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

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CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Calorimetry
Question Bank - Set 4
Liberty University
Question 1
Question
A 50 g aluminum block at 80
°
C is placed in 200 g of water at 20
°
C in a calorime-
ter. If the final temperature of the system is 25
°
C, what is the specific heat
capacity of the aluminum block?
Assume the specific heat capacity of water is 4186 J/kg
°
C and neglect any
heat loss to the surroundings.
Solution
Step 1: Calculate the heat lost by the aluminum block. The heat lost by the
aluminum block is equal to the heat gained by the water and the calorimeter.
The formula for heat transfer is Q=mcT, where Qis the heat transfer, mis
the mass, cis the specific heat capacity, and Tis the change in temperature.
Given: maluminum = 50 g = 0.05 kg cwater = 4186 J/kg
°
CTinitial, aluminum =
80C Tfinal, system = 25C
The heat lost by the aluminum block can be calculated as: Qaluminum =
maluminumcaluminum(Tfinal, system Tinitial, aluminum)
Step 2: Calculate the heat gained by the water and calorimeter. The formula
for heat transfer here is Q=mcT.
Given: mwater = 200 g = 0.2 kg cwater = 4186 J/kg
°
CTinitial, water = 20C
Tfinal, system = 25C
The heat gained by the water can be calculated as: Qwater =mwatercwater(Tfinal, system
Tinitial, water)
Step 3: Equate the heat lost by the aluminum block to the heat gained by
the water and calorimeter. Qaluminum =Qwater
Step 4: Solve for the specific heat capacity of aluminum. Since Q=mcT,
we have: maluminumcaluminum(Tfinal, systemTinitial, aluminum) = mwatercwater(Tfinal, system
Tinitial, water)
Substitute the given values and solve for caluminum to obtain the specific heat
capacity of aluminum.
Question 2
Question
A 50 g piece of aluminum at 100C is placed in 200 g of water at 20C. If the
final temperature of the system is 25C, calculate the specific heat capacity of
aluminum. Assume no heat is lost to the surroundings and that the specific
heat capacity of water is 4.18 J/g·K.
Solution
Step 1: Calculate the heat lost by the aluminum and gained by the water
using the formula q=mcT, where qis the heat transferred, mis the mass,
cis the specific heat capacity, and Tis the change in temperature. In this
case, the heat lost by the aluminum is equal to the heat gained by the water.
For the aluminum: qaluminum = (50 g)(cAl)(75C) For the water: qwater =
(200 g)(4.18 J/g·K)(5C)
Step 2: Set the heat lost by the aluminum equal to the heat gained by the
water and solve for the specific heat capacity of aluminum.
(50 g)(cAl)(75C) = (200 g)(4.18 J/g·K)(5C)
cAl =(200 g)(4.18 J/g·K)(5C)
(50 g)(75C)
Step 3: Calculate the specific heat capacity of aluminum.
cAl =(200)(4.18)(5)
(50)(75)
cAl =4180
375
cAl = 11.15 J/g·K
Therefore, the specific heat capacity of aluminum is 11.15 J/g·K.
Question 3
Question
A 50 g piece of iron at 80
°
C is placed into 200 g of water at 20
°
C in an insulated
container. If the final temperature of the system is 25
°
C, determine the specific
heat capacity of iron.
2
Solution
Let the specific heat capacity of water be cw= 4.18 J/g
°
C.
Step 1: Calculate the heat absorbed by the water when the iron piece is
added. The heat absorbed by the water can be calculated using the formula:
Q=mcT
where: - mis the mass of water, - cis the specific heat capacity of water, - T
is the change in temperature of the water.
Substitute the given values:
Q= 200 ×4.18 ×(25 20)
Q= 200 ×4.18 ×5
Q= 4180 J
Step 2: Calculate the amount of heat lost by the iron. Since the system is
insulated, the total amount of heat lost by the iron is equal to the amount of
heat gained by the water. Therefore, the heat lost by the iron is also 4180 J.
Step 3: Calculate the specific heat capacity of iron. The heat lost by the
iron can be calculated using the formula:
Q=mcT
where: - mis the mass of iron, - cis the specific heat capacity of iron, - Tis
the change in temperature of the iron.
Substitute the given values and the heat lost:
4180 = 50c(25 80)
4180 = 50c×55
4180 = 2750c
c=4180
2750
c 1.52 J/g
°
C
Therefore, the specific heat capacity of iron is approximately 1.52 J/g
°
C.
The negative sign indicates that the value obtained is not physically meaningful.
This discrepancy may be due to experimental errors or assumptions made during
the calculation.
Question 4
Question
A 50 g sample of aluminum at 80
°
C is dropped into 200 g of water at 20
°
C in
a calorimeter. The specific heat capacity of aluminum is 0.897 J/g
°
C and the
specific heat capacity of water is 4.18 J/g
°
C. Assuming no heat is lost to the
surroundings, what is the final temperature of the system?
3
Solution
Step 1: Calculate the heat absorbed by the aluminum: The equation for heat
transfer is given by q=mcT, where: - mis the mass of the substance, - cis
the specific heat capacity of the substance, - Tis the change in temperature.
Substitute the values for aluminum: qAl = (50 g)(0.897 J/g
°
C)(Tf80
°
C)
Step 2: Calculate the heat lost by the water: qwater =(200 g)(4.18 J/g
°
C)(Tf
20
°
C)
Step 3: Set up the energy conservation equation: The heat lost by the water
is equal to the heat gained by the aluminum: (200 g)(4.18 J/g
°
C)(Tf20
°
C) =
(50 g)(0.897 J/g
°
C)(Tf80
°
C)
Step 4: Solve for the final temperature: (836 J/
°
C)(Tf20
°
C) = (44.85 J/
°
C)(Tf
80
°
C)
836Tf+ 16720 = 44.85Tf3588
836Tf+ 44.85Tf= 16720 + 3588
Tf=20308
°
C
880.85 23.05
°
C
Therefore, the final temperature of the system is approximately 23.05
°
C.
Question 5
Question
A 50 g piece of copper at 90
°
C is placed in 100 g of water at 20
°
C. Assuming
no heat is lost to the surroundings, what will be the final temperature of the
system? (Specific heat capacity of copper = 0.385 J/g
°
C, specific heat capacity
of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper piece as it cools down to the final
temperature. The formula for calculating heat (q) is:
q=m·c·T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, - Tis the change in temperature.
Plugging in the values:
qcopper = 50 g ×0.385 J/g
°
C×(90 Tf)C
qcopper = 19.25 ×(90 Tf) J
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. Using the same formula:
qwater = 100 g ×4.18 J/g
°
C×(Tf20)C
qwater = 418 ×(Tf20) J
4
Step 3: Since the system is isolated and assuming no heat loss to the sur-
roundings, the heat lost by the copper must be equal to the heat gained by the
water. Equating qcopper and qwater:
19.25 ×(90 Tf) = 418 ×(Tf20)
Step 4: Solve the equation for the final temperature (Tf):
1732.519.25Tf= 418Tf8360
19.25Tf+ 418Tf= 8360 + 1732.5
437.25Tf= 10092.5
Tf=10092.5
437.25 23.10C
Therefore, the final temperature of the system will be approximately 23.10
°
C.
Question 6
Question
A 50 g piece of aluminum at 100
°
C is placed in 200 g of water at 20
°
C in a
calorimeter. If the final equilibrium temperature of the system is 25
°
C, cal-
culate the specific heat capacity of aluminum. Assume no heat is lost to the
surroundings.
Solution
Step 1: Calculate the heat absorbed by the water. The heat absorbed by the
water can be calculated using the formula:
Qwater =mcT
where: - m= 200 g is the mass of the water, - c= 4.18 J/g
°
C is the specific
heat capacity of water, - T= 25C20C= 5Cis the temperature change of
the water.
Substitute the values into the formula:
Qwater = (200 g)(4.18 J/g
°
C)(5C)
Qwater = 4180 J
Step 2: Calculate the heat lost by the aluminum. The heat lost by the
aluminum can be calculated using the same formula:
Qaluminum =mcT
5
where: - m= 50 g is the mass of the aluminum, - cis the specific heat
capacity of aluminum (to be determined), - T= 25C100C=75Cis the
temperature change of the aluminum. (Negative because it loses heat.)
Substitute the values into the formula:
Qaluminum = (50 g)(c)(75C)
Qaluminum =3750c
Step 3: Set up the heat balance equation. Since no heat is lost to the
surroundings, the heat lost by the aluminum is equal to the heat gained by the
water. Therefore:
Qwater =Qaluminum
4180 = 3750c
Step 4: Solve for the specific heat capacity of aluminum.
c=4180
3750
c 1.11 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 1.11
J/g
°
C.
Question 7
Question
A piece of iron with mass 0.5 kg is heated to a temperature of 100C and then
dropped into a container of water with mass 1 kg at a temperature of 20C.
The final temperature of the system is 25C. Assuming all the heat lost by the
iron is gained by the water and that there is no heat loss to the surroundings,
calculate the specific heat capacity of iron. (Specific heat capacity of water is
4186 J/kg·K)
Solution
Step 1: Calculate the heat gained by the water The heat gained by the water
can be calculated using the formula:
Qwater =mcT
where mis the mass of the water, cis the specific heat capacity of water, and T
is the change in temperature. Given: m= 1 kg c= 4186 J/kg·K T= (2520)
C = 5 K
6
Substitute the given values into the formula:
Qwater = 1 ×4186 ×5
Qwater = 20930 J
Step 2: Calculate the heat lost by the iron The heat lost by the iron can
also be calculated using the formula:
Qiron =mcT
where mis the mass of the iron, cis the specific heat capacity of iron, and T
is the change in temperature. Given: m= 0.5 kg T= (100 25) C = 75 K
Now, we know that the heat lost by the iron is equal to the heat gained by
the water:
Qiron =Qwater
0.5c75 = 20930
c=20930
0.5×75
c558.67 J/kg·K
Therefore, the specific heat capacity of iron is approximately 558.67 J/kg·K.
Question 8
Question
A 50 g piece of copper at 150
°
C is dropped into 200 g of water at 20
°
C in a
calorimeter. Assuming no heat is lost to the surroundings, what will be the final
temperature of the system? Given specific heat capacities: water = 4.18 J/g
°
C,
copper = 0.385 J/g
°
C.
Solution
Step 1: Calculate the heat lost by the copper piece as it cools down to the final
temperature. We can use the formula Q=mcTwhere Qis the heat energy, m
is the mass, cis the specific heat capacity, and Tis the change in temperature.
Heat lost by copper piece = (50 g)(0.385 J/g
°
C)(150 T) J
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature.
Heat gained by water = (200 g)(4.18 J/g
°
C)(T20) J
Step 3: Since the total heat lost by the copper piece is equal to the total
heat gained by the water (assuming no heat is lost to the surroundings), we
have the equation
(50)(0.385)(150 T) = (200)(4.18)(T20)
7
Step 4: Solve for Tto find the final temperature of the system.
(50)(0.385)(150 T) = (200)(4.18)(T20)
19.25(150 T) = 836(T20)
2887.519.25T= 836T16720
1755.5 = 855.25T
T2.05
°
C
The final temperature of the system is approximately 2.05
°
C.
Question 9
Question
A 50 g piece of aluminum at 80
°
C is placed into 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 25
°
C. Assuming no heat
is lost to the surroundings, what is the specific heat capacity of aluminum?
(Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the aluminum using the formula:
Qaluminum =maluminum ×caluminum ×Taluminum
where: - maluminum = mass of aluminum = 50 g - caluminum = specific heat
capacity of aluminum (unknown) - Taluminum = change in temperature of
aluminum = final temperature - initial temperature = (25C80C) = 55C
Qaluminum = 50 g ×caluminum × 55C
Step 2: Calculate the heat gained or lost by the water using the formula:
Qwater =mwater ×cwater ×Twater
where: - mwater = 200 g - cwater = 4.18 J/g
°
C - Twater = 25C20C= 5C
Qwater = 200 g ×4.18 J/g
°
C×5C
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum must equal the heat gained by the water. Therefore:
Qaluminum =Qwater
Step 4: Set the equations equal to each other and solve for caluminum:
50 g ×caluminum × 55 = 200 g ×4.18 J/g
°
C×5
8
caluminum =200 ×4.18 ×5
50 ×55
caluminum =4180
550
caluminum = 7.6 J/g
°
C
Therefore, the specific heat capacity of aluminum is 7.6 J/g
°
C.
Question 10
Question
A 50 g piece of copper initially at 150C is placed in 100 g of water initially at
25C. If the final temperature of the system is 28C, calculate the specific heat
capacity of copper. Assume the specific heat capacity of water is 4.18 J/(g·C).
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water
using the formula:
qcopper =qwater
The heat lost by the copper is given by
qcopper =mcT
where mis the mass of the copper, cis the specific heat capacity of copper,
and Tis the change in temperature of the copper. Since the final temperature
of the system is 28C, the change in temperature for the copper is
Tcopper = 28 150 = 122
Substituting the values, we get
qcopper = (0.05 kg)(c)(122)
The heat gained by the water is given by
qwater =mcT
where mis the mass of the water (100 g), cis the specific heat capacity
of water, and Tis the change in temperature of the water. The change in
temperature for the water is
Twater = 28 25 = 3
9
Substituting the values, we get
qwater = (0.1 kg)(4.18 J/(g ·C))(3 C)
Step 2: Set up the equation using the conservation of energy principle:
mcTcopper =mcTwater
Substitute the expressions for qcopper and qwater to get
(0.05c)(122) = (0.1)(4.18)(3)
Step 3: Solve for the specific heat capacity of copper:
6.1c=1.254
c=1.254
6.10.205 J/(g ·C)
Therefore, the specific heat capacity of copper is approximately 0.205 J/(g·C).
Question 11
Question
A 50 g piece of copper at 500 K is placed in 200 g of water at 300 K. If the
final temperature of the system is 400 K, calculate the specific heat capacity of
copper. Assume no heat is lost to the surroundings. Specific heat capacity of
water = 4186 J/kgC, specific heat capacity of copper = 387 J/kgC.
Solution
Step 1: Calculate the heat lost by copper and gained by water. Let CCbe the
specific heat capacity of copper, CWbe the specific heat capacity of water, mC
be the mass of copper, mWbe the mass of water, and Tbe the change in
temperature of the system. The heat lost by copper is equal to the heat gained
by water, so we have:
mCCCT=mWCWT
Substitute the given values:
50 g ×CC×(400 500) = 200 g ×4186 J/kgC×(400 300)
Simplify the equation:
5000CC= 418600
CC=418600
5000 = 83.72 J/kgC
Therefore, the specific heat capacity of copper is 83.72 J/kgC .
10
Question 12
Question
A 250-g block of copper at 95
°
C is placed in 600 g of water at 25
°
C. The specific
heat capacity of copper is 0.385 J/g
°
C and water is 4.18 J/g
°
C. Assuming no
heat is lost to the surroundings, calculate the final temperature when thermal
equilibrium is reached.
Solution
Step 1: Calculate the heat gained by the water. The formula for heat gained or
lost is Q=mcT, where mis the mass of the substance, cis the specific heat
capacity, and Tis the change in temperature. In this case, the water is losing
heat, so we have: Qwater =mwatercwaterTwater
Substitute the values: Qwater = 600 g ×4.18 J/g
°
C×(Tf25C) where Tf
is the final temperature.
Step 2: Calculate the heat lost by the copper block. Similar to the water,
the copper block is gaining heat: Qcopper =mcopperccopperTcopper
Substitute the values: Qcopper = 250 g ×0.385 J/g
°
C×(Tf95C)
Step 3: Set up the heat gained by water equal to the heat lost by the copper
block. Qwater =Qcopper
600 ×4.18 ×(Tf25) = 250 ×0.385 ×(Tf95)
Step 4: Solve the equation for Tf, the final temperature. 2508(Tf25) =
96.25(Tf95)
Expand and simplify: 2508Tf62700 = 96.25Tf9163.75
2411.75Tf= 53536.25
Tf=53536.25
2411.75 22.2C
Therefore, the final temperature when thermal equilibrium is reached is ap-
proximately 22.2
°
C.
Question 13
Question
A chemistry student wants to determine the enthalpy of combustion of a new
compound. To do this, they burn 2.50 g of the compound in a bomb calorimeter.
The temperature of the calorimeter increases by 6.75
°
C. If the heat capacity
of the calorimeter is 18.2 J/
°
C, calculate the enthalpy of combustion of the
compound in kJ/g.
Solution
Step 1: Calculate the heat absorbed by the calorimeter.
Heat absorbed by calorimeter = C×T
11
where Cis the heat capacity of the calorimeter and Tis the change in tem-
perature.
= 18.2 J/
°
C×6.75
°
C
= 122.85 J
Step 2: Calculate the heat released by the compound during combustion.
Since the heat released by the compound is equal in magnitude but opposite in
sign to the heat absorbed by the calorimeter,
Heat released by compound = 122.85 J
Step 3: Convert the mass of the compound to moles. First, find the molar
mass of the compound by adding the atomic masses of the elements in the
compound. Let’s say the molar mass is 100 g/mol. Then, the moles of the
compound burned would be:
2.50 g
100 g/mol = 0.025 mol
Step 4: Calculate the enthalpy of combustion of the compound per gram.
Enthalpy of combustion per mole = Heat released by compound
moles of compound burned
=122.85 J
0.025 mol
=4,914 J/mol
Step 5: Convert the enthalpy of combustion to kJ/g.
Enthalpy of combustion per gram = 4,914 J/mol
1000 J/kJ ×2.50 g
=0.785 kJ/g
Therefore, the enthalpy of combustion of the compound is -0.785 kJ/g.
Question 14
Question
A 200 g piece of copper at 70C is placed into 400 g of water at 20C. The
final temperature of the mixture is 25C. Assuming no heat is lost to the sur-
roundings, what is the specific heat capacity of copper? Take the specific heat
capacity of water to be 4186 J/kg ·K and the specific heat capacity of copper to
be 386 J/kg ·K.
12
Solution
Step 1: Calculate the heat lost by the copper piece.
The heat lost by the copper piece can be calculated using the formula:
Qlost =mcT,
where: - mis the mass of the copper piece (200 g = 0.2 kg), - cis the specific
heat capacity of copper (386 J/kg ·K), - Tis the change in temperature of
the copper piece.
Given that the initial temperature of the copper piece is 70C and the final
temperature of the mixture is 25C:
T= (25C) (70C) = 45C.
Therefore, the heat lost by the copper piece is:
Qlost = (0.2 kg)(386 J/kg ·K)(45 K) = 3474 J.
Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
Qgained =mcT,
where: - mis the mass of the water (400 g = 0.4 kg), - cis the specific heat
capacity of water (4186 J/kg ·K), - Tis the change in temperature of the
water.
Given that the initial temperature of the water is 20C and the final tem-
perature of the mixture is 25C:
T= (25C) (20C) = 5C.
Therefore, the heat gained by the water is:
Qgained = (0.4 kg)(4186 J/kg ·K)(5 K) = 8368 J.
Step 3: Set up the energy conservation equation.
Since no heat is lost to the surroundings, the heat lost by the copper piece
is equal to the heat gained by the water:
Qlost =Qgained.
Therefore,
3474 J = 8368 J.
This equation has no solution, indicating that there may be an error in the
calculations. Let’s recheck the calculations.
13
Question 15
Question
A 50 g aluminum block at 100
°
C is dropped into 100 g of water at 20
°
C in a
calorimeter. If the final temperature of the system is 30
°
C, what is the specific
heat capacity of the aluminum block? Assume no heat is lost to the surroundings
and that the specific heat capacity of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat absorbed by the water using the formula:
Qwater =mwater ×cwater ×T
where: - mwater = 100 g (mass of water), - cwater = 4.18 J/g
°
C (specific heat
capacity of water), - T= (Tfinal Tinitial) = (30C20C) = 10C.
Calculating:
Qwater = 100 ×4.18 ×10 = 4180 J
Step 2: Use the heat absorbed by the water to find the specific heat capacity
of aluminum using the formula:
Qwater =maluminum ×caluminum ×T
Since no heat is lost to the surroundings, the heat absorbed by the aluminum
is equal to the heat lost by the water:
4180 J = 50 ×caluminum ×(30 100)
Step 3: Solve for the specific heat capacity of aluminum:
caluminum =4180
50 ×(30 100) =4180
50 × 70 =1.194 J/g
°
C
Therefore, the specific heat capacity of the aluminum block is 1.194 J/g
°
C.
Question 16
Question
A 50 g aluminum block is heated to 100
°
C and then placed into 200 g of water
at 25
°
C contained in a calorimeter. The final temperature of the system is 30
°
C.
Assume no heat is lost to the surroundings. Calculate the specific heat capacity
of aluminum. The specific heat capacity of water is 4.18 J/g
°
C.
14
Solution
Step 1: Calculate the heat energy absorbed by the aluminum block as it cools
down from 100
°
C to 30
°
C. The specific heat capacity of aluminum is 0.897 J/g
°
C.
Given: Mass of aluminum block (m) = 50 g Initial temperature of aluminum
block (Tinitial) = 100
°
C Final temperature of aluminum block (Tfinal) = 30
°
C
Specific heat capacity of aluminum (cAl) = 0.897 J/g
°
C
The formula to calculate heat energy (Q) is:
Q=mcAlT
where T=Tfinal Tinitial.
Substitute the values:
Q= (50 g)(0.897 J/g
°
C)(30 100)C
Q= (50 g)(0.897 J/g
°
C)(70)C=3141 J
Therefore, the heat energy absorbed by the aluminum block is 3141 J.
Step 2: Calculate the heat energy released by the aluminum block to the
water and the calorimeter. Since the system is isolated, the heat released by
the aluminum block will be absorbed by the water and the calorimeter.
The heat energy released by the aluminum block is equal to the heat energy
gained by the water and the calorimeter. Therefore:
QAluminum =QWater +QCalorimeter
Given: Mass of water (mwater) = 200 g Specific heat capacity of water (cwater)
= 4.18 J/g
°
C Change in temperature of water (∆Twater) = 30
°
C - 25
°
C = 5
°
C
Change in temperature of calorimeter (∆Tcalorimeter) = 30
°
C - 25
°
C = 5
°
C
Substitute the values into the equation:
3141 J = (200 g)(4.18 J/g
°
C)(5C)+(mcalorimeter)(ccalorimeter)(5C)
Since no information about the mass and specific heat capacity of the calorime-
ter is given, we cannot solve for the specific heat capacity of aluminum.
Question 17
Question
A 50 g piece of aluminum at 90
°
C is dropped into 200 g of water at 20
°
C in an
insulated calorimeter. Assuming no heat is lost to the surroundings, what will
be the final temperature of the system? (Specific heat capacity of aluminum is
0.897 J/g
°
C and specific heat capacity of water is 4.18 J/g
°
C.)
15
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water using the formula Q=mcT, where Qis the heat transferred, mis the
mass of the substance, cis the specific heat capacity, and Tis the change in
temperature. For aluminum: Qaluminum = (50 g)(0.897 J/g
°
C)(Tf90), where
Tfis the final temperature.
Step 2: For water: Qwater = (200 g)(4.18 J/g
°
C)(Tf20)
Step 3: Since the heat lost by the aluminum is equal to the heat gained by the
water (assuming no heat is lost to the surroundings), we can set them equal to
each other and solve for Tf: (50 g)(0.897 J/g
°
C)(Tf90) = (200 g)(4.18 J/g
°
C)(Tf
20)
Step 4: Now, solve for Tf: 44.85(Tf90) = 836(Tf20) 44.85Tf4036.5 =
836Tf16720
Step 5: Rearrange the equation and solve for Tf: 836Tf44.85Tf= 16720
4036.5 791.15Tf= 12683.5Tf=12683.5
791.15 Tf16
°
C
Therefore, the final temperature of the system will be approximately 16
°
C.
Question 18
Question
A 50.0 g sample of iron at 85.0
°
C is placed in 200.0 g of water at 20.0
°
C. The
specific heat capacity of iron is 0.449 J/(g
°
C) and the specific heat capacity of
water is 4.18 J/(g
°
C). Assuming no heat is lost to the surroundings, what will
be the final equilibrium temperature of the system?
Solution
Step 1: Calculate the heat absorbed by the iron as it cools down to the final
temperature. The heat absorbed by the iron is given by the formula:
qiron =m·ciron ·Tiron
where: - m= mass of the iron sample = 50.0 g - ciron = specific heat capac-
ity of iron = 0.449 J/(g
°
C) - Tiron = change in temperature of iron = final
temperature - initial temperature
Tiron =Tfinal 85.0
qiron = 50.0·0.449 ·(Tfinal 85.0)
Step 2: Calculate the heat absorbed by the water as it warms up to the final
temperature. The heat absorbed by the water is given by the formula:
qwater =m·cwater ·Twater
16
where: - m= mass of the water = 200.0 g - cwater = specific heat capacity of
water = 4.18 J/(g
°
C) - Twater =Tfinal 20.0
qwater = 200.0·4.18 ·(Tfinal 20.0)
Step 3: Since there is no heat lost to the surroundings, the heat lost by the
iron is equal to the heat gained by the water.
qiron =qwater
50.0·0.449 ·(Tfinal 85.0) = 200.0·4.18 ·(Tfinal 20.0)
Step 4: Solve for the final equilibrium temperature Tfinal.
50.0·0.449 ·(Tfinal 85.0) = 200.0·4.18 ·(Tfinal 20.0)
22.45 ·(Tfinal 85.0) = 836.0·(Tfinal 20.0)
22.45 ·Tfinal 22.45 ·85.0 = 836.0·Tfinal 836.0·20.0
22.45 ·Tfinal 1902.25 = 836.0·Tfinal 16720.0
813.55 = 813.55Tfinal
Tfinal =813.55
813.55 = 1.0
Therefore, the final equilibrium temperature of the system is 1.0
°
C.
Question 19
Question
A student conducts an experiment to determine the specific heat capacity of a
metal sample using a calorimeter. The student places a 200 g aluminum sample
at 100C into 400 g of water at 20C. After thermal equilibrium is reached, the
final temperature of the system is 25C. Calculate the specific heat capacity of
the metal.
Solution
Step 1: Calculate the heat transfer from the aluminum to the water. The heat
lost by the aluminum sample is equal to the heat gained by the water:
malTal=mwcwTw
where mal= mass of aluminum (kg), Tal= change in temperature of alu-
minum (K), mw= mass of water (kg), cw= specific heat capacity of water
(J/(kg·K)), Tw= change in temperature of water (K).
Given: mal= 0.2 kg, Tal= 100 25 = 75 K, mw= 0.4 kg, cw= 4186
J/(kg·K), and Tw= 25 20 = 5 K.
17
Substitute the values and solve for l:
0.2×l×75 = 0.4×4186 ×5
l=0.4×4186 ×5
0.2×75
l= 4186 J/kg
Step 2: Calculate the specific heat capacity of the metal. The heat lost by
the metal is equal to the heat gained by the water:
malcalTal=mwcwTreservoir
where cal= specific heat capacity of the metal (J/(kg·K)), Treservoir = change
in temperature of the water after reaching equilibrium (K).
Given: mal= 0.2 kg, cal=l= 4186 J/(kg·K), Tal= 75 K, mw= 0.4 kg,
cw= 4186 J/(kg·K), and Treservoir = 25 20 = 5 K.
Substitute the values and solve for cal:
0.2×4186 ×75 = 0.4×4186 ×5
cal=0.4×4186 ×5
0.2×75
cal= 4186 J/kg
Therefore, the specific heat capacity of the metal sample is 4186 J/(kg·K).
Question 20
Question
A 50 g piece of iron at 150
°
C is dropped into 200 g of water at 25
°
C in a 100
g aluminum cup. The final temperature of the system is 40
°
C. Assuming no
heat is lost to the surroundings, determine the specific heat capacity of iron.
The specific heat capacities of iron, water, and aluminum are 0.45 J/g
°
C, 4.18
J/g
°
C, and 0.897 J/g
°
C, respectively.
Solution
Step 1: Calculate the heat gained by the water and the aluminum cup. The
heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·T
Where: - mwater = 200 g (mass of water) - cwater = 4.18 J/g
°
C (specific heat
capacity of water) - Tf= 40
°
C (final temperature) - Ti= 25
°
C (initial temper-
ature)
Twater =TfTi= 40 25 = 15
°
C
18
Qwater = 200 g ×4.18 J/g
°
C×15
°
C
Qwater = 12540 J
The heat gained by the aluminum cup can be calculated using the same
formula:
Qaluminum =maluminum ·caluminum ·Taluminum
Where: - maluminum = 100 g (mass of aluminum) - caluminum = 0.897 J/g
°
C
(specific heat capacity of aluminum) - Taluminum = 40 25 = 15
°
C
Qaluminum = 100 g ×0.897 J/g
°
C×15
°
C
Qaluminum = 1345.5 J
Step 2: Calculate the heat lost by the iron to the water and the aluminum
cup. Since no heat is lost to the surroundings, the heat lost by the iron is equal
to the heat gained by the water and the aluminum cup.
Qiron =Qwater +Qaluminum
Qiron = 12540 J + 1345.5 J
Qiron = 13885.5 J
Step 3: Use the formula for heat
Qiron =miron ·ciron ·Tiron
Since the mass of the iron is 50 g, we can solve for the specific heat capacity of
iron.
ciron =Qiron
miron ·Tiron
ciron =13885.5 J
50 g ×(40 150)
°
C
ciron =13885.5
50 × 110
ciron =2.523 J/g
°
C
Therefore, the specific heat capacity of iron is 2.523 J/g
°
C .
Question 21
Question
A 50 g piece of aluminum at 100
°
C is dropped into 100 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Assuming no heat is
lost to the surroundings, calculate the specific heat capacity of aluminum.
19
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water.
The heat lost by the aluminum can be calculated using the formula:
QAluminum =m×cAluminum ×TAluminum
Where: m= 50 g (mass of aluminum), cAluminum = 0.903 J/g
°
C (specific
heat capacity of aluminum), and TAluminum =TfTi= 22 100 =
78
°
C (change in temperature of the aluminum).
Substituting the values, we get:
QAluminum = 50 ×0.903 ×(78) = 3525.6 J
The heat gained by the water can be calculated using the formula:
QWater =m×cWater ×TWater
Where: m= 100 g (mass of water), cWater = 4.18 J/g
°
C (specific heat
capacity of water), and TWater =TfTi= 22 20 = 2
°
C (change in
temperature of the water).
Substituting the values, we get:
QWater = 100 ×4.18 ×2 = 836 J
Question 22
Question
A 100 g piece of aluminum at 80
°
C is placed in a calorimeter containing 200
g of water at 20
°
C. If the final temperature of the system is 25
°
C, what is the
heat capacity of the calorimeter? The specific heat capacities of aluminum and
water are 0.90 J/g
°
C and 4.18 J/g
°
C, respectively.
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water. The heat lost by the aluminum can be calculated using the formula:
QAl =mAl ·cAl ·T
where: - mAl is the mass of aluminum - cAl is the specific heat capacity of
aluminum - Tis the temperature change of aluminum
Substitute the given values into the formula:
QAl = 100 g ·0.90 J/g
°
C·(25 80)
°
C
20
QAl = 100 ·0.90 ·(55)
QAl =4950 J
The heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·T
where: - mwater is the mass of water - cwater is the specific heat capacity of water
- Tis the temperature change of water
Substitute the given values into the formula:
Qwater = 200 g ·4.18 J/g
°
C·(25 20)
°
C
Qwater = 200 ·4.18 ·5
Qwater = 4180 J
Step 2: Use the principle of conservation of energy to find the heat capacity
of the calorimeter. The heat lost by the aluminum is equal to the heat gained
by the water and calorimeter:
QAl =Qwater +Qcalorimeter
Substitute the calculated values:
4950 = 4180 + Qcalorimeter
Qcalorimeter =4950 4180
Qcalorimeter =9130 J
Step 3: Calculate the heat capacity of the calorimeter using the formula:
Qcalorimeter =Ccalorimeter ·T
where: - Ccalorimeter is the heat capacity of the calorimeter - T=Tfinal
Tinitial = 25 20 = 5
°
C
Substitute the known values:
9130 = Ccalorimeter ·5
Ccalorimeter =9130
5
Ccalorimeter =1826 J/
°
C
Therefore, the heat capacity of the calorimeter is 1826 J/
°
C .
Question 23
Question
A 50 g piece of iron at 80
°
C is dropped into a calorimeter containing 100 g of
water at 20
°
C. If the final temperature of the system is 30
°
C, what is the heat
capacity of the calorimeter? Assume no heat is lost to the surroundings.
21
Solution
Step 1: Calculate the heat gained by the water to warm up from 20
°
C to 30
°
C.
Step 2: Calculate the heat lost by the iron to cool down from 80
°
C to 30
°
C.
Step 3: Use the fact that the total heat lost by the iron is equal to the total
heat gained by the water and the calorimeter to find the heat capacity of the
calorimeter.
Step 1: The specific heat capacity of water is 4.18 J/g
°
C.
The heat gained by the water can be calculated using the formula:
qwater =m·c·T
where: - mis the mass of the water (100 g), - cis the specific heat capacity of
water (4.18 J/g
°
C), - Tis the change in temperature (from 20
°
C to 30
°
C).
Substitute the values into the formula:
qwater = 100 g ×4.18 J/g
°
C×(30 20)
°
C
qwater = 100 g ×4.18 J/g
°
C×10
°
C
qwater = 4180 J
Therefore, the heat gained by the water is 4180 J.
Step 2: The specific heat capacity of iron is 0.45 J/g
°
C.
The heat lost by the iron can be calculated using the formula:
qiron =m·c·T
where: - mis the mass of the iron (50 g), - cis the specific heat capacity of iron
(0.45 J/g
°
C), - Tis the change in temperature (from 80
°
C to 30
°
C).
Substitute the values into the formula:
qiron = 50 g ×0.45 J/g
°
C×(30 80)
°
C
qiron = 50 g ×0.45 J/g
°
C× 50
°
C
qiron =1125 J
Therefore, the heat lost by the iron is -1125 J. Note the negative sign indi-
cates heat loss.
Step 3: The total heat gained by the water and the calorimeter is equal
to the absolute value of the total heat lost by the iron. Therefore, |qwater +
qcalorimeter|=|qiron|.
Substitute the values of qwater and qiron into the equation:
|4180 J + qcalorimeter|=| 1125 J|
4180 J + qcalorimeter = 1125 J
qcalorimeter = 1125 J 4180 J
qcalorimeter =3055 J
Therefore, the heat capacity of the calorimeter is 3055 J.
22
Question 24
Question
A 50 g piece of copper at 150
°
C is placed in 200 g of water at 20
°
C. Assuming
no heat is lost to the surroundings, what will be the final temperature of the
system? (Specific heat capacity of copper = 0.386 J/g
°
C, specific heat capacity
of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by copper and the heat gained by water using
the formula Q=mcTwhere Qis heat, mis mass, cis specific heat capacity,
and Tis the change in temperature.
For copper: Qcopper = (50 g)(0.386 J/g
°
C)(Tf150C)
For water: Qwater = (200 g)(4.18 J/g
°
C)(Tf20C)
Since no heat is lost or gained from the surroundings, we have Qcopper =
Qwater
Step 2: Set up the equation: (50 g)(0.386 J/g
°
C)(Tf150C) = (200 g)(4.18 J/g
°
C)(Tf
20C)
Step 3: Solve for the final temperature Tf: 50(0.386)(Tf150) = 200(4.18)(Tf
20)
19.3Tf2895 = 836Tf+ 16720
855Tf= 19615
Tf22.96C
Therefore, the final temperature of the system will be approximately 22.96C.
Question 25
Question
A 50 g piece of aluminum at 100
°
C is placed in 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Assuming no heat
is lost to the surroundings, calculate the specific heat capacity of aluminum.
(Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water. The heat lost by the aluminum is equal to the heat gained by the water.
Therefore, we have:
mAl ·cAl ·(TfTi) = mw·cw·(TfTi)
Substitute the given values:
50 g ·cAl ·(22 100) = 200 g ·4.18 J/g
°
C·(22 20)
23
Simplify and solve for cAl.
Step 2: Calculate the specific heat capacity of aluminum.
50 g ·cAl ·(78) = 200 g ·4.18 J/g
°
C·2
cAl =200 g ·4.18 J/g
°
C·2
50 g ·78
cAl =1672 J
3900 g ·
°
C
cAl 0.43 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 0.43 J/g
°
C.
Question 26
Question
A student wants to determine the specific heat capacity of a metal using calorime-
try. They take a 150 g block of the metal at an initial temperature of 200
°
C
and place it in 200 g of water at an initial temperature of 20
°
C contained in a
calorimeter. The final temperature of the system is 23
°
C. What is the specific
heat capacity of the metal? Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat absorbed by the water. The heat absorbed by the
water can be calculated using the formula:
Qwater =m·c·T
where: - m= 200 g is the mass of water, - c= 4.18 J/g
°
C is the specific heat
capacity of water, - T=TfTi= 23C20C= 3Cis the change in temper-
ature.
Substitute the values into the formula:
Qwater = 200 g ×4.18 J/g
°
C×3C= 2508 J
Step 2: Calculate the heat lost by the metal. The heat lost by the metal
is equal to the heat absorbed by the water, as there is no heat lost to the
surroundings. We can use the formula:
Qmetal =Qwater
Qmetal =2508 J
Step 3: Calculate the specific heat capacity of the metal. The heat lost by
the metal can be calculated using the formula:
Qmetal =m·cmetal ·T
24
where: - m= 150 g is the mass of the metal and - T=TfTi= 200C23C=
177Cis the change in temperature.
Substitute the known values and solve for cmetal:
2508 J = 150 g ×cmetal ×177C
cmetal =2508 J
150 g ×177C
cmetal 0.094 J/g
°
C
Therefore, the specific heat capacity of the metal is approximately -0.094
J/g
°
C.
Question 27
Question
A piece of metal with mass 0.25 kg is heated to 100C and then placed in 0.5
kg of water at 20C. The final temperature of the system is 25C. If the specific
heat capacity of the metal is 450 J/kg·K, determine the specific heat capacity
of the metal.
Solution
Step 1: Calculate the heat lost by the metal as it cools down from 100C to
the final temperature. The heat lost by the metal can be calculated using the
formula:
Qmetal =mcT
where: - mis the mass of the metal, - cis the specific heat capacity of the metal,
- Tis the change in temperature of the metal.
Substitute the values:
Qmetal = 0.25 kg ×450 J/kg·K×(100 25) K
Qmetal = 0.25 ×450 ×75
Qmetal = 8437.5 J
Step 2: Calculate the heat gained by the water as it heats up from 20C to
the final temperature. The heat gained by the water can be calculated using
the formula:
Qwater =mcT
where: - mis the mass of the water, - cis the specific heat capacity of water, -
Tis the change in temperature of the water.
Substitute the values:
Qwater = 0.5 kg ×4186 J/kg·K×(25 20) K
25
Qwater = 0.5×4186 ×5
Qwater = 10465 J
Step 3: Since energy is conserved, the heat lost by the metal is equal to the
heat gained by the water.
Qmetal =Qwater
8437.5 = 10465
8437.5 = 10465
Note: We reached an error in our calculation, please check the heat lost by
the metal and heat gained by the water and make sure they are equal.
Question 28
Question
A piece of metal of mass 0.5 kg at 150
°
C is placed in 1 kg of water at 20
°
C
contained in a 1 kg aluminum calorimeter cup at 20
°
C. The final temperature
of the system is 30
°
C. Given the specific heat capacities of the metal and water
are 0.5 J/g
°
C and 4.18 J/g
°
C respectively, calculate the specific heat capacity
of the metal.
Solution
Step 1: Calculate the heat gained by the water and the calorimeter cup.
The heat gained by the water and calorimeter cup can be calculated using the
formula:
Q=mcT
where Qis the heat gained, mis the mass, cis the specific heat capacity, and
Tis the change in temperature.
For water:
Qwater = (1 kg ×1000 g/kg ×4.18 J/g
°
C×(30 20)
°
C)
Qwater = 41800 J
For the calorimeter cup (assume specific heat capacity of aluminum is 0.9
J/g
°
C):
Qcup = (1 kg ×1000 g/kg ×0.9 J/g
°
C×(30 20)
°
C)
Qcup = 900 J
Step 2: Calculate the heat lost by the metal.
The heat lost by the metal is equal to the sum of the heat gained by the water
and calorimeter cup:
Qmetal =Qwater +Qcup
26
Qmetal = 41800 J + 900 J
Qmetal = 42700 J
Step 3: Calculate the specific heat capacity of the metal.
We know that the heat lost by the metal is given by:
Qmetal =mcT
Substitute the values we have:
42700 = 0.5×c×(30 150)
42700 = 60c
c=42700
60
c 711.67 J/kg
°
C
Therefore, the specific heat capacity of the metal is approximately 711.67
J/kg
°
C.
Question 29
Question
A piece of copper with a mass of 200 g is heated to a temperature of 100
°
C
and then dropped into a calorimeter containing 500 g of water at an initial
temperature of 20
°
C. If the final temperature of the system is 25
°
C, calculate
the specific heat capacity of copper. Assume no heat is lost to the surroundings
and the specific heat capacity of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
Heat lost by copper = Heat gained by water
Step 2: The heat lost by the copper is given by the formula:
Qcopper =mcT
where: - mis the mass of copper (200 g) - cis the specific heat capacity of copper
(unknown) - Tis the change in temperature of the copper (final temperature
- initial temperature)
Qcopper = (200 g)(c)(25 100)
Qcopper =15000cJ
27
Step 3: The heat gained by the water is given by the formula:
Qwater =mcT
where: - mis the mass of water (500 g) - cis the specific heat capacity of water
(4.18 J/g
°
C) - Tis the change in temperature of the water (final temperature
- initial temperature)
Qwater = (500 g)(4.18 J/g
°
C)(25 20)
Qwater = 2090 J
Step 4: Set the heat lost by the copper equal to the heat gained by the water:
15000c= 2090
Step 5: Solve for the specific heat capacity of copper:
c=2090
15000 =0.139 J/g
°
C
Therefore, the specific heat capacity of copper is -0.139 J/g
°
C. The negative
Question 30
Question
A 50 g piece of aluminum initially at 80
°
C is dropped into 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 capacity of aluminum = 0.9 J/g
°
C, specific heat
capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the aluminum piece. The heat lost by the
aluminum can be calculated using the formula:
qaluminum =mcT
where: - m= 50 g (mass of aluminum) - c= 0.9 J/g
°
C (specific heat capacity
of aluminum) - T=Tfinal Tinitial =Tfinal 80
°
C
Step 2: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the formula:
qwater =mcT
where: - m= 200 g (mass of water) - c= 4.18 J/g
°
C (specific heat capacity of
water) - T=Tfinal Tinitial =Tfinal 20
°
C
28
Step 3: Set up the conservation of energy equation: Since the system is
isolated and no heat is lost to the surroundings, the heat lost by the aluminum
must be equal to the heat gained by the water, so we have:
qaluminum =qwater
Step 4: Substitute the calculated values and solve for the final temperature,
Tfinal.
maluminumcaluminum(Tfinal 80) = mwatercwater(Tfinal 20)
Step 5: Simplify the equation and solve for Tfinal.
50 ×0.9×(Tfinal 80) = 200 ×4.18 ×(Tfinal 20)
45(Tfinal 80) = 836(Tfinal 20)
45Tfinal 3600 = 836Tfinal 16720
791Tfinal = 13120
Tfinal =13120
791 16.58
°
C
Therefore, the final temperature of the system is approximately 16.58
°
C.
Question 31
Question
A 50 g aluminum block is heated to 100
°
C and then placed in 200 g of water
at 20
°
C inside a calorimeter. The final temperature of the system is 25
°
C.
Assuming no heat is lost to the surroundings, calculate the specific heat capacity
of the aluminum block. (Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat absorbed by the water during the cooling process.
The heat absorbed by the water can be calculated using the formula:
qwater =mwater ×cwater ×T
where: - mwater = 200 g is the mass of water, - cwater = 4.18 J/g
°
C is the specific
heat capacity of water, and - T= 25 20 = 5
°
C is the temperature change
of the water. Substitute these values into the formula:
qwater = 200 ×4.18 ×5 = 4180 J
Step 2: Calculate the heat lost by the aluminum block. The heat lost by the
aluminum block is equal in magnitude to the heat gained by the water, so we
have:
qaluminum =qwater =4180 J
29
Step 3: Calculate the heat lost by the aluminum block using the formula:
qaluminum =maluminum ×caluminum ×T
where: - maluminum = 50 g is the mass of the aluminum block, - caluminum is
the specific heat capacity of aluminum (which we want to find), and - T=
100 25 = 75
°
C is the temperature change of the aluminum block.
Equating the heat lost by the aluminum block in step 2 and step 3, we have:
50 ×caluminum ×75 = 4180
caluminum =4180
50 ×75
caluminum =1.40 J/g
°
C
Therefore, the specific heat capacity of the aluminum block is 1.40 J/g
°
C.
Question 32
Question
A piece of metal with mass 0.2 kg is heated to 100
°
C and then placed in 0.5 kg
of water at 25
°
C in an insulated container. The final temperature of the system
is 30
°
C. If the specific heat capacity of the metal is 450 J/kg
°
C and the specific
heat capacity of water is 4186 J/kg
°
C, calculate the specific heat capacity of the
container if none of the heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the metal and gained by the water using the
formula:
Qmetal =Qwater
Step 2: Calculate the heat lost by the metal using the formula:
Qmetal =mcT
where: m= 0.2 kg (mass of the metal), c= 450 J/kg
°
C (specific heat capacity
of the metal), T= 100 30 = 70
°
C (final temperature - initial temperature).
Step 3: Substitute the values into the formula to find Qmetal:
Qmetal = 0.2×450 ×70
Step 4: Calculate the heat gained by the water using the formula:
Qwater =mcT
where: m= 0.5 kg (mass of the water), c= 4186 J/kg
°
C (specific heat capacity
of water), T= 30 25 = 5
°
C (final temperature - initial temperature).
30
Step 5: Substitute the values into the formula to find Qwater:
Qwater = 0.5×4186 ×5
Step 6: Set Qmetal =Qwater and solve for the specific heat capacity of the
container:
0.2×450 ×70 = 0.5×4186 ×5
Step 7: Solve for the specific heat capacity of the container.
Question 33
Question
A 50 g piece of aluminum initially at 100C is placed in a calorimeter containing
200 g of water at 20C. The final temperature of the system is 25C. Assuming
no heat is lost to the surroundings, what is the specific heat capacity of the
aluminum? (Specific heat capacity of water is 4.18 J/gC)
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water using the formula:
q=m·c·T
where: q= heat energy in Joules (J), m= mass in grams (g), c= specific
heat capacity in J/gC, T= change in temperature in C.
The heat lost by aluminum is equal to the heat gained by water, since no
heat is lost to the surroundings.
For the aluminum:
qAl = (50 g) ·cAl ·(25 100)C
For the water:
qH2O = (200 g) ·4.18 J/gC·(25 20)C
Since qAl =qH2O:
(50 g) ·cAl ·(75)C = (200 g) ·4.18 J/gC·5C
Simplify and solve for cAl in terms of the given values.
Question 34
Question
A 100 g piece of copper at 200
°
C is placed into a calorimeter containing 200 g of
water at 20
°
C. The final temperature of the system is 25
°
C. Assuming no heat
is lost to the surroundings, what is the specific heat capacity of the calorimeter?
31
Solution
Given: Mass of copper (mc) = 100 g = 0.1 kg
Initial temperature of copper (Tc,i) = 200
°
C
Specific heat capacity of copper (cc) = 390 J/kg ·
°
C
Mass of water in calorimeter (mw) = 200 g = 0.2 kg
Initial temperature of water (Tw,i) = 20
°
C
Final temperature of the system (Tf) = 25
°
C
Specific heat capacity of water (cw) = 4186 J/kg ·
°
C
We will assume that no heat is lost to the surroundings, so the total heat
lost by the copper is equal to the total heat gained by the water:
mc·cc·(Tc,i Tf) = mw·cw·(TfTw,i)
Step 1: Calculate the heat lost by the copper (Qc) and the heat
gained by the water (Qw).
Qc=mc·cc·(Tc,i Tf)
Qw=mw·cw·(TfTw,i)
Substitute the given values:
Qc= 0.1 kg ·390 J/kg ·
°
C·(200 25)
°
C
Qw= 0.2 kg ·4186 J/kg ·
°
C·(25 20)
°
C
Step 2: Equate the heat lost and gained equations. Solve for the
specific heat capacity of the calorimeter.
mc·cc·(Tc,i Tf) = mw·cw·(TfTw,i)
0.1·390 ·(200 25) = 0.2·4186 ·(25 20)
0.1·390 ·175 = 0.2·4186 ·5
6825 = 4186 ·1·c
c=6825
4186
Therefore, the specific heat capacity of the calorimeter is 6825
4186 1.63 J/g·
°
C.
Question 35
Question
A 50 g aluminum block at 80
°
C is dropped into 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Given the specific
heat capacity of water is 4.18 J/g
°
C, determine the specific heat capacity of
aluminum. Assume no heat is lost to the surroundings and that the specific
heat capacity of aluminum is constant over the temperature range involved.
32
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the equation:
qwater =m·c·T
where: - mis the mass of water (200 g) - cis the specific heat capacity of water
(4.18 J/g
°
C) - Tis the change in temperature of the water (22C20C= 2C)
Substitute the given values to find qwater:
qwater = 200 g ×4.18 J/g
°
C×2C= 1672 J
Step 2: Calculate the heat lost by the aluminum. The heat lost by the
aluminum block must be equal to the heat gained by the water, since no heat
is lost to the surroundings. We can express this as:
qaluminum =qwater
Since the heat lost by the aluminum block can be calculated using its specific
heat capacity (caluminum), mass (maluminum = 50 g), and temperature change
(∆Taluminum =TfTi= 22C80C=58C):
qaluminum =maluminum ·caluminum ·Taluminum
Substitute the known values to find caluminum:
1672 J = 50 g ×caluminum × 58C
Solve for caluminum:
caluminum =1672 J
50 g × 58C0.57 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 0.57 J/g
°
C.
33
Step 4: Solve for the specific heat capacity of aluminum. Since Q=mcT,
we have: maluminumcaluminum(Tfinal, systemTinitial, aluminum) = mwatercwater(Tfinal, system
Tinitial, water)
Substitute the given values and solve for caluminum to obtain the specific heat
capacity of aluminum.
Question 2
Question
A 50 g piece of aluminum at 100C is placed in 200 g of water at 20C. If the
final temperature of the system is 25C, calculate the specific heat capacity of
aluminum. Assume no heat is lost to the surroundings and that the specific
heat capacity of water is 4.18 J/g·K.
Solution
Step 1: Calculate the heat lost by the aluminum and gained by the water
using the formula q=mcT, where qis the heat transferred, mis the mass,
cis the specific heat capacity, and Tis the change in temperature. In this
case, the heat lost by the aluminum is equal to the heat gained by the water.
For the aluminum: qaluminum = (50 g)(cAl)(75C) For the water: qwater =
(200 g)(4.18 J/g·K)(5C)
Step 2: Set the heat lost by the aluminum equal to the heat gained by the
water and solve for the specific heat capacity of aluminum.
(50 g)(cAl)(75C) = (200 g)(4.18 J/g·K)(5C)
cAl =(200 g)(4.18 J/g·K)(5C)
(50 g)(75C)
Step 3: Calculate the specific heat capacity of aluminum.
cAl =(200)(4.18)(5)
(50)(75)
cAl =4180
375
cAl = 11.15 J/g·K
Therefore, the specific heat capacity of aluminum is 11.15 J/g·K.
Question 3
Question
A 50 g piece of iron at 80
°
C is placed into 200 g of water at 20
°
C in an insulated
container. If the final temperature of the system is 25
°
C, determine the specific
heat capacity of iron.
2
Solution
Let the specific heat capacity of water be cw= 4.18 J/g
°
C.
Step 1: Calculate the heat absorbed by the water when the iron piece is
added. The heat absorbed by the water can be calculated using the formula:
Q=mcT
where: - mis the mass of water, - cis the specific heat capacity of water, - T
is the change in temperature of the water.
Substitute the given values:
Q= 200 ×4.18 ×(25 20)
Q= 200 ×4.18 ×5
Q= 4180 J
Step 2: Calculate the amount of heat lost by the iron. Since the system is
insulated, the total amount of heat lost by the iron is equal to the amount of
heat gained by the water. Therefore, the heat lost by the iron is also 4180 J.
Step 3: Calculate the specific heat capacity of iron. The heat lost by the
iron can be calculated using the formula:
Q=mcT
where: - mis the mass of iron, - cis the specific heat capacity of iron, - Tis
the change in temperature of the iron.
Substitute the given values and the heat lost:
4180 = 50c(25 80)
4180 = 50c×55
4180 = 2750c
c=4180
2750
c 1.52 J/g
°
C
Therefore, the specific heat capacity of iron is approximately 1.52 J/g
°
C.
The negative sign indicates that the value obtained is not physically meaningful.
This discrepancy may be due to experimental errors or assumptions made during
the calculation.
Question 4
Question
A 50 g sample of aluminum at 80
°
C is dropped into 200 g of water at 20
°
C in
a calorimeter. The specific heat capacity of aluminum is 0.897 J/g
°
C and the
specific heat capacity of water is 4.18 J/g
°
C. Assuming no heat is lost to the
surroundings, what is the final temperature of the system?
3
Solution
Step 1: Calculate the heat absorbed by the aluminum: The equation for heat
transfer is given by q=mcT, where: - mis the mass of the substance, - cis
the specific heat capacity of the substance, - Tis the change in temperature.
Substitute the values for aluminum: qAl = (50 g)(0.897 J/g
°
C)(Tf80
°
C)
Step 2: Calculate the heat lost by the water: qwater =(200 g)(4.18 J/g
°
C)(Tf
20
°
C)
Step 3: Set up the energy conservation equation: The heat lost by the water
is equal to the heat gained by the aluminum: (200 g)(4.18 J/g
°
C)(Tf20
°
C) =
(50 g)(0.897 J/g
°
C)(Tf80
°
C)
Step 4: Solve for the final temperature: (836 J/
°
C)(Tf20
°
C) = (44.85 J/
°
C)(Tf
80
°
C)
836Tf+ 16720 = 44.85Tf3588
836Tf+ 44.85Tf= 16720 + 3588
Tf=20308
°
C
880.85 23.05
°
C
Therefore, the final temperature of the system is approximately 23.05
°
C.
Question 5
Question
A 50 g piece of copper at 90
°
C is placed in 100 g of water at 20
°
C. Assuming
no heat is lost to the surroundings, what will be the final temperature of the
system? (Specific heat capacity of copper = 0.385 J/g
°
C, specific heat capacity
of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper piece as it cools down to the final
temperature. The formula for calculating heat (q) is:
q=m·c·T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, - Tis the change in temperature.
Plugging in the values:
qcopper = 50 g ×0.385 J/g
°
C×(90 Tf)C
qcopper = 19.25 ×(90 Tf) J
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. Using the same formula:
qwater = 100 g ×4.18 J/g
°
C×(Tf20)C
qwater = 418 ×(Tf20) J
4
Step 3: Since the system is isolated and assuming no heat loss to the sur-
roundings, the heat lost by the copper must be equal to the heat gained by the
water. Equating qcopper and qwater:
19.25 ×(90 Tf) = 418 ×(Tf20)
Step 4: Solve the equation for the final temperature (Tf):
1732.519.25Tf= 418Tf8360
19.25Tf+ 418Tf= 8360 + 1732.5
437.25Tf= 10092.5
Tf=10092.5
437.25 23.10C
Therefore, the final temperature of the system will be approximately 23.10
°
C.
Question 6
Question
A 50 g piece of aluminum at 100
°
C is placed in 200 g of water at 20
°
C in a
calorimeter. If the final equilibrium temperature of the system is 25
°
C, cal-
culate the specific heat capacity of aluminum. Assume no heat is lost to the
surroundings.
Solution
Step 1: Calculate the heat absorbed by the water. The heat absorbed by the
water can be calculated using the formula:
Qwater =mcT
where: - m= 200 g is the mass of the water, - c= 4.18 J/g
°
C is the specific
heat capacity of water, - T= 25C20C= 5Cis the temperature change of
the water.
Substitute the values into the formula:
Qwater = (200 g)(4.18 J/g
°
C)(5C)
Qwater = 4180 J
Step 2: Calculate the heat lost by the aluminum. The heat lost by the
aluminum can be calculated using the same formula:
Qaluminum =mcT
5
where: - m= 50 g is the mass of the aluminum, - cis the specific heat
capacity of aluminum (to be determined), - T= 25C100C=75Cis the
temperature change of the aluminum. (Negative because it loses heat.)
Substitute the values into the formula:
Qaluminum = (50 g)(c)(75C)
Qaluminum =3750c
Step 3: Set up the heat balance equation. Since no heat is lost to the
surroundings, the heat lost by the aluminum is equal to the heat gained by the
water. Therefore:
Qwater =Qaluminum
4180 = 3750c
Step 4: Solve for the specific heat capacity of aluminum.
c=4180
3750
c 1.11 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 1.11
J/g
°
C.
Question 7
Question
A piece of iron with mass 0.5 kg is heated to a temperature of 100C and then
dropped into a container of water with mass 1 kg at a temperature of 20C.
The final temperature of the system is 25C. Assuming all the heat lost by the
iron is gained by the water and that there is no heat loss to the surroundings,
calculate the specific heat capacity of iron. (Specific heat capacity of water is
4186 J/kg·K)
Solution
Step 1: Calculate the heat gained by the water The heat gained by the water
can be calculated using the formula:
Qwater =mcT
where mis the mass of the water, cis the specific heat capacity of water, and T
is the change in temperature. Given: m= 1 kg c= 4186 J/kg·K T= (2520)
C = 5 K
6
Substitute the given values into the formula:
Qwater = 1 ×4186 ×5
Qwater = 20930 J
Step 2: Calculate the heat lost by the iron The heat lost by the iron can
also be calculated using the formula:
Qiron =mcT
where mis the mass of the iron, cis the specific heat capacity of iron, and T
is the change in temperature. Given: m= 0.5 kg T= (100 25) C = 75 K
Now, we know that the heat lost by the iron is equal to the heat gained by
the water:
Qiron =Qwater
0.5c75 = 20930
c=20930
0.5×75
c558.67 J/kg·K
Therefore, the specific heat capacity of iron is approximately 558.67 J/kg·K.
Question 8
Question
A 50 g piece of copper at 150
°
C is dropped into 200 g of water at 20
°
C in a
calorimeter. Assuming no heat is lost to the surroundings, what will be the final
temperature of the system? Given specific heat capacities: water = 4.18 J/g
°
C,
copper = 0.385 J/g
°
C.
Solution
Step 1: Calculate the heat lost by the copper piece as it cools down to the final
temperature. We can use the formula Q=mcTwhere Qis the heat energy, m
is the mass, cis the specific heat capacity, and Tis the change in temperature.
Heat lost by copper piece = (50 g)(0.385 J/g
°
C)(150 T) J
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature.
Heat gained by water = (200 g)(4.18 J/g
°
C)(T20) J
Step 3: Since the total heat lost by the copper piece is equal to the total
heat gained by the water (assuming no heat is lost to the surroundings), we
have the equation
(50)(0.385)(150 T) = (200)(4.18)(T20)
7
Step 4: Solve for Tto find the final temperature of the system.
(50)(0.385)(150 T) = (200)(4.18)(T20)
19.25(150 T) = 836(T20)
2887.519.25T= 836T16720
1755.5 = 855.25T
T2.05
°
C
The final temperature of the system is approximately 2.05
°
C.
Question 9
Question
A 50 g piece of aluminum at 80
°
C is placed into 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 25
°
C. Assuming no heat
is lost to the surroundings, what is the specific heat capacity of aluminum?
(Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the aluminum using the formula:
Qaluminum =maluminum ×caluminum ×Taluminum
where: - maluminum = mass of aluminum = 50 g - caluminum = specific heat
capacity of aluminum (unknown) - Taluminum = change in temperature of
aluminum = final temperature - initial temperature = (25C80C) = 55C
Qaluminum = 50 g ×caluminum × 55C
Step 2: Calculate the heat gained or lost by the water using the formula:
Qwater =mwater ×cwater ×Twater
where: - mwater = 200 g - cwater = 4.18 J/g
°
C - Twater = 25C20C= 5C
Qwater = 200 g ×4.18 J/g
°
C×5C
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum must equal the heat gained by the water. Therefore:
Qaluminum =Qwater
Step 4: Set the equations equal to each other and solve for caluminum:
50 g ×caluminum × 55 = 200 g ×4.18 J/g
°
C×5
8
caluminum =200 ×4.18 ×5
50 ×55
caluminum =4180
550
caluminum = 7.6 J/g
°
C
Therefore, the specific heat capacity of aluminum is 7.6 J/g
°
C.
Question 10
Question
A 50 g piece of copper initially at 150C is placed in 100 g of water initially at
25C. If the final temperature of the system is 28C, calculate the specific heat
capacity of copper. Assume the specific heat capacity of water is 4.18 J/(g·C).
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water
using the formula:
qcopper =qwater
The heat lost by the copper is given by
qcopper =mcT
where mis the mass of the copper, cis the specific heat capacity of copper,
and Tis the change in temperature of the copper. Since the final temperature
of the system is 28C, the change in temperature for the copper is
Tcopper = 28 150 = 122
Substituting the values, we get
qcopper = (0.05 kg)(c)(122)
The heat gained by the water is given by
qwater =mcT
where mis the mass of the water (100 g), cis the specific heat capacity
of water, and Tis the change in temperature of the water. The change in
temperature for the water is
Twater = 28 25 = 3
9
Substituting the values, we get
qwater = (0.1 kg)(4.18 J/(g ·C))(3 C)
Step 2: Set up the equation using the conservation of energy principle:
mcTcopper =mcTwater
Substitute the expressions for qcopper and qwater to get
(0.05c)(122) = (0.1)(4.18)(3)
Step 3: Solve for the specific heat capacity of copper:
6.1c=1.254
c=1.254
6.10.205 J/(g ·C)
Therefore, the specific heat capacity of copper is approximately 0.205 J/(g·C).
Question 11
Question
A 50 g piece of copper at 500 K is placed in 200 g of water at 300 K. If the
final temperature of the system is 400 K, calculate the specific heat capacity of
copper. Assume no heat is lost to the surroundings. Specific heat capacity of
water = 4186 J/kgC, specific heat capacity of copper = 387 J/kgC.
Solution
Step 1: Calculate the heat lost by copper and gained by water. Let CCbe the
specific heat capacity of copper, CWbe the specific heat capacity of water, mC
be the mass of copper, mWbe the mass of water, and Tbe the change in
temperature of the system. The heat lost by copper is equal to the heat gained
by water, so we have:
mCCCT=mWCWT
Substitute the given values:
50 g ×CC×(400 500) = 200 g ×4186 J/kgC×(400 300)
Simplify the equation:
5000CC= 418600
CC=418600
5000 = 83.72 J/kgC
Therefore, the specific heat capacity of copper is 83.72 J/kgC .
10
Question 12
Question
A 250-g block of copper at 95
°
C is placed in 600 g of water at 25
°
C. The specific
heat capacity of copper is 0.385 J/g
°
C and water is 4.18 J/g
°
C. Assuming no
heat is lost to the surroundings, calculate the final temperature when thermal
equilibrium is reached.
Solution
Step 1: Calculate the heat gained by the water. The formula for heat gained or
lost is Q=mcT, where mis the mass of the substance, cis the specific heat
capacity, and Tis the change in temperature. In this case, the water is losing
heat, so we have: Qwater =mwatercwaterTwater
Substitute the values: Qwater = 600 g ×4.18 J/g
°
C×(Tf25C) where Tf
is the final temperature.
Step 2: Calculate the heat lost by the copper block. Similar to the water,
the copper block is gaining heat: Qcopper =mcopperccopperTcopper
Substitute the values: Qcopper = 250 g ×0.385 J/g
°
C×(Tf95C)
Step 3: Set up the heat gained by water equal to the heat lost by the copper
block. Qwater =Qcopper
600 ×4.18 ×(Tf25) = 250 ×0.385 ×(Tf95)
Step 4: Solve the equation for Tf, the final temperature. 2508(Tf25) =
96.25(Tf95)
Expand and simplify: 2508Tf62700 = 96.25Tf9163.75
2411.75Tf= 53536.25
Tf=53536.25
2411.75 22.2C
Therefore, the final temperature when thermal equilibrium is reached is ap-
proximately 22.2
°
C.
Question 13
Question
A chemistry student wants to determine the enthalpy of combustion of a new
compound. To do this, they burn 2.50 g of the compound in a bomb calorimeter.
The temperature of the calorimeter increases by 6.75
°
C. If the heat capacity
of the calorimeter is 18.2 J/
°
C, calculate the enthalpy of combustion of the
compound in kJ/g.
Solution
Step 1: Calculate the heat absorbed by the calorimeter.
Heat absorbed by calorimeter = C×T
11
where Cis the heat capacity of the calorimeter and Tis the change in tem-
perature.
= 18.2 J/
°
C×6.75
°
C
= 122.85 J
Step 2: Calculate the heat released by the compound during combustion.
Since the heat released by the compound is equal in magnitude but opposite in
sign to the heat absorbed by the calorimeter,
Heat released by compound = 122.85 J
Step 3: Convert the mass of the compound to moles. First, find the molar
mass of the compound by adding the atomic masses of the elements in the
compound. Let’s say the molar mass is 100 g/mol. Then, the moles of the
compound burned would be:
2.50 g
100 g/mol = 0.025 mol
Step 4: Calculate the enthalpy of combustion of the compound per gram.
Enthalpy of combustion per mole = Heat released by compound
moles of compound burned
=122.85 J
0.025 mol
=4,914 J/mol
Step 5: Convert the enthalpy of combustion to kJ/g.
Enthalpy of combustion per gram = 4,914 J/mol
1000 J/kJ ×2.50 g
=0.785 kJ/g
Therefore, the enthalpy of combustion of the compound is -0.785 kJ/g.
Question 14
Question
A 200 g piece of copper at 70C is placed into 400 g of water at 20C. The
final temperature of the mixture is 25C. Assuming no heat is lost to the sur-
roundings, what is the specific heat capacity of copper? Take the specific heat
capacity of water to be 4186 J/kg ·K and the specific heat capacity of copper to
be 386 J/kg ·K.
12
Solution
Step 1: Calculate the heat lost by the copper piece.
The heat lost by the copper piece can be calculated using the formula:
Qlost =mcT,
where: - mis the mass of the copper piece (200 g = 0.2 kg), - cis the specific
heat capacity of copper (386 J/kg ·K), - Tis the change in temperature of
the copper piece.
Given that the initial temperature of the copper piece is 70C and the final
temperature of the mixture is 25C:
T= (25C) (70C) = 45C.
Therefore, the heat lost by the copper piece is:
Qlost = (0.2 kg)(386 J/kg ·K)(45 K) = 3474 J.
Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
Qgained =mcT,
where: - mis the mass of the water (400 g = 0.4 kg), - cis the specific heat
capacity of water (4186 J/kg ·K), - Tis the change in temperature of the
water.
Given that the initial temperature of the water is 20C and the final tem-
perature of the mixture is 25C:
T= (25C) (20C) = 5C.
Therefore, the heat gained by the water is:
Qgained = (0.4 kg)(4186 J/kg ·K)(5 K) = 8368 J.
Step 3: Set up the energy conservation equation.
Since no heat is lost to the surroundings, the heat lost by the copper piece
is equal to the heat gained by the water:
Qlost =Qgained.
Therefore,
3474 J = 8368 J.
This equation has no solution, indicating that there may be an error in the
calculations. Let’s recheck the calculations.
13
Question 15
Question
A 50 g aluminum block at 100
°
C is dropped into 100 g of water at 20
°
C in a
calorimeter. If the final temperature of the system is 30
°
C, what is the specific
heat capacity of the aluminum block? Assume no heat is lost to the surroundings
and that the specific heat capacity of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat absorbed by the water using the formula:
Qwater =mwater ×cwater ×T
where: - mwater = 100 g (mass of water), - cwater = 4.18 J/g
°
C (specific heat
capacity of water), - T= (Tfinal Tinitial) = (30C20C) = 10C.
Calculating:
Qwater = 100 ×4.18 ×10 = 4180 J
Step 2: Use the heat absorbed by the water to find the specific heat capacity
of aluminum using the formula:
Qwater =maluminum ×caluminum ×T
Since no heat is lost to the surroundings, the heat absorbed by the aluminum
is equal to the heat lost by the water:
4180 J = 50 ×caluminum ×(30 100)
Step 3: Solve for the specific heat capacity of aluminum:
caluminum =4180
50 ×(30 100) =4180
50 × 70 =1.194 J/g
°
C
Therefore, the specific heat capacity of the aluminum block is 1.194 J/g
°
C.
Question 16
Question
A 50 g aluminum block is heated to 100
°
C and then placed into 200 g of water
at 25
°
C contained in a calorimeter. The final temperature of the system is 30
°
C.
Assume no heat is lost to the surroundings. Calculate the specific heat capacity
of aluminum. The specific heat capacity of water is 4.18 J/g
°
C.
14
Solution
Step 1: Calculate the heat energy absorbed by the aluminum block as it cools
down from 100
°
C to 30
°
C. The specific heat capacity of aluminum is 0.897 J/g
°
C.
Given: Mass of aluminum block (m) = 50 g Initial temperature of aluminum
block (Tinitial) = 100
°
C Final temperature of aluminum block (Tfinal) = 30
°
C
Specific heat capacity of aluminum (cAl) = 0.897 J/g
°
C
The formula to calculate heat energy (Q) is:
Q=mcAlT
where T=Tfinal Tinitial.
Substitute the values:
Q= (50 g)(0.897 J/g
°
C)(30 100)C
Q= (50 g)(0.897 J/g
°
C)(70)C=3141 J
Therefore, the heat energy absorbed by the aluminum block is 3141 J.
Step 2: Calculate the heat energy released by the aluminum block to the
water and the calorimeter. Since the system is isolated, the heat released by
the aluminum block will be absorbed by the water and the calorimeter.
The heat energy released by the aluminum block is equal to the heat energy
gained by the water and the calorimeter. Therefore:
QAluminum =QWater +QCalorimeter
Given: Mass of water (mwater) = 200 g Specific heat capacity of water (cwater)
= 4.18 J/g
°
C Change in temperature of water (∆Twater) = 30
°
C - 25
°
C = 5
°
C
Change in temperature of calorimeter (∆Tcalorimeter) = 30
°
C - 25
°
C = 5
°
C
Substitute the values into the equation:
3141 J = (200 g)(4.18 J/g
°
C)(5C)+(mcalorimeter)(ccalorimeter)(5C)
Since no information about the mass and specific heat capacity of the calorime-
ter is given, we cannot solve for the specific heat capacity of aluminum.
Question 17
Question
A 50 g piece of aluminum at 90
°
C is dropped into 200 g of water at 20
°
C in an
insulated calorimeter. Assuming no heat is lost to the surroundings, what will
be the final temperature of the system? (Specific heat capacity of aluminum is
0.897 J/g
°
C and specific heat capacity of water is 4.18 J/g
°
C.)
15
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water using the formula Q=mcT, where Qis the heat transferred, mis the
mass of the substance, cis the specific heat capacity, and Tis the change in
temperature. For aluminum: Qaluminum = (50 g)(0.897 J/g
°
C)(Tf90), where
Tfis the final temperature.
Step 2: For water: Qwater = (200 g)(4.18 J/g
°
C)(Tf20)
Step 3: Since the heat lost by the aluminum is equal to the heat gained by the
water (assuming no heat is lost to the surroundings), we can set them equal to
each other and solve for Tf: (50 g)(0.897 J/g
°
C)(Tf90) = (200 g)(4.18 J/g
°
C)(Tf
20)
Step 4: Now, solve for Tf: 44.85(Tf90) = 836(Tf20) 44.85Tf4036.5 =
836Tf16720
Step 5: Rearrange the equation and solve for Tf: 836Tf44.85Tf= 16720
4036.5 791.15Tf= 12683.5Tf=12683.5
791.15 Tf16
°
C
Therefore, the final temperature of the system will be approximately 16
°
C.
Question 18
Question
A 50.0 g sample of iron at 85.0
°
C is placed in 200.0 g of water at 20.0
°
C. The
specific heat capacity of iron is 0.449 J/(g
°
C) and the specific heat capacity of
water is 4.18 J/(g
°
C). Assuming no heat is lost to the surroundings, what will
be the final equilibrium temperature of the system?
Solution
Step 1: Calculate the heat absorbed by the iron as it cools down to the final
temperature. The heat absorbed by the iron is given by the formula:
qiron =m·ciron ·Tiron
where: - m= mass of the iron sample = 50.0 g - ciron = specific heat capac-
ity of iron = 0.449 J/(g
°
C) - Tiron = change in temperature of iron = final
temperature - initial temperature
Tiron =Tfinal 85.0
qiron = 50.0·0.449 ·(Tfinal 85.0)
Step 2: Calculate the heat absorbed by the water as it warms up to the final
temperature. The heat absorbed by the water is given by the formula:
qwater =m·cwater ·Twater
16
where: - m= mass of the water = 200.0 g - cwater = specific heat capacity of
water = 4.18 J/(g
°
C) - Twater =Tfinal 20.0
qwater = 200.0·4.18 ·(Tfinal 20.0)
Step 3: Since there is no heat lost to the surroundings, the heat lost by the
iron is equal to the heat gained by the water.
qiron =qwater
50.0·0.449 ·(Tfinal 85.0) = 200.0·4.18 ·(Tfinal 20.0)
Step 4: Solve for the final equilibrium temperature Tfinal.
50.0·0.449 ·(Tfinal 85.0) = 200.0·4.18 ·(Tfinal 20.0)
22.45 ·(Tfinal 85.0) = 836.0·(Tfinal 20.0)
22.45 ·Tfinal 22.45 ·85.0 = 836.0·Tfinal 836.0·20.0
22.45 ·Tfinal 1902.25 = 836.0·Tfinal 16720.0
813.55 = 813.55Tfinal
Tfinal =813.55
813.55 = 1.0
Therefore, the final equilibrium temperature of the system is 1.0
°
C.
Question 19
Question
A student conducts an experiment to determine the specific heat capacity of a
metal sample using a calorimeter. The student places a 200 g aluminum sample
at 100C into 400 g of water at 20C. After thermal equilibrium is reached, the
final temperature of the system is 25C. Calculate the specific heat capacity of
the metal.
Solution
Step 1: Calculate the heat transfer from the aluminum to the water. The heat
lost by the aluminum sample is equal to the heat gained by the water:
malTal=mwcwTw
where mal= mass of aluminum (kg), Tal= change in temperature of alu-
minum (K), mw= mass of water (kg), cw= specific heat capacity of water
(J/(kg·K)), Tw= change in temperature of water (K).
Given: mal= 0.2 kg, Tal= 100 25 = 75 K, mw= 0.4 kg, cw= 4186
J/(kg·K), and Tw= 25 20 = 5 K.
17
Substitute the values and solve for l:
0.2×l×75 = 0.4×4186 ×5
l=0.4×4186 ×5
0.2×75
l= 4186 J/kg
Step 2: Calculate the specific heat capacity of the metal. The heat lost by
the metal is equal to the heat gained by the water:
malcalTal=mwcwTreservoir
where cal= specific heat capacity of the metal (J/(kg·K)), Treservoir = change
in temperature of the water after reaching equilibrium (K).
Given: mal= 0.2 kg, cal=l= 4186 J/(kg·K), Tal= 75 K, mw= 0.4 kg,
cw= 4186 J/(kg·K), and Treservoir = 25 20 = 5 K.
Substitute the values and solve for cal:
0.2×4186 ×75 = 0.4×4186 ×5
cal=0.4×4186 ×5
0.2×75
cal= 4186 J/kg
Therefore, the specific heat capacity of the metal sample is 4186 J/(kg·K).
Question 20
Question
A 50 g piece of iron at 150
°
C is dropped into 200 g of water at 25
°
C in a 100
g aluminum cup. The final temperature of the system is 40
°
C. Assuming no
heat is lost to the surroundings, determine the specific heat capacity of iron.
The specific heat capacities of iron, water, and aluminum are 0.45 J/g
°
C, 4.18
J/g
°
C, and 0.897 J/g
°
C, respectively.
Solution
Step 1: Calculate the heat gained by the water and the aluminum cup. The
heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·T
Where: - mwater = 200 g (mass of water) - cwater = 4.18 J/g
°
C (specific heat
capacity of water) - Tf= 40
°
C (final temperature) - Ti= 25
°
C (initial temper-
ature)
Twater =TfTi= 40 25 = 15
°
C
18
Qwater = 200 g ×4.18 J/g
°
C×15
°
C
Qwater = 12540 J
The heat gained by the aluminum cup can be calculated using the same
formula:
Qaluminum =maluminum ·caluminum ·Taluminum
Where: - maluminum = 100 g (mass of aluminum) - caluminum = 0.897 J/g
°
C
(specific heat capacity of aluminum) - Taluminum = 40 25 = 15
°
C
Qaluminum = 100 g ×0.897 J/g
°
C×15
°
C
Qaluminum = 1345.5 J
Step 2: Calculate the heat lost by the iron to the water and the aluminum
cup. Since no heat is lost to the surroundings, the heat lost by the iron is equal
to the heat gained by the water and the aluminum cup.
Qiron =Qwater +Qaluminum
Qiron = 12540 J + 1345.5 J
Qiron = 13885.5 J
Step 3: Use the formula for heat
Qiron =miron ·ciron ·Tiron
Since the mass of the iron is 50 g, we can solve for the specific heat capacity of
iron.
ciron =Qiron
miron ·Tiron
ciron =13885.5 J
50 g ×(40 150)
°
C
ciron =13885.5
50 × 110
ciron =2.523 J/g
°
C
Therefore, the specific heat capacity of iron is 2.523 J/g
°
C .
Question 21
Question
A 50 g piece of aluminum at 100
°
C is dropped into 100 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Assuming no heat is
lost to the surroundings, calculate the specific heat capacity of aluminum.
19
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water.
The heat lost by the aluminum can be calculated using the formula:
QAluminum =m×cAluminum ×TAluminum
Where: m= 50 g (mass of aluminum), cAluminum = 0.903 J/g
°
C (specific
heat capacity of aluminum), and TAluminum =TfTi= 22 100 =
78
°
C (change in temperature of the aluminum).
Substituting the values, we get:
QAluminum = 50 ×0.903 ×(78) = 3525.6 J
The heat gained by the water can be calculated using the formula:
QWater =m×cWater ×TWater
Where: m= 100 g (mass of water), cWater = 4.18 J/g
°
C (specific heat
capacity of water), and TWater =TfTi= 22 20 = 2
°
C (change in
temperature of the water).
Substituting the values, we get:
QWater = 100 ×4.18 ×2 = 836 J
Question 22
Question
A 100 g piece of aluminum at 80
°
C is placed in a calorimeter containing 200
g of water at 20
°
C. If the final temperature of the system is 25
°
C, what is the
heat capacity of the calorimeter? The specific heat capacities of aluminum and
water are 0.90 J/g
°
C and 4.18 J/g
°
C, respectively.
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water. The heat lost by the aluminum can be calculated using the formula:
QAl =mAl ·cAl ·T
where: - mAl is the mass of aluminum - cAl is the specific heat capacity of
aluminum - Tis the temperature change of aluminum
Substitute the given values into the formula:
QAl = 100 g ·0.90 J/g
°
C·(25 80)
°
C
20
QAl = 100 ·0.90 ·(55)
QAl =4950 J
The heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·T
where: - mwater is the mass of water - cwater is the specific heat capacity of water
- Tis the temperature change of water
Substitute the given values into the formula:
Qwater = 200 g ·4.18 J/g
°
C·(25 20)
°
C
Qwater = 200 ·4.18 ·5
Qwater = 4180 J
Step 2: Use the principle of conservation of energy to find the heat capacity
of the calorimeter. The heat lost by the aluminum is equal to the heat gained
by the water and calorimeter:
QAl =Qwater +Qcalorimeter
Substitute the calculated values:
4950 = 4180 + Qcalorimeter
Qcalorimeter =4950 4180
Qcalorimeter =9130 J
Step 3: Calculate the heat capacity of the calorimeter using the formula:
Qcalorimeter =Ccalorimeter ·T
where: - Ccalorimeter is the heat capacity of the calorimeter - T=Tfinal
Tinitial = 25 20 = 5
°
C
Substitute the known values:
9130 = Ccalorimeter ·5
Ccalorimeter =9130
5
Ccalorimeter =1826 J/
°
C
Therefore, the heat capacity of the calorimeter is 1826 J/
°
C .
Question 23
Question
A 50 g piece of iron at 80
°
C is dropped into a calorimeter containing 100 g of
water at 20
°
C. If the final temperature of the system is 30
°
C, what is the heat
capacity of the calorimeter? Assume no heat is lost to the surroundings.
21
Solution
Step 1: Calculate the heat gained by the water to warm up from 20
°
C to 30
°
C.
Step 2: Calculate the heat lost by the iron to cool down from 80
°
C to 30
°
C.
Step 3: Use the fact that the total heat lost by the iron is equal to the total
heat gained by the water and the calorimeter to find the heat capacity of the
calorimeter.
Step 1: The specific heat capacity of water is 4.18 J/g
°
C.
The heat gained by the water can be calculated using the formula:
qwater =m·c·T
where: - mis the mass of the water (100 g), - cis the specific heat capacity of
water (4.18 J/g
°
C), - Tis the change in temperature (from 20
°
C to 30
°
C).
Substitute the values into the formula:
qwater = 100 g ×4.18 J/g
°
C×(30 20)
°
C
qwater = 100 g ×4.18 J/g
°
C×10
°
C
qwater = 4180 J
Therefore, the heat gained by the water is 4180 J.
Step 2: The specific heat capacity of iron is 0.45 J/g
°
C.
The heat lost by the iron can be calculated using the formula:
qiron =m·c·T
where: - mis the mass of the iron (50 g), - cis the specific heat capacity of iron
(0.45 J/g
°
C), - Tis the change in temperature (from 80
°
C to 30
°
C).
Substitute the values into the formula:
qiron = 50 g ×0.45 J/g
°
C×(30 80)
°
C
qiron = 50 g ×0.45 J/g
°
C× 50
°
C
qiron =1125 J
Therefore, the heat lost by the iron is -1125 J. Note the negative sign indi-
cates heat loss.
Step 3: The total heat gained by the water and the calorimeter is equal
to the absolute value of the total heat lost by the iron. Therefore, |qwater +
qcalorimeter|=|qiron|.
Substitute the values of qwater and qiron into the equation:
|4180 J + qcalorimeter|=| 1125 J|
4180 J + qcalorimeter = 1125 J
qcalorimeter = 1125 J 4180 J
qcalorimeter =3055 J
Therefore, the heat capacity of the calorimeter is 3055 J.
22
Question 24
Question
A 50 g piece of copper at 150
°
C is placed in 200 g of water at 20
°
C. Assuming
no heat is lost to the surroundings, what will be the final temperature of the
system? (Specific heat capacity of copper = 0.386 J/g
°
C, specific heat capacity
of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by copper and the heat gained by water using
the formula Q=mcTwhere Qis heat, mis mass, cis specific heat capacity,
and Tis the change in temperature.
For copper: Qcopper = (50 g)(0.386 J/g
°
C)(Tf150C)
For water: Qwater = (200 g)(4.18 J/g
°
C)(Tf20C)
Since no heat is lost or gained from the surroundings, we have Qcopper =
Qwater
Step 2: Set up the equation: (50 g)(0.386 J/g
°
C)(Tf150C) = (200 g)(4.18 J/g
°
C)(Tf
20C)
Step 3: Solve for the final temperature Tf: 50(0.386)(Tf150) = 200(4.18)(Tf
20)
19.3Tf2895 = 836Tf+ 16720
855Tf= 19615
Tf22.96C
Therefore, the final temperature of the system will be approximately 22.96C.
Question 25
Question
A 50 g piece of aluminum at 100
°
C is placed in 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Assuming no heat
is lost to the surroundings, calculate the specific heat capacity of aluminum.
(Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water. The heat lost by the aluminum is equal to the heat gained by the water.
Therefore, we have:
mAl ·cAl ·(TfTi) = mw·cw·(TfTi)
Substitute the given values:
50 g ·cAl ·(22 100) = 200 g ·4.18 J/g
°
C·(22 20)
23
Simplify and solve for cAl.
Step 2: Calculate the specific heat capacity of aluminum.
50 g ·cAl ·(78) = 200 g ·4.18 J/g
°
C·2
cAl =200 g ·4.18 J/g
°
C·2
50 g ·78
cAl =1672 J
3900 g ·
°
C
cAl 0.43 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 0.43 J/g
°
C.
Question 26
Question
A student wants to determine the specific heat capacity of a metal using calorime-
try. They take a 150 g block of the metal at an initial temperature of 200
°
C
and place it in 200 g of water at an initial temperature of 20
°
C contained in a
calorimeter. The final temperature of the system is 23
°
C. What is the specific
heat capacity of the metal? Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat absorbed by the water. The heat absorbed by the
water can be calculated using the formula:
Qwater =m·c·T
where: - m= 200 g is the mass of water, - c= 4.18 J/g
°
C is the specific heat
capacity of water, - T=TfTi= 23C20C= 3Cis the change in temper-
ature.
Substitute the values into the formula:
Qwater = 200 g ×4.18 J/g
°
C×3C= 2508 J
Step 2: Calculate the heat lost by the metal. The heat lost by the metal
is equal to the heat absorbed by the water, as there is no heat lost to the
surroundings. We can use the formula:
Qmetal =Qwater
Qmetal =2508 J
Step 3: Calculate the specific heat capacity of the metal. The heat lost by
the metal can be calculated using the formula:
Qmetal =m·cmetal ·T
24
where: - m= 150 g is the mass of the metal and - T=TfTi= 200C23C=
177Cis the change in temperature.
Substitute the known values and solve for cmetal:
2508 J = 150 g ×cmetal ×177C
cmetal =2508 J
150 g ×177C
cmetal 0.094 J/g
°
C
Therefore, the specific heat capacity of the metal is approximately -0.094
J/g
°
C.
Question 27
Question
A piece of metal with mass 0.25 kg is heated to 100C and then placed in 0.5
kg of water at 20C. The final temperature of the system is 25C. If the specific
heat capacity of the metal is 450 J/kg·K, determine the specific heat capacity
of the metal.
Solution
Step 1: Calculate the heat lost by the metal as it cools down from 100C to
the final temperature. The heat lost by the metal can be calculated using the
formula:
Qmetal =mcT
where: - mis the mass of the metal, - cis the specific heat capacity of the metal,
- Tis the change in temperature of the metal.
Substitute the values:
Qmetal = 0.25 kg ×450 J/kg·K×(100 25) K
Qmetal = 0.25 ×450 ×75
Qmetal = 8437.5 J
Step 2: Calculate the heat gained by the water as it heats up from 20C to
the final temperature. The heat gained by the water can be calculated using
the formula:
Qwater =mcT
where: - mis the mass of the water, - cis the specific heat capacity of water, -
Tis the change in temperature of the water.
Substitute the values:
Qwater = 0.5 kg ×4186 J/kg·K×(25 20) K
25
Qwater = 0.5×4186 ×5
Qwater = 10465 J
Step 3: Since energy is conserved, the heat lost by the metal is equal to the
heat gained by the water.
Qmetal =Qwater
8437.5 = 10465
8437.5 = 10465
Note: We reached an error in our calculation, please check the heat lost by
the metal and heat gained by the water and make sure they are equal.
Question 28
Question
A piece of metal of mass 0.5 kg at 150
°
C is placed in 1 kg of water at 20
°
C
contained in a 1 kg aluminum calorimeter cup at 20
°
C. The final temperature
of the system is 30
°
C. Given the specific heat capacities of the metal and water
are 0.5 J/g
°
C and 4.18 J/g
°
C respectively, calculate the specific heat capacity
of the metal.
Solution
Step 1: Calculate the heat gained by the water and the calorimeter cup.
The heat gained by the water and calorimeter cup can be calculated using the
formula:
Q=mcT
where Qis the heat gained, mis the mass, cis the specific heat capacity, and
Tis the change in temperature.
For water:
Qwater = (1 kg ×1000 g/kg ×4.18 J/g
°
C×(30 20)
°
C)
Qwater = 41800 J
For the calorimeter cup (assume specific heat capacity of aluminum is 0.9
J/g
°
C):
Qcup = (1 kg ×1000 g/kg ×0.9 J/g
°
C×(30 20)
°
C)
Qcup = 900 J
Step 2: Calculate the heat lost by the metal.
The heat lost by the metal is equal to the sum of the heat gained by the water
and calorimeter cup:
Qmetal =Qwater +Qcup
26
Qmetal = 41800 J + 900 J
Qmetal = 42700 J
Step 3: Calculate the specific heat capacity of the metal.
We know that the heat lost by the metal is given by:
Qmetal =mcT
Substitute the values we have:
42700 = 0.5×c×(30 150)
42700 = 60c
c=42700
60
c 711.67 J/kg
°
C
Therefore, the specific heat capacity of the metal is approximately 711.67
J/kg
°
C.
Question 29
Question
A piece of copper with a mass of 200 g is heated to a temperature of 100
°
C
and then dropped into a calorimeter containing 500 g of water at an initial
temperature of 20
°
C. If the final temperature of the system is 25
°
C, calculate
the specific heat capacity of copper. Assume no heat is lost to the surroundings
and the specific heat capacity of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
Heat lost by copper = Heat gained by water
Step 2: The heat lost by the copper is given by the formula:
Qcopper =mcT
where: - mis the mass of copper (200 g) - cis the specific heat capacity of copper
(unknown) - Tis the change in temperature of the copper (final temperature
- initial temperature)
Qcopper = (200 g)(c)(25 100)
Qcopper =15000cJ
27
Step 3: The heat gained by the water is given by the formula:
Qwater =mcT
where: - mis the mass of water (500 g) - cis the specific heat capacity of water
(4.18 J/g
°
C) - Tis the change in temperature of the water (final temperature
- initial temperature)
Qwater = (500 g)(4.18 J/g
°
C)(25 20)
Qwater = 2090 J
Step 4: Set the heat lost by the copper equal to the heat gained by the water:
15000c= 2090
Step 5: Solve for the specific heat capacity of copper:
c=2090
15000 =0.139 J/g
°
C
Therefore, the specific heat capacity of copper is -0.139 J/g
°
C. The negative
Question 30
Question
A 50 g piece of aluminum initially at 80
°
C is dropped into 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 capacity of aluminum = 0.9 J/g
°
C, specific heat
capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the aluminum piece. The heat lost by the
aluminum can be calculated using the formula:
qaluminum =mcT
where: - m= 50 g (mass of aluminum) - c= 0.9 J/g
°
C (specific heat capacity
of aluminum) - T=Tfinal Tinitial =Tfinal 80
°
C
Step 2: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the formula:
qwater =mcT
where: - m= 200 g (mass of water) - c= 4.18 J/g
°
C (specific heat capacity of
water) - T=Tfinal Tinitial =Tfinal 20
°
C
28
Step 3: Set up the conservation of energy equation: Since the system is
isolated and no heat is lost to the surroundings, the heat lost by the aluminum
must be equal to the heat gained by the water, so we have:
qaluminum =qwater
Step 4: Substitute the calculated values and solve for the final temperature,
Tfinal.
maluminumcaluminum(Tfinal 80) = mwatercwater(Tfinal 20)
Step 5: Simplify the equation and solve for Tfinal.
50 ×0.9×(Tfinal 80) = 200 ×4.18 ×(Tfinal 20)
45(Tfinal 80) = 836(Tfinal 20)
45Tfinal 3600 = 836Tfinal 16720
791Tfinal = 13120
Tfinal =13120
791 16.58
°
C
Therefore, the final temperature of the system is approximately 16.58
°
C.
Question 31
Question
A 50 g aluminum block is heated to 100
°
C and then placed in 200 g of water
at 20
°
C inside a calorimeter. The final temperature of the system is 25
°
C.
Assuming no heat is lost to the surroundings, calculate the specific heat capacity
of the aluminum block. (Specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat absorbed by the water during the cooling process.
The heat absorbed by the water can be calculated using the formula:
qwater =mwater ×cwater ×T
where: - mwater = 200 g is the mass of water, - cwater = 4.18 J/g
°
C is the specific
heat capacity of water, and - T= 25 20 = 5
°
C is the temperature change
of the water. Substitute these values into the formula:
qwater = 200 ×4.18 ×5 = 4180 J
Step 2: Calculate the heat lost by the aluminum block. The heat lost by the
aluminum block is equal in magnitude to the heat gained by the water, so we
have:
qaluminum =qwater =4180 J
29
Step 3: Calculate the heat lost by the aluminum block using the formula:
qaluminum =maluminum ×caluminum ×T
where: - maluminum = 50 g is the mass of the aluminum block, - caluminum is
the specific heat capacity of aluminum (which we want to find), and - T=
100 25 = 75
°
C is the temperature change of the aluminum block.
Equating the heat lost by the aluminum block in step 2 and step 3, we have:
50 ×caluminum ×75 = 4180
caluminum =4180
50 ×75
caluminum =1.40 J/g
°
C
Therefore, the specific heat capacity of the aluminum block is 1.40 J/g
°
C.
Question 32
Question
A piece of metal with mass 0.2 kg is heated to 100
°
C and then placed in 0.5 kg
of water at 25
°
C in an insulated container. The final temperature of the system
is 30
°
C. If the specific heat capacity of the metal is 450 J/kg
°
C and the specific
heat capacity of water is 4186 J/kg
°
C, calculate the specific heat capacity of the
container if none of the heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the metal and gained by the water using the
formula:
Qmetal =Qwater
Step 2: Calculate the heat lost by the metal using the formula:
Qmetal =mcT
where: m= 0.2 kg (mass of the metal), c= 450 J/kg
°
C (specific heat capacity
of the metal), T= 100 30 = 70
°
C (final temperature - initial temperature).
Step 3: Substitute the values into the formula to find Qmetal:
Qmetal = 0.2×450 ×70
Step 4: Calculate the heat gained by the water using the formula:
Qwater =mcT
where: m= 0.5 kg (mass of the water), c= 4186 J/kg
°
C (specific heat capacity
of water), T= 30 25 = 5
°
C (final temperature - initial temperature).
30
Step 5: Substitute the values into the formula to find Qwater:
Qwater = 0.5×4186 ×5
Step 6: Set Qmetal =Qwater and solve for the specific heat capacity of the
container:
0.2×450 ×70 = 0.5×4186 ×5
Step 7: Solve for the specific heat capacity of the container.
Question 33
Question
A 50 g piece of aluminum initially at 100C is placed in a calorimeter containing
200 g of water at 20C. The final temperature of the system is 25C. Assuming
no heat is lost to the surroundings, what is the specific heat capacity of the
aluminum? (Specific heat capacity of water is 4.18 J/gC)
Solution
Step 1: Calculate the heat lost by the aluminum and the heat gained by the
water using the formula:
q=m·c·T
where: q= heat energy in Joules (J), m= mass in grams (g), c= specific
heat capacity in J/gC, T= change in temperature in C.
The heat lost by aluminum is equal to the heat gained by water, since no
heat is lost to the surroundings.
For the aluminum:
qAl = (50 g) ·cAl ·(25 100)C
For the water:
qH2O = (200 g) ·4.18 J/gC·(25 20)C
Since qAl =qH2O:
(50 g) ·cAl ·(75)C = (200 g) ·4.18 J/gC·5C
Simplify and solve for cAl in terms of the given values.
Question 34
Question
A 100 g piece of copper at 200
°
C is placed into a calorimeter containing 200 g of
water at 20
°
C. The final temperature of the system is 25
°
C. Assuming no heat
is lost to the surroundings, what is the specific heat capacity of the calorimeter?
31
Solution
Given: Mass of copper (mc) = 100 g = 0.1 kg
Initial temperature of copper (Tc,i) = 200
°
C
Specific heat capacity of copper (cc) = 390 J/kg ·
°
C
Mass of water in calorimeter (mw) = 200 g = 0.2 kg
Initial temperature of water (Tw,i) = 20
°
C
Final temperature of the system (Tf) = 25
°
C
Specific heat capacity of water (cw) = 4186 J/kg ·
°
C
We will assume that no heat is lost to the surroundings, so the total heat
lost by the copper is equal to the total heat gained by the water:
mc·cc·(Tc,i Tf) = mw·cw·(TfTw,i)
Step 1: Calculate the heat lost by the copper (Qc) and the heat
gained by the water (Qw).
Qc=mc·cc·(Tc,i Tf)
Qw=mw·cw·(TfTw,i)
Substitute the given values:
Qc= 0.1 kg ·390 J/kg ·
°
C·(200 25)
°
C
Qw= 0.2 kg ·4186 J/kg ·
°
C·(25 20)
°
C
Step 2: Equate the heat lost and gained equations. Solve for the
specific heat capacity of the calorimeter.
mc·cc·(Tc,i Tf) = mw·cw·(TfTw,i)
0.1·390 ·(200 25) = 0.2·4186 ·(25 20)
0.1·390 ·175 = 0.2·4186 ·5
6825 = 4186 ·1·c
c=6825
4186
Therefore, the specific heat capacity of the calorimeter is 6825
4186 1.63 J/g·
°
C.
Question 35
Question
A 50 g aluminum block at 80
°
C is dropped into 200 g of water at 20
°
C in a
calorimeter. The final temperature of the system is 22
°
C. Given the specific
heat capacity of water is 4.18 J/g
°
C, determine the specific heat capacity of
aluminum. Assume no heat is lost to the surroundings and that the specific
heat capacity of aluminum is constant over the temperature range involved.
32
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the equation:
qwater =m·c·T
where: - mis the mass of water (200 g) - cis the specific heat capacity of water
(4.18 J/g
°
C) - Tis the change in temperature of the water (22C20C= 2C)
Substitute the given values to find qwater:
qwater = 200 g ×4.18 J/g
°
C×2C= 1672 J
Step 2: Calculate the heat lost by the aluminum. The heat lost by the
aluminum block must be equal to the heat gained by the water, since no heat
is lost to the surroundings. We can express this as:
qaluminum =qwater
Since the heat lost by the aluminum block can be calculated using its specific
heat capacity (caluminum), mass (maluminum = 50 g), and temperature change
(∆Taluminum =TfTi= 22C80C=58C):
qaluminum =maluminum ·caluminum ·Taluminum
Substitute the known values to find caluminum:
1672 J = 50 g ×caluminum × 58C
Solve for caluminum:
caluminum =1672 J
50 g × 58C0.57 J/g
°
C
Therefore, the specific heat capacity of aluminum is approximately 0.57 J/g
°
C.
33