Therefore, the final equilibrium temperature of the system is approximately

13.72

Question 11

Question

A 50 g piece of copper metal is heated to 100◦C and then dropped into 200

g of water at 20◦C in a calorimeter. If the final temperature of the system

is 25◦C, calculate the specific heat capacity of the copper metal. The specific

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

calorimeter is negligible.

Solution

Step 1: Calculate the heat transfer from the copper piece to the water using the

formula:

Qcopper =−Qwater

Where Qcopper is the heat lost by the copper piece and Qwater is the heat

gained by the water. The formula for heat transfer is:

Q=mc∆T

For the copper piece, the heat lost is:

Qcopper =mcopperccopper∆T

Substitute the values mcopper = 50 g, cwater = 0.387 J/(g·◦C) (specific heat

capacity of copper), ∆T= 100 −25 = 75:

Qcopper = 50 ×ccopper ×75

For the water, the heat gained is:

Qwater =mwatercwater∆T

Substitute the values mwater = 200 g, cwater = 4.18 J/(g·◦C) (specific heat

capacity of water), ∆T= 25 −20 = 5:

Qwater = 200 ×4.18 ×5

Set Qcopper equal to −Qwater and solve for ccopper.

50 ×ccopper ×75 = −200 ×4.18 ×5

ccopper =−200 ×4.18 ×5

50 ×75

ccopper =−0.557 J/(g ·◦C)

So, the specific heat capacity of copper is 0.557 J/(g·◦C).

Question 12

Question

A piece of copper weighing 200 g at a temperature of 100◦Cis placed in a

calorimeter containing 500 g of water at 20◦C. The final temperature of the

system is 25◦C. Assuming no heat loss to the surroundings, what is the specific

heat capacity of copper? (Specific heat capacities: copper = 0.385 J/g◦C, water

= 4.18 J/g◦C)

Solution

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

water can be calculated using the formula:

Q=mc∆T

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

water (4.18 J/g◦C), - ∆Tis the temperature change of the water (25◦C−20◦C=

5◦C).

Substitute these values into the formula:

Q= (500 g)(4.18 J/g◦C)(5◦C) = 10450 J

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

is equal to the heat absorbed by the water (by the principle of conservation of

energy), so Qcopper =−Qwater.

Substitute the value of Qwater into the formula:

Qcopper =−10450 J

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

copper can be calculated using the formula:

Qcopper =mc∆T

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

copper (0.385 J/g◦C), - ∆Tis the temperature change of the copper (25◦C−

100◦C=−75◦C).

Substitute the known values into the formula and solve for c:

−10450 = (200 g)(0.385 J/g◦C)(−75◦C)

−10450 = −5782.5 Jg×(−75◦C)

−10450 = 433875 J

c=−10450

433875

c≈ −0.024 J/g◦C

Therefore, the specific heat capacity of copper is approximately 0.024 J/g◦C.

Question 13

Question

A 50 g piece of copper 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, determine the specific heat capacity of copper. (Specific

heat capacity of water = 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 =mwater ·cwater ·∆Twater

where: - mwater = 200 g = 0.2 kg (mass of water), - cwater = 4186 J/kg·K

(specific heat capacity of water), and - ∆Twater =Tf−Ti= 22 −20 = 2 K

(change in temperature of water). Therefore,

Qwater = 0.2 kg ×4186 J/kg ·K×2 K = 1674.4 J

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

can be calculated using the formula:

Qcopper =−mcopper ·ccopper ·∆Tcopper

where: - mcopper = 50 g = 0.05 kg (mass of copper), - ccopper (specific heat

capacity of copper), and - ∆Tcopper =Tf−Ti= 22 −100 = −78 K (change in

temperature of copper). We need to find ccopper.

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

surroundings, the total heat lost by the copper must equal the total heat gained

by the water:

Qcopper =Qwater

−0.05 kg ×ccopper × −78 K = 1674.4 J

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

0.05 kg ×ccopper ×78 K = 1674.4 J

ccopper =1674.4 J

0.05 kg ×78 K = 429.69 J/kg ·K

Therefore, the specific heat capacity of copper is approximately 429.69 J/kg·K.

Question 14

Question

A piece of metal weighing 150 g at a temperature of 180

C is placed in 300 g of

water at 20

C. The final temperature of the system is 30

C. Assuming no heat

loss to the surroundings, calculate the specific heat capacity of the metal. The

specific heat capacity of water is 4.18 J/g

Solution

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

qwater =m·c·∆T

where: - m= 300 g (mass of water) - c= 4.18 J/g

C (specific heat capacity

of water) - ∆T= 30 −20 = 10

C (change in temperature)

Calculating:

qwater = 300 g ×4.18 J/g

C×10

qwater = 12540 J

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

qmetal =m·cm·∆T

where: - m= 150 g (mass of metal) - cmis the specific heat capacity of the

metal (unknown) - ∆T= 30 −180 = −150

C (change in temperature)

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

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

qwater =−qmetal

Substitute the known values and solve for the specific heat capacity of the

metal:

12540 J = 150 g ×cm× −150

cm=12540 J

150 g × −150

cm=−0.56 J/g

Therefore, the specific heat capacity of the metal is −0.56 J/g

Question 15

Question

A 200 g aluminum block at 100◦C is dropped into 400 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, calculate the specific heat capacity of aluminum. The

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

Solution

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

can be calculated using the formula:

Q=mc∆T

where: - m= 400 g is the mass of water, - c= 4.18 J/(g·◦C) is the specific heat

capacity of water, - ∆T= 25 −20 = 5 ◦C is the change in temperature.

Substitute the values into the formula:

Qwater = 400 g ×4.18 J/(g·◦C) ×5◦C = 400 ×4.18 ×5 = 8360 J

Step 2: Find the heat lost by the aluminum block: The heat lost by the

aluminum block will be equal to the heat absorbed by the water, since no heat

is lost to the surroundings. Therefore,

Qaluminum =−Qwater =−8360 J

Step 3: Find the specific heat capacity of aluminum: The heat lost by the

aluminum block can be calculated using the formula:

Q=mc∆T

where: - m= 200 g is the mass of aluminum, - cis the specific heat capacity

of aluminum (to be determined), - ∆T= 25 −100 = −75 ◦C is the change in

temperature.

Substitute the values into the formula and solve for c:

−8360 J = 200 g ×c× −75 ◦C

8360 = 15000c

c=8360

15000 = 0.557 J/(g·◦C)

Therefore, the specific heat capacity of aluminum is 0.557 J/(g·◦C).

Question 16

Question

A piece of aluminum (specific heat, c= 0.897 J/g

C) at 100

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? (specific heat of water, c= 4.18 J/g

Solution

Step 1: First, calculate the heat gained or lost by the aluminum as it cools down

to the final temperature. Given: - Mass of aluminum, mAl = 200 g - Initial

temperature of aluminum, TAl,i = 100

C - Final temperature of the system,

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

qAl =−mAlcAl(TAl,i −Tfinal)

qAl =−200 ×0.897 ×(100 −Tfinal)

Step 2: Next, calculate the heat gained or lost by the water as it warms up to

the final temperature. Given: - Mass of water, mw= 200 g - Initial temperature

of water, Tw,i = 20

C - Final temperature of the system, Tfinal The heat gained

by the water can be calculated using the formula:

qw=mwcw(Tfinal −Tw,i)

qw= 200 ×4.18 ×(Tfinal −20)

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

minum is equal to the heat gained by the water. Set qAl =qw.

−200 ×0.897 ×(100 −Tfinal) = 200 ×4.18 ×(Tfinal −20)

Step 4: Solve the equation for Tfinal to find the final temperature of the

system. This involves algebraic manipulation and solving for Tfinal in

Question 17

Question

A 50 g piece of iron 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, calculate the specific heat capacity of iron. (Specific heat capacity

of water = 4.18 J/g

C).

Solution

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

lost by the iron is equal to the heat gained by the water. Using the formula

Q=mc∆T, where - Qis the heat energy absorbed or released, - mis the mass

of the substance, - cis the specific heat capacity of the substance, - ∆Tis the

change in temperature.

Let’s denote the specific heat capacity of iron as cFe. The heat lost by the

iron is QFe =mFe ·cFe ·∆TFe, where - mFe = 50 g, - ∆TFe =Tfinal −Tinitial =

25 −80 = −55

The heat gained by the water is Qwater =mwater ·cwater ·∆Twater, where -

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

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

Since the heat lost by the iron is equal to the heat gained by the water, we

have: QFe =Qwater

Step 2: Set up the equation and solve for cFe.mFe ·cFe ·∆TFe =mwater ·

cwater ·∆Twater

Substitute the given values: 50 ·cFe ·(−55) = 200 ·4.18 ·5

Solving for cFe:cFe =200·4.18·5

50·55 cFe = 0.381 J/g

Therefore, the specific heat capacity of iron is 0.381 J/g

Question 18

Question

A 50.0 g piece of iron at 150.0

C is placed in a calorimeter containing 200.0 g of

water at 20.0

C. The final temperature of the iron and water mixture is 25.0

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

of the iron?

Solution

Step 1: First, calculate the heat lost by the iron and the heat gained by the

water using the formula:

Q=mc∆T

where Qis the heat energy, mis the mass, cis the specific heat capacity,

and ∆Tis the change in temperature.

For the iron:

Qiron =mironciron∆Tiron

Qiron = (0.050 kg)(ciron)(25 −150)

Qiron = (0.050 kg)(ciron)(−125)

For the water:

Qwater =mwatercwater∆Twater

Qwater = (0.200 kg)(4190 J/kg ·

C)(25 −20)

Qwater = (0.200 kg)(4190 J/kg ·

C)(5)

Step 2: Since the system is isolated:

Qiron =−Qwater

(0.050 kg)(ciron)(−125) = (0.200 kg)(4190 J/kg ·

C)(5)

Step 3: Solve for ciron:

ciron =(0.200 kg)(4190 J/kg ·

C)(5)

0.050 kg(−125)

ciron =((0.200)(4190)(5))

(0.050)(−125)

ciron =4190

−25 =−167.6 J/kg ·

Therefore, the specific heat capacity of iron is −167.6 J/kg·

C. The negative

sign indicates that heat is lost by the iron.

Question 19

Question

A 50.0 g piece of iron at 95.0

C is placed in a calorimeter containing 100.0 g of

water at 20.0

C. The final temperature of the system is 25.0

C. Assuming no

heat is transferred to the surroundings, calculate the specific heat capacity of

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

Solution

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

can be calculated using the formula:

Qlost =m·c·∆T

Where: - m= 50.0 g (mass of iron) - cis the specific heat capacity of iron

(unknown) - ∆T= 95.0−25.0 = 70.0

C (change in temperature of the iron)

Plugging in the values, we get:

Qlost = 50.0 g ×c×70.0

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

water can be calculated using the formula:

Qgain =m·c·∆T

Where: - m= 100.0 g (mass of water) - c= 4.18 J/g◦C (specific heat capacity

of water) - ∆T= 25.0−20.0=5.0

C (change in temperature of the water)

Plugging in the values, we get:

Qgain = 100.0 g ×4.18 J/g◦C×5.0

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

According to the law of conservation of energy, the heat lost by the iron must

be equal to the heat gained by the water. Therefore:

Qlost =Qgain

Step 4: Solve for the specific heat capacity of iron. Equating Qlost and Qgain,

we have:

50.0 g ·c·70.0

C = 100.0 g ·4.18 J/g◦C·5.0

c=100.0 g ·4.18 J/g◦C·5.0

50.0 g ·70.0

Question 20

Question

A piece of metal of mass 0.2 kg is heated to 100◦C and then dropped into 0.5

kg of water at 20◦C. The final temperature of the mixture is 30◦C. Given that

the specific heat capacity of the metal is 500 J/kg◦C, calculate the specific heat

capacity of the water.

Solution

Step 1: First, we need to calculate the heat absorbed by the metal as it cools

down to the final temperature. The heat absorbed by the metal is given by the

formula:

Qmetal =mcmetal∆T

where: - m= 0.2 kg is the mass of the metal, - cmetal = 500 J/kg◦C is the

specific heat capacity of the metal, and - ∆T=Tf−Ti= 30◦C−100◦C = −70◦C

is the change in temperature of the metal.

Therefore,

Qmetal = 0.2×500 ×(−70)

Qmetal =−7000 J

Step 2: Next, calculate the heat lost by the metal is gained by the water

in order for the system to reach thermal equilibrium. Let’s assume the specific

heat capacity of water is cwater (in J/kg◦C). The heat lost by the metal is equal

to the heat gained by the water:

mcmetal∆T=mcwater∆T

Substitute the known values:

(−7000) = (0.5)(cwater)(10)

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

cwater =−7000

0.5×10

cwater =−1400 J/kg◦C

Therefore, the specific heat capacity of the water is 1400 J/kg◦C.

Question 21

Question

A piece of metal with mass 200 g is heated to 100

C and then transferred

to a calorimeter of mass 150 g containing 400 g of water at 20

C. The final

temperature of the system is 25

C. If the specific heat capacity of the metal is

0.2 J/g

C, calculate the specific heat capacity of the calorimeter. Assume no

heat is lost to the surroundings.

Solution

Step 1: Calculate the heat transfer from the metal to the water in the calorime-

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

calorimeter:

mmetal ·cmetal ·∆T= (mwater +mcalorimeter)·cwater ·∆T

Substitute the given values:

200 g ·0.2 J/g

C·(100 −25)C= (400 + 150) g ·cwater ·(25 −20)C

3500 J = 550 g ·cwater ·5C

cwater =3500

2750 ≈1.27 J/g

Step 2: Calculate the specific heat capacity of the calorimeter. To find

the specific heat capacity of the calorimeter, we can rewrite the heat transfer

equation in terms of the calorimeter:

mcalorimeter ·ccalorimeter ·∆T= (mwater +mmetal)·cwater ·∆T

Substitute the known values:

150 g ·ccalorimeter ·(25 −20)C= 550 g ·1.27 J/g

C·5C

30 g ·ccalorimeter = 3183.5

ccalorimeter =3183.5

30 ≈106.12 J/g

Therefore, the specific heat capacity of the calorimeter is approximately

106.12 J/g

Question 22

Question

A student carries out an experiment to determine the specific heat capacity of

a metal using a calorimeter. The student places 200 g of the metal at an initial

temperature of 120

C into 400 g of water at an initial temperature of 20

C. The

final temperature of the mixture is 30

C. If the specific heat capacity of water

is 4.18 J/g ·

C, determine the specific heat capacity of the metal.

Solution

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

can be calculated using the equation:

Qwater =mwater ·cwater ·∆T

where: - mwater = 400 g is the mass of the water, - cwater = 4.18 J/g ·

C is the

specific heat capacity of water, and - T=Tfinal −Tinitial = 30

C−20

C = 10

is the temperature change.

Therefore:

Qwater = 400 g ×4.18 J/g ·

C×10

C = 16720 J

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

is equal in magnitude but opposite in sign to the heat gained by the water.

Therefore, the heat lost by the metal can be calculated as:

Qmetal =−Qwater =−16720 J

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

the metal can be calculated using the equation:

Qmetal =mmetal ·cmetal ·∆T

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

capacity of the metal, and - T=Tfinal −Tinitial = 30

C−120

C = −90

C is

the temperature change.

Substitute in the values:

−16720 J = 200 g ×cmetal × −90

Solve for cmetal:

cmetal =−16720 J

200 g × −90

C= 9.29 J/g ·

Therefore, the specific heat capacity of the metal is 9.29 J/g ·

Question 23

Question

A 50 g aluminum sample 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, determine the

specific heat capacity of the material of the calorimeter. Assume no heat is

lost to the surroundings. (Specific heat capacity of aluminum is 0.9 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 sample.

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

qaluminum =m·c·∆T

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

capacity of aluminum (0.9 J/g◦C) - ∆Tis the change in temperature of the

aluminum sample (80◦C−25◦C = 55◦C)

Substitute the values into the formula:

qaluminum = 50 g ·0.9 J/g◦C·55◦C

qaluminum = 2475 J

Step 2: Calculate the heat gained by the water in the calorimeter.

The heat gained by the water in the calorimeter can be calculated using the

formula:

qwater =m·c·∆T

where: - mis the mass of the water in the calorimeter (200 g) - cis the specific

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

water (25◦C−20◦C=5◦C)

Substitute the values into the formula:

qwater = 200 g ·4.18 J/g◦C·5◦C

qwater = 4180 J

Step 3: Determine the specific heat capacity of the calorimeter material.

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

sample is equal to the heat gained by the water:

qaluminum =qwater

2475 J = 4180 J + m·ccalorimeter ·∆T

m·ccalorimeter = 2475 J −4180 J

m·ccalorimeter =−1705 J

Since the mass of the calorimeter material is not given, we can solve for the

specific heat capacity of the calorimeter material:

ccalorimeter =−1705 J

m·∆T

Therefore, the specific heat capacity of the material of the calorimeter is

ccalorimeter =−1705 J

m·∆T.

Question 24

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? The specific heat capacity of copper is 0.385 J/g

C and that of water

is 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 is Q=m·c·∆T, where: - Qis

the heat energy, - mis the mass of the substance, - cis the specific heat capacity

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

Given that the initial temperature of the copper piece is 200

C and the final

temperature is T(in

C), the change in temperature is ∆T= 200 −T

Therefore, the heat lost by the copper piece is:

Qcopper = 50 g ·0.385 J/g

C·(200 −T)

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

temperature. Following the same formula as above and with the same reasoning,

the heat gained by the water is:

Qwater = 100 g ·4.18 J/g

C·(T−20)

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

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

the water. Therefore,

Qcopper =Qwater

Step 4: Set the two expressions for heat equal to each other and solve for T.

50 ·0.385 ·(200 −T) = 100 ·4.18 ·(T−20)

Step 5: Solve the equation for T to find the final temperature of the system.

Question 25

Question

A 50 g gold coin at a temperature of 85

C is dropped into 200 g of water at

C. If the final temperature of the mixture is 25

C, calculate the specific heat

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

Solution

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

temperature of the mixture. The heat lost by the gold coin can be calculated

using the formula:

Qlost, gold =mgold ·cgold ·(Tf−Ti)

where: - mgold = 50 g = 0.05 kg is the mass of the gold coin, - cgold is the specific

heat capacity of gold (to be determined), - Tf= 25◦C = 25 + 273 = 298 K is

the final temperature of the mixture, - Ti= 85◦C = 85 + 273 = 358 K is the

initial temperature of the gold coin.

Substitute the given values into the formula:

Qlost, gold = 0.05 kg ·cgold ·(298 K −358 K)

Qlost, gold =−15 J ·cgold

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

temperature of the mixture. The heat gained by the water can be calculated

using the formula:

Qgained, water =mwater ·cwater ·(Tf−Ti)

where: - mwater = 200 g = 0.2 kg is the mass of the water, - cwater = 4186 J/kg·K

is the specific heat capacity of water, - Tf= 298 K is the final temperature of

the mixture, - Ti= 15◦C = 15 + 273 = 288 K is the initial temperature of the

water.

Substitute the given values into the formula:

Qgained, water = 0.2 kg ·4186 J/kg ·K·(298 K −288 K)

Qgained, water = 4186 J

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

coin must equal the heat gained by the water. Set the expressions for heat lost

and heat gained equal to each other:

−15 J ·cgold = 1846 J

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

cgold =4186 J

0.05 kg ·(358 K −298 K)

cgold =4186 J

0.05 kg ·60 K

cgold =4186 J

3 J/kg ·K

cgold = 1395.33 J/kg ·K

Therefore, the specific heat capacity of gold is approximately 1395.33 J/kg·K.

Question 26

Question

A 100 g piece of aluminum at 80

C is placed in 200 g of water at 20

C in an

insulated container. The final equilibrium temperature of the system is 30

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◦C)

Solution

Step 1: First, we need to 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×c×∆T

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

aluminum (to be calculated) - ∆Tis the change in temperature of aluminum

Given that the initial temperature of aluminum (Tinitial, Al) is 80

C and the

final temperature of the system is 30

C, the change in temperature of aluminum

is:

∆TAluminum =Tfinal −Tinitial, Al = 30C−80C=−50C

(We use a negative sign because the temperature of aluminum is decreasing)

Step 2: Now, we calculate the heat gained by the water. This can be calcu-

lated using the formula:

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 water

Given that the initial temperature of water (Tinitial, Water) is 20

C and the

final temperature of the system is 30

C, the change in temperature of water is:

∆TWater =Tfinal −Tinitial, Water = 30C−20C= 10C

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

we can assume that the heat lost by the aluminum is equal to the heat gained

by the water:

QAluminum =QWater

mAl ×cAl ×∆TAl =mWater ×cWater ×∆TWater

Substitute the given values:

100 ×cAl ×(−50) = 200 ×4.18 ×10

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

cAl =200 ×4.18 ×10

100 ×(−50)

cAl =8360

−5000

cAl =−1.672 J/g◦C

Therefore, the specific heat capacity of aluminum is −1.672 J/g◦C.

Question 27

Question

A 50 g piece of iron at 80

C is placed into 200 g of water at 20

C. Assuming no

heat is lost to the surroundings, calculate the final temperature of the system.

The specific heat capacity of iron is 0.45 J/g

C and the specific heat capacity

of water is 4.18 J/g

Solution

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

or lost is given by: Q=mc∆T, where Qis the heat, mis the mass, cis the

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

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

C, Tinitial, water = 20

C, Tfinal, water =

x(unknown).

Qwater =mwater ×cwater ×∆Twater

Qwater = 200 ×4.18 ×(x−20)

Step 2: Calculate the heat lost by the iron. Given: miron = 50 g, ciron = 0.45

J/g

C, Tinitial, iron = 80

C, Tfinal, iron =x(unknown).

Qiron =miron ×ciron ×∆Tiron

Qiron = 50 ×0.45 ×(x−80)

Step 3: Since the total heat gained by the water should be equal to the total

heat lost by the iron (neglecting any heat loss to the surroundings), we can set

Qwater equal to Qiron and solve for the final temperature, x.

200 ×4.18 ×(x−20) = 50 ×0.45 ×(x−80)

Question 28

Question

A 50 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 30

C, what is the heat

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

Solution

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

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

QAl =mAl ·cAl ·∆TAl

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

of aluminum (0.903 J/g

C), - ∆TAl is the change in temperature of aluminum

(final temperature - initial temperature).

Substitute the given values:

QAl = 50 g ×0.903 J/g

C×(30 −80)

QAl = 50 ×0.903 ×(−50)

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

water can be calculated using the formula:

QH2O =mH2O ·cH2O ·∆TH2O

where: - mH2O is the mass of water (200 g), - cH2O is the specific heat capacity

of water (4.18 J/g

C), - ∆TH2O is the change in temperature of water.

Substitute the given values:

QH2O = 200 g ×4.18 J/g

C×(30 −20)

Step 3: Recall that in an isolated system, heat lost equals heat gained. Set

up an equation: QAl =QH2O and solve for the heat capacity of the calorimeter,

denoted as Ccal.

50 ×0.903 ×(−50) = 200 ×4.18 ×(30 −20) + Ccal ×(30 −20)

This equation can be solved to find the heat capacity of the calorimeter.

Question 29

Question

A student wants to determine the specific heat capacity of an unknown metal.

To do this, the student heats a 150 g metal sample to 100

C and then drops it

into 200 g of water initially at 20

C in a calorimeter. The final temperature of

the system is 25

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

specific heat capacity of the metal.

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

C is the specific

heat capacity of water, - ∆T= (Tf−Ti) = (25C−20C)=5Cis the change in

temperature.

Substitute the values into the formula:

Qwater = (200 g)(4.18 J/g

C)(5C)

Qwater = 4180 J

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

Step 2: Calculate the heat lost by the metal.

The heat lost by the metal is equal to the heat gained by the water, as no

heat is lost to the surroundings (assuming an insulated calorimeter). Therefore,

the heat lost by the metal can be calculated as:

Qmetal = 4180 J

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

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

Qmetal =mc∆T

where: - m= 150 g is the mass of the metal, - cis the specific heat capacity

of the metal, - ∆T= 100C−25C= 75Cis the change in temperature.

Substitute the values into the formula and solve for c:

4180 J = (150 g)c(75C)

c=4180 J

(150 g)(75C)

c≈37.07 J/g

Therefore, the specific heat capacity of the metal is approximately 37.07

J/g

Question 30

Question

A 50 g piece of aluminum at 100

C is dropped into 200 g of water at 20

C in a

calorimeter. The final temperature of the system is 25

C. Assuming no heat is

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

calculate the specific heat capacity of aluminum.

Solution

Step 1: Calculate the heat lost by aluminum. The formula for calculating heat

is:

Q=mc∆T

where: - Q is the heat absorbed or lost, - m is the mass of the material, - c is

the specific heat capacity of the material, - T is the change in temperature.

Substitute the given values:

QAl = (50 g)(cAl)(25 −100)

QAl =−50cAl

Step 2: Calculate the heat gained by water.

QH2O= (200 g)(4.18 J/g

C)(25 −20)

QH2O= 4180 J

Step 3: Since the system is isolated, the heat lost by aluminum is equal to

the heat gained by water.

−50cAl = 4180

cAl =−4180

cAl =−83.6 J/g

Therefore, the specific heat capacity of aluminum is 83.6 J/g

Question 31

Question

A 50.0 g piece of aluminum at 90.0◦C is added to 100.0 g of water at 20.0◦C

in a calorimeter. The final temperature of the mixture is 27.0◦C. Assume all

the heat lost by the aluminum is gained by the water and the calorimeter itself.

The specific heat capacity of aluminum is 0.900 J/g◦C, and the specific heat

capacity of water is 4.184 J/g◦C. What is the heat capacity (in J◦C) of the

calorimeter?

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: m= mass of water in grams c= specific heat capacity of water in J/g◦C

∆T= change in temperature of the water

Given that m= 100.0 g, c= 4.184 J/g◦C, and ∆T= 27.0−20.0=7.0◦C,

we can calculate:

qwater = 100.0 g ×4.184 J/g◦C×7.0◦C

qwater = 2928.8 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= mass of aluminum in grams c= specific heat capacity of aluminum

in J/g◦C ∆T= change in temperature of the aluminum

Given that m= 50.0 g, c= 0.900 J/g◦C, and ∆T= 27.0−90.0 = −63.0◦C

(negative because the aluminum is losing heat), we can calculate:

qaluminum = 50.0 g ×0.900 J/g◦C× −63.0◦C

qaluminum =−2835.0 J

Step 3: Calculate the heat capacity of the calorimeter. Since the heat lost

by the aluminum is equal to the heat gained by the water and the calorimeter,

we can write:

qaluminum =qwater +qcalorimeter

Substitute the calculated values for qaluminum and qwater:

−2835.0 = 2928.8 + qcalorimeter

qcalorimeter =−2835.0−2928.8

qcalorimeter =−5763.8 J

Therefore, the heat capacity of the calorimeter is 5763.8 J◦C .

Question 32

Question

A 100 g piece of copper at 200

C is dropped into 200 g of water at 20

C in a

calorimeter. If the final temperature of the system is 22.5

C, find the specific

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

specific heat capacity of water is 4.18 J/g

Solution

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

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

(mcopper)(ccopper)(∆Tcopper) = (mwater)(cwater)(∆Twater)

where: - mcopper is the mass of copper (100 g) - ccopper is the specific heat

capacity of copper (to be found) - ∆Tcopper is the change in temperature of

copper (final temp - initial temp) - mwater is the mass of water (200 g) - cwater

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

C) - ∆Twater is the change in

temperature of water (final temp - initial temp)

Step 2: Calculate the change in temperature for copper and water.

∆Tcopper = final temp −initial temp = 22.5−200 = −177.5C

∆Twater = final temp −initial temp = 22.5−20 = 2.5C

Step 3: Substitute the known values into the heat equation and solve for

ccopper.

(100)(ccopper)(−177.5) = (200)(4.18)(2.5)

−17750ccopper = 2095

ccopper =2095

−17750 =−0.118 J/g

Therefore, the specific heat capacity of copper is −0.118 J/g

C .

Question 33

Question

A 50 g block of copper at 95

C is dropped into 200 g of water at 25

C in a

calorimeter. If the final temperature of the copper-water mixture is 30

C, what

is the specific heat capacity of the copper block? The specific heat capacity of

water is 4.18 J/g

Solution

Step 1: Calculate the heat absorbed by the water. Given: Mass of water,

mw= 200 g Initial temperature of water, Twi = 25

C Final temperature of

water and copper mixture, Tf= 30

C Specific heat capacity of water, cw= 4.18

J/g

Using the formula for heat absorbed or released:

Q=mc∆T

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

the change in temperature.

The heat absorbed by the water can be calculated as:

Qw=mwcw∆T

Qw= (200 g)(4.18 J/g

C)(30 −25)

Qw= 4180 J

Step 2: Calculate the heat lost by the copper block. Given: Mass of copper

block, mc= 50 g Initial temperature of copper block, Tci = 95

C Final temper-

ature of copper-water mixture, Tf= 30

C Specific heat capacity of copper, cc

(to be determined)

The heat lost by the copper block can also be calculated using the formula

for heat:

Qc=mc∆T

Step 3: Set up the conservation of energy equation. According to the prin-

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

heat gained by the water:

Qc=Qw

Step 4: Solve for the specific heat capacity of the copper block. Substitute

the known values and solve for cc:

mccc∆T=mwcw∆T

(50 g)cc(30 −95)

C = 4180 J

cc(−65) = 4180

cc=4180

−65

cc≈ −64.31 J/g

Therefore, the specific heat capacity of the copper block is approximately

−64.31 J/g

C. Note the negative sign indicates that the direction of heat transfer

for the copper block is opposite to that of the water.

Question 34

Question

A 50 g piece of metal at 150

C is placed into a calorimeter containing 200 g of

water at 20

C. The final temperature of the system is 25

C. If the specific heat

capacity of the metal is 0.5 J/g

C and the specific heat capacity of water is 4.18

J/g

C, calculate the initial temperature of the metal before it was added to the

calorimeter.

Solution

Step 1: Calculate the heat gained by the water Let the initial temperature of

the metal be Tm

C. The heat lost by the metal = heat gained by the water

mm·cm·(Tf−Tm) = mw·cw·(Tf−Tw)

50 ·0.5·(25 −Tm) = 200 ·4.18 ·(25 −20)

25(25 −Tm) = 836

625 −25Tm= 836

25Tm=−211

Tm=−8.44

Therefore, the initial temperature of the metal was -8.44

C before it was

added to the calorimeter.

Question 35

Question

A 50 g cube of iron at 80

C is placed in 200 g of water at 20

C in 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 = 450 J/kg·

K, specific heat capacity of water = 4186 J/kg ·K)

Solution

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

can be expressed as Q=mc∆T, where mis the mass of the water, cis the

specific heat capacity of water, and ∆Tis the change in temperature. Given:

m= 200 g = 0.2 kg, c= 4186 J/kg ·K, and ∆T=Tf−20 (since the water is

originally at 20

C). Therefore, Q= (0.2)(4186)(Tf−20).

Step 2: Calculate the heat lost by the iron: Similarly, the heat lost by the

iron can be calculated as Q=mc∆T, where mis the mass of the iron, cis

the specific heat capacity of iron, and ∆Tis the change in temperature. Given:

m= 50 g = 0.05 kg, c= 450 J/kg ·K, and ∆T= 80 −Tf(since the iron is

originally at 80

C). Therefore, Q= (0.05)(450)(80 −Tf).

Step 3: Set up the energy balance equation: Since the system is perfectly

insulated, the heat gained by the water must be equal to the heat lost by the

iron:

(0.2)(4186)(Tf−20) = (0.05)(450)(80 −Tf)

Step 4: Solve for the final temperature (Tf): Expanding and simplifying the

equation from step 3:

837.2Tf−8372 = 2025 −22.5Tf

837.2Tf+ 22.5Tf= 8372 + 2025

859.7Tf= 10397

Tf≈12.1C

Therefore, the final temperature of the system will be approximately 12.1C.

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Calorimetry Question Bank Set 3

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