PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 5
Liberty University
Question 1
Question
A copper sphere of radius 10 cm is heated to a temperature of 100◦C. Calculate
the amount of heat required to increase the temperature of the sphere to 300◦C.
Assume the specific heat capacity of copper is 0.385 J/g◦C.
Solution
Step 1: Calculate the mass of the copper sphere using its density and volume.
Given that the density of copper is 8.96 g/cm3, the volume of the sphere is:
V=4
3πr3=4
3π(10 cm)3
The mass of the sphere is then:
m= density ×volume = 8.96 g/cm3×4
3π(10 cm)3
Step 2: Calculate the initial temperature of the copper sphere. The initial
temperature is 100◦C.
Step 3: Calculate the final temperature of the copper sphere. The final
temperature is 300◦C.
Step 4: Calculate the change in temperature.
∆T= final temperature −initial temperature = 300◦C−100◦C
Step 5: Calculate the amount of heat required using the formula:
Q=mc∆T
where mis the mass of the sphere, cis the specific heat capacity of copper, and
∆Tis the change in temperature.
Now, substitute the values calculated in the previous steps into the formula
to find the amount of heat required.
Question 2
Question
A copper rod of length 2 m and diameter 1 cm is heated from 20
°
C to 80
°
C. If
the rod has thermal conductivity k= 380 W/(m·K) and specific heat capacity
c= 390 J/(kg·K), calculate the amount of heat transferred to the rod during
this process. Assume the density of copper is 8900 kg/m3.
Solution
Step 1: Calculate the cross-sectional area of the copper rod. The radius of the
rod is given by r=1
2×0.01 m = 0.005 m. Therefore, the cross-sectional area
A=πr2.
Step 2: Calculate the volume of the copper rod. The volume of the rod is
given by V=A×length = π×(0.005)2×2 m3.
Step 3: Calculate the mass of the copper rod. The mass of the rod can be
calculated using the formula m= density ×volume.
Step 4: Calculate the energy required to increase the temperature of the
copper rod. The energy required is given by Q=mc∆T, where ∆T= 80 −20
and mis the mass of the rod.
Step 5: Substitute the values into the formula. Substitute the values of m,
c, ∆Tinto the formula Q=mc∆Tto find the amount of heat transferred to
the rod. Remember to convert the temperature from Celsius to Kelvin.
Question 3
Question
A steel rod with a length of 2 meters and a diameter of 1 cm is initially at a
temperature of 100◦C. If the rod is placed in a water bath with a temperature
of 20◦C, how long will it take for the rod to reach a temperature of 30◦C? The
thermal conductivity of steel is 50 W/(m ·K), the heat capacity of steel is 500
J/kg·K, and the density of steel is 8000 kg/m3.
Solution
Step 1: Calculate the cross-sectional area of the steel rod. The cross-sectional
area of the rod can be calculated using the formula for the area of a circle:
A=πr2, where ris the radius of the rod. Given that the diameter of the rod
is 1 cm, the radius rcan be calculated as 0.5 cm or 0.005 m. Therefore, the
cross-sectional area of the rod is:
A=π(0.005)2≈7.85 ×10−5m2
Step 2: Calculate the volume of the steel rod. The volume of the rod can
be calculated using the formula for the volume of a cylinder: V=A×L, where
2
Lis the length of the rod. Given that the length of the rod is 2 meters, the
volume of the rod is:
V= 7.85 ×10−5×2=1.57 ×10−4m3
Step 3: Calculate the mass of the steel rod. The mass of the rod can be
calculated using the formula m= density ×volume. Given that the density of
steel is 8000 kg/m3, the mass of the rod is:
m= 8000 ×1.57 ×10−4= 1.256 kg
Step 4: Calculate the thermal energy required to heat the rod from 20◦C
to 30◦C. The thermal energy Qrequired can be calculated using the formula
Q=mc∆T, where mis the mass of the rod, cis the specific heat capacity
of steel, and ∆Tis the change in temperature. Given that the specific heat
capacity of steel is 500 J/kg·K, the change in temperature is 10◦C, and the
mass of the rod is 1.256 kg, the thermal energy required is:
Q= 1.256 ×500 ×10 = 6280 J
Step 5: Calculate the rate of heat transfer through the rod. The rate of
heat transfer Pcan be calculated using the formula P=kA∆T
L, where kis the
thermal conductivity of steel, Ais the cross-sectional area of the rod, ∆Tis the
temperature difference between the rod and the water bath, and Lis the length
of the rod. Given that the thermal conductivity of steel is 50 W/(m·K), the
temperature difference is 80◦C, the cross-sectional area is 7.85 ×10−5m2, and
the length of the rod is 2 meters, the rate of heat transfer is:
P= 50 ×7.85 ×10−5×80
2= 0.314 W
Step 6: Calculate the time taken for the rod to reach 30◦C. The time taken
tcan be calculated using the formula Q=P t, where Qis the thermal energy
required and Pis the rate of heat transfer. Substitute the values of Qand P
into the formula:
6280 = 0.314t
Solving for t:
t≈6280
0.314 ≈20000 s
Therefore, it will take approximately 20000 seconds for the rod to reach a tem-
perature of 30◦C.
Question 4
Question
A copper bar with a length of 2 meters and a cross-sectional area of 10 cm2is
initially at a temperature of 100◦C. If 5000 J of heat energy is added to the bar,
calculate the final temperature of the bar given that the specific heat capacity
of copper is 390 J/kg◦C and its density is 8900 kg/m3.
3
Solution
Step 1: Calculate the mass of the copper bar. The formula to calculate mass is
given by:
mass = density ×volume
The volume of the copper bar can be calculated using its length and cross-
sectional area:
volume = length ×cross-sectional area
Substitute the given values:
volume = 2 m ×10 ×10−4m2= 2 ×10−3m3
mass = 8900 kg/m3×2×10−3m3= 17.8 kg
Step 2: Calculate the change in temperature. The heat energy added to
the copper bar causes a temperature rise. The change in temperature can be
calculated using the formula:
Q=mc∆T
where: - Q= 5000 J (heat energy added) - m= 17.8 kg (mass of the copper
bar) - c= 390 J/kg◦C (specific heat capacity of copper) - ∆Tis the change in
temperature (unknown)
Solve for ∆T:
5000 = 17.8×390 ×∆T
∆T=5000
17.8×390 ≈7.95◦C
Step 3: Calculate the final temperature. The final temperature of the copper
bar is the initial temperature plus the change in temperature:
Final temperature = 100 + 7.95 = 107.95◦C
Therefore, the final temperature of the copper bar is 107.95◦C after 5000 J
of heat energy is added.
Question 5
Question
A metal rod of length 50 cm and thermal conductivity 200 W/(m2K) is being
heated at one end with a constant heat flux of 500 W/m2. The ambient tem-
perature surrounding the rod is 20
°
C. If the rod has a uniform cross-sectional
area of 2 cm2, calculate the temperature at a distance of 20 cm from the heated
end after 5 minutes.
Given: Thermal conductivity k= 200 W/(m2K), Heat flux q= 500 W/m2,
Rod length L= 0.5 m, Cross-sectional area A= 2 ×10−4m2, Time t= 5
minutes, Ambient temperature Ta= 20
°
C.
4
Solution
Step 1: Calculate the heat transfer rate through the metal rod using Fourier’s
law:
q=−k·AdT
dx
Where: q= Heat flux = 500 W/m2,k= Thermal conductivity = 200
W/(m2K), A= Cross-sectional area = 2×10−4m2,dT
dx = Temperature gradient.
Given that the heat flux q= 500 W/m2is constant, we can rewrite the
equation as:
−500 = −200 ·2×10−4·dT
dx
Solving for dT
dx gives:
dT
dx =500
200 ·2×10−4
dT
dx = 125,000 K/m
Step 2: Integrate to find the temperature distribution in the rod:
ZdT =Z125,000dx
T= 125,000x+C
Where Cis the integration constant.
Step 3: Apply the boundary condition that the temperature at the heated
end (x= 0) is Th(unknown) and at the ambient end (x= 0.5 m) is Ta= 20
°
C:
Th= 125,000 ×0 + C
Ta= 125,000 ×0.5 + C
20 = 62,500 + C
C=−62,480
Step 4: Substitute back into the temperature distribution equation to find
the temperature profile:
T= 125,000x−62,480
Step 5: Calculate the temperature at x= 0.2 m:
5
T(0.2) = 125,000 ×0.2−62,480
T(0.2) = 25,000 −62,480 = −37,480 K
Therefore, the temperature at a distance of 20 cm from the heated end after
5 minutes is −37,480 K.
Question 6
Question
A copper block at 100
°
C is dropped into a calorimeter containing 200 g of water
at 20
°
C. The final temperature of the system is found to be 30
°
C. If the specific
heat capacity of water is 4.18 J/g
°
C, determine the mass of the copper block.
Assume no heat is lost to the surroundings.
Solution
Step 1: Determine the heat gained by the water and calorimeter. The heat
gained by the water and calorimeter is equal to the heat lost by the copper block.
The formula for heat transfer is Q=mc∆T, where Qis the heat transferred, m
is the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Let mcbe the mass of the copper block. The heat lost by the copper block
is equal to the heat gained by the water and calorimeter:
mc·cc·∆T= (mw+mcal)·cw·∆T
Step 2: Substitute the known values into the equation. Substitute the given
mass and specific heat capacity values into the equation:
mc·cc·(100 −30) = (200 + 200) ·4.18 ·(30 −20)
Step 3: Solve for the mass of the copper block. Simplify the equation and
solve for mc:
mc·0.387 = 800 ·4.18
mc=800 ·4.18
0.387
mc≈8678.27 g
Therefore, the mass of the copper block is approximately 8678.27 g.
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Question 7
Question
A steel rod of length 2.0 m and diameter 1.0 cm is initially at a temperature
of 30
°
C. If the rod is heated until its temperature reaches 100
°
C, determine the
amount of heat transferred to the rod. Assume the linear expansion coefficient
of steel is 1.2×10−5K−1and the specific heat capacity of steel is 450 J/kg ·K.
Solution
Step 1: Calculate the initial volume of the steel rod. The initial volume of
the steel rod can be calculated using the formula for the volume of a cylinder:
V=πr2h, where ris the radius and his the length. Given that r= 0.5 cm
(since the diameter is 1 cm), we have r= 0.005 m and h= 2.0 m. Plugging
these values into the formula, we get:
V=π(0.005 m)2·2.0 m
Step 2: Calculate the final volume of the steel rod after heating. When
heated, the steel rod will expand due to the increase in temperature. The
change in length ∆Lof the rod can be calculated using the formula:
∆L=αL0∆T
where αis the linear expansion coefficient, L0is the initial length (2.0 m), and
∆Tis the change in temperature (100
°
C - 30
°
C). Using this information, we can
calculate the final length Lfas:
Lf=L0+ ∆L
Step 3: Calculate the final volume of the steel rod. Once we have the final
length of the rod, we can calculate the final volume using the formula for the
volume of a cylinder:
Vf=πr2Lf
Step 4: Calculate the mass of the steel rod. The mass of the steel rod can
be calculated using the formula:
m= density ×V
Given that the density of steel is approximately 8000 kg/m3, we can calculate
the mass of the rod.
Step 5: Calculate the heat transferred to the rod. The amount of heat
transferred to the rod can be calculated using the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity of steel, and ∆T
is the change in temperature. Substitute the values into the formula to find the
heat transferred.
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Question 8
Question
A copper rod of length Land cross-sectional area Ais used to connect two
large metal blocks at different temperatures. The ends of the rod are in thermal
contact with the blocks. If the rod is initially at a uniform temperature T0and
the steady-state temperature distribution is linear, find the rate at which heat
flows through the rod. Assume that the thermal conductivity of copper is k.
Solution
The rate at which heat flows through the rod can be found using Fourier’s Law of
Heat Conduction, which states that the rate of heat transfer through a material
is directly proportional to the negative gradient of temperature in the direction
of heat flow.
Step 1: Determine the temperature distribution along the rod.
Let T(x) be the temperature at a distance xfrom the end in contact with
the hotter block. Since the steady-state temperature distribution is linear, we
have
T(x) = Th−Th−T0
Lx,
where This the temperature of the hotter block.
Step 2: Calculate the temperature gradient.
The temperature gradient ∆T
∆xcan be found as
dT
dx =−Th−T0
L.
Step 3: Apply Fourier’s Law of Heat Conduction.
According to Fourier’s Law, the rate at which heat flows through the rod
(˙
Q) is given by
˙
Q=−kAdT
dx .
Substitute the expression for dT
dx into the equation above to obtain
˙
Q=kATh−T0
L.
Step 4: Finalize the answer.
Therefore, the rate at which heat flows through the rod is
˙
Q=kATh−T0
L.
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Question 9
Question
A copper block of mass 0.2 kg is heated to a temperature of 100◦C. The block
is then placed in 0.5 kg of water at 20◦C. Assuming no heat is lost to the
surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of copper is 390 J/kg◦Cand specific heat capacity of
water is 4186 J/kg◦C)
Solution
Step 1: Calculate the heat gained by the copper block. The formula to calculate
heat is:
Q=mc∆T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, - ∆Tis the change in temperature.
Given: - m= 0.2 kg, - c= 390 J/kg◦C, - Initial temperature of the copper
block = 100◦C, - Final temperature to be determined.
The heat gained by the copper block is:
Qcopper = 0.2×390 ×(Tf−100)
Step 2: Calculate the heat lost by the water. Given that the heat lost by the
copper block is equal to the heat gained by the water (assuming no heat loss to
the surroundings), the formula is:
Qcopper =Qwater
Substitute in the known values:
0.2×390 ×(Tf−100) = 0.5×4186 ×(Tf−20)
Step 3: Solve for the final equilibrium temperature, Tf.
78(Tf−100) = 2093(Tf−20)
78Tf−7800 = 2093Tf−41860
2015Tf= 34060
Tf= 16.91◦C
Therefore, the final equilibrium temperature of the system will be 16.91◦C.
9
Question 10
Question
A 0.5 kg aluminum block at 30
°
C is submerged in 2 kg of water at 20
°
C. As-
suming no heat is lost to the surroundings, what will be the final temperature
of the system when thermal equilibrium is reached? (Specific heat capacity of
aluminum = 900 J/kg
°
C, specific heat capacity of water = 4186 J/kg
°
C, and
latent heat of fusion of water = 334 kJ/kg)
Solution
Step 1: Calculate the heat gained or lost by the aluminum block using the
formula:
QAl =mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Given: mAl = 0.5 kg cAl = 900 J/kg
°
CTinitial, Al = 30C Tfinal =Tfinal, system
Calculating the heat lost by the aluminum block:
QAl = 0.5×900 ×(Tfinal −30)
Step 2: Calculate the heat gained or lost by the water through temperature
change using the formula:
Qw=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Given: mw= 2 kg cw= 4186 J/kg
°
CTinitial, w = 20C Tfinal =Tfinal, system
Calculating the heat gained by the water:
Qw= 2 ×4186 ×(Tfinal −20)
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block is equal to the heat gained by the water:
0.5×900 ×(Tfinal −30) = 2 ×4186 ×(Tfinal −20)
Solving this equation will give us the final temperature of the system.
Question 11
Question
A cylinder filled with an ideal gas is placed on a stove. The gas occupies a
volume of 2.0 m3at a pressure of 1.0 atm and a temperature of 20◦C. The
stove is turned on and the gas is rapidly heated, causing the volume of the gas
to increase to 4.0 m3. If the pressure remains constant during this process,
calculate the final temperature of the gas in degrees Celsius.
10
Solution
Step 1: Convert the initial temperature to Kelvin. Since the temperature is
given in degrees Celsius, we need to convert it to Kelvin using the formula:
T(K) = T(◦C) + 273.15
Plugging in the values:
Tinitial = 20◦C+ 273.15 = 293.15K
Step 2: Use the ideal gas law to calculate the initial number of moles of gas.
The ideal gas law is given by:
P V =nRT
where: P= 1.0 atm = 1.013 ×105Pa (since 1 atm = 1.013 ×105Pa)
V= 2.0 m3R= 8.314 J/(mol K) (universal gas constant)
Substitute the values into the equation to solve for the initial number of
moles (n):
n=P V
RT
n=(1.013 ×105Pa)(2.0 m3)
(8.314 J/(mol K))(293.15 K)
n≈817.73 mol
Step 3: Calculate the final temperature of the gas in Kelvin. Since the
pressure remains constant, we can use the formula:
P1V1
T1
=P2V2
T2
Substitute the initial and final conditions into the equation:
(1.013 ×105Pa)(2.0 m3)
293.15 K =(1.013 ×105Pa)(4.0 m3)
T2
Solve for T2:
T2=(1.013 ×105Pa)(4.0 m3)
(1.013 ×105Pa)(2.0 m3)(293.15 K)
T2= 586.3 K
Step 4: Convert the final temperature back to degrees Celsius.
Tfinal = 586.3 K −273.15 = 313.15◦C
Therefore, the final temperature of the gas is 313.15◦C.
11
Question 12
Question
A piece of copper of mass 250 g at a temperature of 150◦C is placed in a
calorimeter containing 500 g of water at 20◦C. If the final temperature of the
mixture 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 copper and the heat gained by water. The
heat lost by copper is equal to the heat gained by water. Let cCu be the specific
heat capacity of copper. Using the formula:
Qcopper =mcCu∆T
Qwater =mcwater∆T
where mis the mass, and ∆Tis the change in temperature.
Step 2: Calculate the heat lost by copper. Substitute the values into the
formula:
Qcopper = (0.25 kg)(cCu)(150 −25)
Step 3: Calculate the heat gained by water. Substitute the values into the
formula:
Qwater = (0.5 kg)(4.18 J/g◦C)(25 −20)
Step 4: Set the heat lost by copper equal to the heat gained by water and
solve for cCu.
(0.25)(cCu)(125) = (0.5)(4.18)(5)
31.25cCu = 10.45
cCu =10.45
31.25
cCu = 0.3344 J/g◦C
Therefore, the specific heat capacity of copper is 0.3344 J/g◦C.
Question 13
Question
A copper kettle with a mass of 500g contains 1 kg of water at 100
°
C. If the
specific heat capacity of copper is 390 J/kg·
°
C, and that of water is 4186 J/kg·
°
C,
what will be the final temperature of the system once the kettle and water come
to thermal equilibrium?
12
Solution
Step 1: Let the final temperature of the system be T
°
C. The heat lost by the
water will be equal to the heat gained by the copper kettle. We can set up the
following equation:
mwcw(T−100) = mccc(T−100)
where mw= 1 kg (mass of water), cw= 4186 J/kg ·
°
C (specific heat capacity
of water), mc= 0.5 kg (mass of copper), and cc= 390 J/kg ·
°
C (specific heat
capacity of copper).
Step 2: Substituting the values into the equation, we get:
1×4186 ×(T−100) = 0.5×390 ×(T−100)
Step 3: Simplifying the equation, we have:
4186T−418600 = 195T−19500
Step 4: Rearranging the equation, we get:
4186T−195T= 418600 −19500
3991T= 399100
Step 5: Solving for T, we find:
T=399100
3991 ≈100
°
C
Therefore, the final temperature of the system once the kettle and water
come to thermal equilibrium is approximately 100
°
C.
Question 14
Question
A brass bar of length Land cross-sectional area Ais initially at a temperature
T1. If heat is added to the bar at a constant rate P, what is the final temperature
of the bar in terms of its initial temperature, the rate of heat supplied, and its
thermal conductivity?
Solution
Step 1: Determine the rate of heat conduction. Given that the rate of heat
supplied is P, the rate at which heat is conducted through the brass bar is also
P.
13
Step 2: Apply the formula for rate of heat conduction. The rate of heat
conduction through a material is given by
P=kA∆T
L,
where kis the thermal conductivity of the material, Ais the cross-sectional area
of the bar, ∆Tis the temperature difference across the bar, and Lis the length
of the bar.
Step 3: Evaluate the temperature difference. Since the initial temperature
is T1, the final temperature can be denoted as Tf. Therefore, the temperature
difference is ∆T=Tf−T1.
Step 4: Rewrite the formula in terms of the quantities given. Based on Step
2, we can rewrite the formula as
P=kATf−T1
L.
Step 5: Solve for the final temperature, Tf. Rearranging the equation from
Step 4 to solve for Tf, we have
Tf=T1+P L
kA .
Therefore, the final temperature of the brass bar in terms of its initial tem-
perature, the rate of heat supplied, and its thermal conductivity is Tf=T1+P L
kA .
Question 15
Question
A copper container with a mass of 1.5 kg contains 2.0 kg of water at a tem-
perature of 80◦C. The container and water are cooled to a final temperature of
20◦C. If the specific heat capacity of water is 4186 J/kg◦C and the specific heat
capacity of copper is 390 J/kg◦C, calculate the heat lost by the water and the
copper container during this process.
Solution
Step 1: Calculate the heat lost by the water.
The formula for calculating heat lost or gained is Q=mc∆T, where: - Qis
the heat lost or gained, - mis the mass of the object, - cis the specific heat
capacity of the object, - ∆Tis the change in temperature.
Given that m= 2.0 kg, c= 4186 J/kg◦C, and ∆T= 80◦C−20◦C = 60◦C,
we can calculate the heat lost by the water:
Q= (2.0 kg)(4186 J/kg◦C)(60◦C)
Q= 502320 J
14
Step 2: Calculate the heat lost by the copper container.
Using the same formula, Q=mc∆T, for the copper container: - m= 1.5 kg, -
c= 390 J/kg◦C, - ∆T= 60◦C.
Therefore, the heat lost by the copper container is:
Q= (1.5 kg)(390 J/kg◦C)(60◦C)
Q= 35100 J
Thus, the heat lost by the water is 502320 J and the heat lost by the copper
container is 35100 J.
Question 16
Question
A 5 kg block of copper is initially at a temperature of 80
°
C. How much heat
would need to be added to raise the temperature of the block to 200
°
C? (Specific
heat capacity of copper is 386 J/kg◦C)
Solution
Step 1: Find the change in temperature: The change in temperature (∆T) is
given by:
∆T=Tf−Ti
where Tiis the initial temperature and Tfis the final temperature. Given
Ti= 80Cand Tf= 200C,
∆T= 200 −80 = 120C
Step 2: Calculate the heat energy required: The heat energy required to
change the temperature of an object is given by the formula:
Q=mc∆T
where Qis the heat energy, mis the mass of the object, cis the specific heat
capacity of the material, and ∆Tis the change in temperature. Given m= 5 kg,
c= 386 J/kg◦C, and ∆T= 120C,
Q= 5 ×386 ×120 = 231600 J
Therefore, the amount of heat that needs to be added to raise the tempera-
ture of the copper block to 200
°
C is 231600 J.
15
Question 17
Question
A 2 kg block of iron at 200
°
C is dropped into a container of water at 20
°
C. If the
final temperature of the system is 50
°
C, how much heat was transferred from
the iron to the water? (Specific heat capacity of iron = 450 J/kg
°
C, specific heat
capacity of water = 4186 J/kg
°
C)
Solution
Step 1: Calculate the heat lost by the iron block to reach the final temperature.
The heat lost by the iron block can be calculated using the formula:
Q=m·c·∆T
Where: - Qis the heat lost, - mis the mass of the iron block (2 kg), - c
is the specific heat capacity of iron (450 J/kg
°
C), and - ∆Tis the change in
temperature of the iron (200
°
C - 50
°
C).
Plugging in the values:
Qiron = 2 kg ×450 J/kg
°
C×(200C−50C)
Qiron = 2 kg ×450 J/kg
°
C×150C
Qiron = 2 ×450 ×150 J
Qiron = 135000 J
Step 2: Calculate the heat gained by the water to reach the final temperature.
The heat gained by the water can be calculated using the same formula:
Q=m·c·∆T
Where: - Qis the heat gained, - mis the mass of the water (let’s assume it’s the
same as the mass of the iron block for simplicity, i.e., 2 kg), - cis the specific
heat capacity of water (4186 J/kg
°
C), and - ∆Tis the change in temperature
of the water (50
°
C - 20
°
C).
Plugging in the values:
Qwater = 2 kg ×4186 J/kg
°
C×(50C−20C)
Qwater = 2 kg ×4186 J/kg
°
C×30C
Qwater = 2 ×4186 ×30 J
Qwater = 251160 J
Step 3: The heat transferred from the iron to the water is the same as the
heat lost by the iron (since energy is conserved). Hence, the amount of heat
transferred from the iron to the water is:
Qtransfer =Qiron = 135000 J
Therefore, the amount of heat transferred from the iron to the water is
135,000 J.
16
Question 18
Question
A block of iron initially at a temperature of 100
°
C is dropped into a container of
water at 20
°
C. The mass of the iron block is 2 kg and the mass of the water in
the container is 5 kg. If the specific heat capacity of iron is 450 J/kg ·K and the
specific heat capacity of water is 4186 J/kg ·K, calculate the final temperature
of the system when thermal equilibrium is reached. Assume no heat is lost to
the surroundings.
Solution
Step 1: Calculate the heat lost by the iron block using the formula Q=mc∆T,
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature. The initial temperature of the iron block is 100
°
C and the final
temperature is the same as the final temperature of the system.
Qlost =mironciron∆Tiron
Qlost = 2 kg ×450 J/kg ·K×(Tfinal −100)
Step 2: Calculate the heat gained by the water using the same formula.
Qgain =mwatercwater∆Twater
Qgain = 5 kg ×4186 J/kg ·K×(Tfinal −20)
Step 3: Since no heat is lost to the surroundings, the heat lost by the iron
block is equal to the heat gained by the water.
2 kg ×450 J/kg ·K×(Tfinal −100) = 5 kg ×4186 J/kg ·K×(Tfinal −20)
Step 4: Solve the equation to find the final temperature Tfinal.
900 J/K ×(Tfinal −100) = 20930 J/K ×(Tfinal −20)
900Tfinal −90000 = 20930Tfinal −418600
20930Tfinal −900Tfinal = 418600 −90000
20030Tfinal = 328600
Tfinal = 16.4◦C
Therefore, the final temperature of the system when thermal equilibrium is
reached is 16.4◦C.
17
Question 19
Question
A block of copper with a mass of 0.5 kg is heated from 20
°
C to 90
°
C. Calcu-
late the amount of heat required to raise the temperature of the copper block.
(Specific heat of copper = 385 J/kg
°
C)
Solution
Step 1: Determine the change in temperature Given: Initial temperature, T1=
20CFinal temperature, T2= 90CChange in temperature, ∆T=T2−T1
∆T= 90C−20C
∆T= 70C
Step 2: Calculate the amount of heat using the formula Q = mcT Given:
Mass of copper block, m= 0.5 kg Specific heat of copper, c= 385 J/kg
°
C
Change in temperature, ∆T= 70C
Q=mc∆T
Q= 0.5 kg ×385 J/kg
°
C×70C
Q= 0.5×385 ×70 J
Q= 13,475 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 90
°
C is 13,475 J.
Question 20
Question
A 2 kg block of metal is heated to a temperature of 300
°
C. If the specific heat
of the metal is 450 J/kg
°
C, how much heat energy is required to raise the
temperature of the block by 100
°
C?
Solution
Step 1: Identify the given values.
The mass of the metal block, m= 2 kg
The initial temperature of the block, T1= 300◦C
The specific heat of the metal, c= 450 J/kg
°
C
The change in temperature, ∆T= 100◦C
Step 2: Calculate the heat energy using the formula
Q=mc∆T
18
where: Qis the heat energy, mis the mass of the block, cis the specific heat of
the metal, and ∆Tis the change in temperature.
Step 3: Substitute the given values into the formula
Q= (2 kg)(450 J/kg
°
C)(100◦C)
Q= 90000 J
Therefore, the amount of heat energy required to raise the temperature of
the block by 100
°
C is 90,000 Joules.
Question 21
Question
Two identical blocks of aluminum are heated separately to different tempera-
tures. Block A is heated to 200◦C, while block B is heated to 300◦C. If both
blocks are placed in contact with each other and allowed to reach thermal equi-
librium, what will be the final temperature of the system?
Solution
Step 1: Calculate the specific heat capacity of aluminum. The specific heat
capacity of aluminum is approximately 900 J/kg◦C.
Step 2: Determine the mass of the aluminum blocks. Let’s assume the mass
of each block is mkg.
Step 3: Calculate the heat absorbed by each block using the formula: Q=
mc∆T, where mis the mass, cis the specific heat capacity, and ∆Tis the
change in temperature.
For block A: QA=m·900 ·(300 −200) QA= 100m·900 QA= 90000m
Joules
For block B: QB=m·900 ·(300 −200) QB= 100m·900 QB= 90000m
Joules
Step 4: Since no heat is lost to the surroundings, the total heat gained
by block A must be equal to the total heat lost by block B in order to reach
thermal equilibrium. Therefore, we can set QA=QBand solve for the final
temperature.
90000m= 90000m200 ·c·∆TA= 200 ·c·∆TB200 ·∆TA= 300 ·∆TB
∆TB=2
3·∆TA
The final equation we will solve is: 200 + ∆TA= 300 −∆TB.
200 + ∆TA= 300 −2
3·∆TA5
3·∆TA= 100 ∆TA=3
5·100 ∆TA= 60
Therefore, the final temperature of the system is 300 −60 = 240◦C.
19
Question 22
Question
A copper bar of length 2 m and cross-sectional area 0.01 m2is heated from 20
°
C
to 80
°
C. If the thermal conductivity of copper is 400 W/(m K), calculate the
rate at which heat flows through the bar.
Solution
Step 1: Calculate the temperature difference Given that the initial temperature,
T1, is 20
°
C and the final temperature, T2, is 80
°
C, the temperature difference,
∆T, is:
∆T=T2−T1= 80
°
C−20
°
C = 60
°
C
Step 2: Find the rate at which heat flows through the bar The rate at which
heat flows through the bar can be calculated using the formula:
P=kA∆T
L
where: P= rate of heat transfer through the bar, k= thermal conductivity of
copper (400 W/(m K)), A= cross-sectional area of the bar (0.01 m2), ∆T=
temperature difference (60
°
C), and L= length of the bar (2 m).
Plugging in the values:
P=400 ×0.01 ×60
2
P=240
2
P= 120 W
Therefore, the rate at which heat flows through the copper bar is 120 W.
Question 23
Question
A 2 kg block of copper initially at 300 K is heated until it reaches a final
temperature of 600 K. If the specific heat capacity of copper is 386 J/kg-K, how
much heat is added to the block during this process?
Solution
Step 1: Calculate the heat added using the equation Q=mc∆T, where Qis
the heat added, mis the mass of the block, cis the specific heat capacity of
copper, and ∆Tis the change in temperature.
Given: m= 2 kg, c = 386 J/kg-K,∆T= 600 K −300 K = 300 K
20
Q= (2 kg)(386 J/kg-K)(300 K)
Q= 2 ×386 ×300 J
Q= 231600 J
Therefore, 231600 J of heat is added to the block during this process.
Question 24
Question
A block of ice at -10
°
C is initially on a scale and its melting produces water
at 0
°
C. If the block’s mass is 500 grams and no heat is exchanged with the
surroundings, determine:
1. the amount of heat involved in the process,
2. the entropy change for the ice-water process, and
3. the entropy change for the water increase in temperature to 10
°
C.
(The specific heat capacity of ice is cice = 0.5 cal/g
°
C, the specific heat capacity
of water is cwater = 1 cal/g
°
C, and the heat of fusion for ice is 80 cal/g.)
Solution
1. Let’s start by calculating the heat involved in the process.
Step 1: The heat required to raise the temperature of the ice from -10
°
C to
0
°
C is given by the formula:
Q1=mcice∆T
where mis the mass of the ice, cice is the specific heat capacity of ice, and ∆T
is the temperature change. Substitute the given values to obtain:
Q1= 500 ×0.5×(0 −(−10))
Q1= 500 ×0.5×10 = 2500 cal
Step 2: The heat of fusion required to melt the ice into water at 0
°
C is
given by:
Q2=m·heat of fusion
Substitute the values to find:
Q2= 500 ×80 = 40,000 cal
Therefore, the total heat involved in the process is:
Qtotal =Q1+Q2= 2500 + 40,000 = 42,500 cal
21
2. Next, let’s calculate the entropy change for the ice-water process.
Step 3: The entropy change for the ice-water process is given by:
∆S1=Qtotal
T
where Qtotal is the total heat involved in the process and Tis the temperature
of the process. Substitute the values to find:
∆S1=42,500
273
∆S1= 156.02 cal/K
3. Finally, let’s calculate the entropy change for the water increasing in
temperature from 0
°
C to 10
°
C.
Step 4: The entropy change for the water heating process is given by:
∆S2=Q3
T
where Q3is the heat required for the water to increase in temperature and Tis
the temperature of the process.
Q3=mcwater∆T= 500 ×1×(10 −0) = 5000 cal
Therefore,
∆S2=5000
283 = 18.01 cal/K
Question 25
Question
A copper rod of length 2 m and cross-sectional area 2 cm2is initially at a
temperature of 200◦C. It is placed in an ice-water mixture at 0◦C. If the rod
absorbs 5000 J of heat from the surrounding, calculate the final temperature of
the rod. Assume the specific heat capacity of copper is 400 J/(kg·K), density
of copper is 8.96 g/cm3, and latent heat of fusion of ice is 334 kJ/kg.
Solution
Step 1: Calculate the mass of the copper rod.
Given the density of copper is 8.96 g/cm3, we can convert the volume of the rod
(2 m ×2 cm2) into grams:
Volume = length ×cross-sectional area = 2 m ×2 cm2= 4 ×104cm3
Mass = Density×Volume = 8.96 g/cm3×4×104cm3= 3.584×105g = 358.4 kg
22
Step 2: Calculate the initial heat content of the copper rod.
The initial temperature of the copper rod is 200◦C. The specific heat capacity
of copper is 400 J/(kg·K). The formula to calculate heat content is:
Q=mc∆T
where Qis the heat content, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature. Substitute the given values:
Q= 358.4 kg ×400 J/(kg ·K) ×(200 −T)
Step 3: Calculate the heat required to change the temperature of the copper
rod to 0◦C.
Using the same formula as in Step 2, but with the final temperature as 0◦C:
Q2= 358.4 kg ×400 J/(kg ·K) ×(T−0)
Step 4: Calculate the heat required to freeze the water in the ice-water
mixture.
Since the rod absorbs 5000 J of heat, the heat required to freeze the water can
be calculated as:
Q3= 5000 J + Q2+Q
Step 5: Calculate the final temperature of the rod.
Substitute Q3into the equation from Step 4 and solve for T:
5000 J = Q3
Question 26
Question
A copper rod of length 2 meters is initially at a temperature of 100
°
C. If one
end of the rod is placed in a furnace at 300
°
C and the other end is placed in an
ice bath at 0
°
C, what is the temperature at a point 0.5 meters from the end in
the furnace after 10 minutes? Assume that the rod is well insulated and can be
approximated as one-dimensional. The thermal conductivity of copper is 401
W/(m
·
K) and its density is 8.96 ×103kg/m
³
. Specific heat capacity of copper
is 385 J/(kg
·
K).
Solution
Step 1: We first need to calculate the thermal diffusivity of copper, given by
the formula
α=κ
ρcp
,
23
where κis the thermal conductivity, ρis the density, and cpis the specific heat
capacity. Substituting the given values, we have
α=401
8.96 ×103×385 ≈1.41 ×10−4m2/s.
Step 2: Next, we can use the solution to the heat equation in one dimension
with a fixed end to find the temperature distribution along the rod. The general
form of the temperature distribution is given by
T(x, t) = T1+T2−T1
Lx+
∞
X
n=1
Bnsin nπx
Le−α(nπ/L)2t,
where T1= 100◦C, T2= 300◦C, and L= 2 m.
Step 3: To find the coefficients Bn, we use the initial condition T(x, 0) =
100◦C for 0 < x < L. The uniform temperature distribution at t= 0 implies
that Bn= 0 for all nexcept for n= 0. Thus, the temperature distribution
simplifies to
T(x, t) = 100 + 200
2x−400
π
∞
X
n=1
(−1)n
nsin nπx
2e−α(nπ/2)2t.
Step 4: Now, we need to find the temperature at x= 0.5 m and t= 10
minutes. Substituting x= 0.5 m, t= 600 s into the equation above and
evaluating the series numerically, we can find the temperature.
This solution involves solving a partial differential equation and evaluating
an infinite series, which may require numerical techniques.
Question 27
Question
A metal block of mass 0.5 kg is initially at a temperature of 200
°
C. It is then
heated until it reaches a temperature of 400
°
C. If the specific heat capacity
of the metal is 500 J/kg
°
C, calculate the heat energy transferred to the block
during this process.
Solution
Step 1: Identify the given values and the formula for calculating heat energy
transferred. Given: Mass of the metal block, m= 0.5 kg Initial temperature,
Ti= 200CFinal temperature, Tf= 400CSpecific heat capacity, c= 500 J/kg
°
C
The formula for calculating heat energy transferred is:
Q=mc∆T
24
where Qis the heat energy transferred, mis the mass of the object, cis the
specific heat capacity of the material, and ∆Tis the change in temperature
(Tf−Ti).
Step 2: Calculate the change in temperature (∆T).
∆T=Tf−Ti= 400C−200C= 200C
Step 3: Substitute the given values into the formula to find the heat energy
transferred.
Q= (0.5 kg) ×(500 J/kg
°
C) ×200C
Q= 100 J/
°
C×200C= 20000 J
Therefore, the heat energy transferred to the metal block during this process
is 20,000 J.
Question 28
Question
A 500 g block of copper at 100
°
C is dropped into a container of water at 20
°
C.
If the final temperature of the system is 30
°
C, determine the mass of water in
the container. (Specific heat capacity of copper: cCu = 0.385 J/g
°
C, specific
heat capacity of water: cH2O = 4.18 J/g
°
C)
Solution
Step 1: Identify the heat lost by the copper block and the heat gained by the
water. The heat lost by the copper block can be calculated as:
Qlost =mCu ×cCu ×(Tf−TCu)
Where: - mCu is the mass of the copper block (500 g), - cCu is the specific heat
capacity of copper (0.385 J/g
°
C), - Tfis the final temperature of the system
(30
°
C), - TCu is the initial temperature of the copper block (100
°
C).
The heat gained by the water can be calculated as:
Qgain =mH2O ×cH2O ×(Tf−TH2O)
Where: - mH2O is the mass of water in the container, - cH2O is the specific heat
capacity of water (4.18 J/g
°
C), - TH2O is the initial temperature of the water
(20
°
C).
Step 2: Set up the equation with the heat lost equal to the heat gained.
Since the system is isolated, the heat lost by the copper block must equal the
heat gained by the water.
mCu ×cCu ×(Tf−TCu) = mH2O ×cH2O ×(Tf−TH2O)
25
Step 3: Solve for the mass of water in the container. Substitute the given
values into the equation:
500 ×0.385 ×(30 −100) = mH2O ×4.18 ×(30 −20)
−500 ×0.385 ×70 = mH2O ×4.18 ×10
−133.75 = mH2O ×41.8
mH2O =−133.75
41.8
mH2O ≈ −3.194 g
The negative mass indicates there was an error in the calculations or as-
sumptions made. Please review the calculations and assumptions to find the
mistake.
Question 29
Question
A block of copper of mass 0.5 kg at a temperature of 200
°
C is dropped into 1 kg
of water at 20
°
C. If the specific heat capacity of copper is 0.39 J/g
°
C and the
specific heat capacity of water is 4.18 J/g
°
C, determine the final equilibrium
temperature of the system. Assume no heat is lost to the surroundings and
neglect any heat capacity of the container.
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
equilibrium temperature. The formula to calculate heat exchanged is given by:
Q=mc∆T
where: - Qis the heat exchanged, - mis the mass of the substance, - cis the
specific heat capacity, - ∆Tis the change in temperature.
For the copper block: m= 0.5 kg = 500 g c= 0.39 J/g
°
C Initial tempera-
ture, Tcopper = 200CFinal temperature, Tf
The heat lost by the copper block is equal to the heat gained by the water:
500 ×0.39 ×(200 −Tf) = 1000 ×4.18 ×(Tf−20)
Step 2: Solve for the final equilibrium temperature. Solving the equation
from Step 1 will give us the value of Tf:
500 ×0.39 ×(200 −Tf) = 1000 ×4.18 ×(Tf−20)
195 ×(200 −Tf) = 4180 ×(Tf−20)
26
39000 −195Tf= 4180Tf−83600
43780 = 4375Tf
Tf=43780
4375 ≈10C
Therefore, the final equilibrium temperature of the system is approximately
10C.
Question 30
Question
A solid metal cube with a side length of 10 cm is heated from an initial temper-
ature of 20
°
C to a final temperature of 80
°
C. The specific heat capacity of the
metal is 0.4 J/g
°
C. If 1000 J of heat is added to the cube, calculate the mass of
the metal cube.
Solution
Step 1: Calculate the change in temperature using the formula:
∆T=Tf−Ti= 80◦C−20◦C= 60◦C
Step 2: Next, calculate the mass of the metal cube using the formula:
Q=mc∆T
Step 3: Substitute the given values into the formula:
1000 = m×0.4×60
Step 4: Solve for the mass m:
m=1000
0.4×60 =1000
24 = 41.67 g
Therefore, the mass of the metal cube is 41.67 g.
Question 31
Question
A copper rod of length 1 m and cross-sectional area 1 cm2is used to transfer
heat from one end to the other. The temperature at one end is 200
°
C, while the
temperature at the other end is 100
°
C. If the thermal conductivity of copper is
400 W/(m·K), calculate the rate of heat transfer through the rod.
27
Solution
Step 1: Calculate the temperature difference across the rod. Given that the
temperature at one end is 200
°
C and the temperature at the other end is 100
°
C,
the temperature difference across the rod is:
∆T= 200
°
C−100
°
C = 100
°
C
Step 2: Convert the temperature to Kelvin. To convert temperature in
degrees Celsius to Kelvin, we use the formula T(K) = T(
°
C) + 273.15. Thus,
the temperature difference in Kelvin is:
∆T= 100
°
C + 273.15 = 373.15 K
Step 3: Calculate the rate of heat transfer using Fourier’s Law. The rate of
heat transfer through the rod is given by Fourier’s Law:
Q=kA∆T
L
where Qis the rate of heat transfer, kis the thermal conductivity of copper, A
is the cross-sectional area of the rod, ∆Tis the temperature difference across
the rod, and Lis the length of the rod.
Step 4: Substitute the known values into Fourier’s Law. Substitute k= 400
W/(m·K), A= 1 cm2= 0.0001 m2, ∆T= 373.15 K, and L= 1 m into Fourier’s
Law:
Q= (400 W/(m·K)) ×(0.0001 m2)×373.15 K
1 m
Step 5: Calculate the rate of heat transfer.
Q= 400 ×0.0001 ×373.15 W = 14.926 W
Therefore, the rate of heat transfer through the rod is 14.926 W.
Question 32
Question
A copper rod of length 1 m and cross-sectional area 1 ×10−4m2is heated from
20
°
C to 60
°
C. If the thermal conductivity of copper is 400 W/(m·K) and the
density of copper is 8.96 ×103kg/m3, calculate the amount of heat transferred
to the rod during this process.
Solution
Step 1: Calculate the volume of the copper rod. The volume of the copper rod
can be calculated using the formula:
Volume = Length ×Area
28
Volume = 1 m ×1×10−4m2
Volume = 1 ×10−4m3
Step 2: Calculate the mass of the copper rod. The mass of the copper rod
can be calculated using the formula:
Mass = Volume ×Density
Mass = 1 ×10−4m3×8.96 ×103kg/m3
Mass = 8.96 ×10−1kg
Step 3: Calculate the initial temperature of the copper rod in Kelvin.
T1= 20 + 273.15
T1= 293.15 K
Step 4: Calculate the final temperature of the copper rod in Kelvin.
T2= 60 + 273.15
T2= 333.15 K
Step 5: Calculate the change in temperature.
∆T=T2−T1
∆T= 333.15 −293.15
∆T= 40 K
Step 6: Calculate the amount of heat transferred. The amount of heat
transferred can be calculated using the formula:
Heat = Thermal Conductivity ×Area ×∆T×Time
Heat = 400 W/(m ·K) ×1×10−4m2×40 K
Heat = 160 W
Therefore, the amount of heat transferred to the copper rod during this
process is 160 J.
Question 33
Question
A metal object of mass 0.5 kg at a temperature of 100
°
C is placed in a container
of water at 20
°
C. The final temperature of the metal object and water is 30
°
C.
If the specific heat capacity of the metal is 0.5 J/g
°
C, determine the mass of
water in the container. Assume no heat is lost to the surroundings.
29
Solution
Step 1: Calculate the heat absorbed by the metal object. The heat absorbed
by the metal object can be calculated using the formula:
Qmetal =mc∆T
where: - m= 0.5 kg is the mass of the metal object, - c= 0.5 J/g
°
C is the specific
heat capacity of the metal, and - ∆T=Tfinal −Tinitial = 30 −100 = −70
°
C is
the change in temperature of the metal object.
Plugging in the values, we get:
Qmetal = 0.5 kg ×0.5 J/g
°
C× −70
°
C
Step 2: Calculate the heat released by the metal object to the water. Since
the heat lost by the metal object is equal to the heat gained by the water
(assuming no heat is lost to the surroundings), we have:
Qmetal =Qwater
Step 3: Calculate the heat absorbed by the water. The heat absorbed by
the water can be calculated using the formula:
Qwater =mc∆T
where: - cwater = 4.18 J/g
°
C is the specific heat capacity of water, and - ∆T=
Tfinal −Tinitial = 30 −20 = 10
°
C is the change in temperature of the water.
Plugging in the values, we get:
Qwater =mwater ×4.18 J/g
°
C×10
°
C
Step 4: Solve for the mass of water. Since Qmetal =Qwater, we can set the
two expressions equal to each other:
0.5 kg ×0.5 J/g
°
C× −70
°
C = mwater ×4.18 J/g
°
C×10
°
C
Solving for mwater, we find:
mwater =0.5×0.5× −70
4.18 ×10
Therefore, the mass of water in the container is mwater kg.
Question 34
Question
A copper block with a mass of 500 g is initially at a temperature of 100◦C. The
block is placed in 200 g of water at 20◦C. If all the heat lost by the copper block
is absorbed by the water, what will be the final equilibrium temperature of the
system? Assume specific heat capacity of copper is 0.385 J/g◦C and specific
heat capacity of water is 4.18 J/g◦C.
30
Solution
Step 1: Calculate the heat lost by the copper block: The heat lost by the copper
block can be calculated using the formula:
Q=mc∆T
where - mis the mass of the copper block (500 g), - cis the specific heat capacity
of copper (0.385 J/g◦C), - ∆Tis the change in temperature of the copper block.
The change in temperature (∆T) of the copper block is:
∆T= (final temperature) −(initial temperature)
∆T=Tf−100
Step 2: Calculate the heat gained by the water: The heat gained by the
water can also be calculated using the formula:
Q=mc∆T
where - mis the mass of the water (200 g), - cis the specific heat capacity of
water (4.18 J/g◦C), - ∆Tis the change in temperature of the water.
The change in temperature (∆T) of the water is:
∆T= (final temperature) −20
Step 3: Since the heat lost by the copper block is equal to the heat gained
by the water (assuming no heat losses to the surroundings), we can set the two
equations equal to each other and solve for the final temperature (Tf):
mcopperccopper∆Tcopper =mwatercwater∆Twater
Substitute the given values and expressions for ∆Tcopper and ∆Twater into
the equation and solve for Tf.
Question 35
Question
A copper rod of length 1 m and diameter 2 cm is initially at a temperature
of 100
°
C. The rod is then submerged in a water bath at 0
°
C. Given that the
thermal conductivity of copper is 390 W/(m·K), determine how long it will take
for the rod to cool down to 10
°
C.
Solution
Step 1: Calculate the cross-sectional area of the rod. The cross-sectional area
of the rod can be calculated using the formula for the area of a circle: A=πd2
4,
31
Question 2
Question
A copper rod of length 2 m and diameter 1 cm is heated from 20
°
C to 80
°
C. If
the rod has thermal conductivity k= 380 W/(m·K) and specific heat capacity
c= 390 J/(kg·K), calculate the amount of heat transferred to the rod during
this process. Assume the density of copper is 8900 kg/m3.
Solution
Step 1: Calculate the cross-sectional area of the copper rod. The radius of the
rod is given by r=1
2×0.01 m = 0.005 m. Therefore, the cross-sectional area
A=πr2.
Step 2: Calculate the volume of the copper rod. The volume of the rod is
given by V=A×length = π×(0.005)2×2 m3.
Step 3: Calculate the mass of the copper rod. The mass of the rod can be
calculated using the formula m= density ×volume.
Step 4: Calculate the energy required to increase the temperature of the
copper rod. The energy required is given by Q=mc∆T, where ∆T= 80 −20
and mis the mass of the rod.
Step 5: Substitute the values into the formula. Substitute the values of m,
c, ∆Tinto the formula Q=mc∆Tto find the amount of heat transferred to
the rod. Remember to convert the temperature from Celsius to Kelvin.
Question 3
Question
A steel rod with a length of 2 meters and a diameter of 1 cm is initially at a
temperature of 100◦C. If the rod is placed in a water bath with a temperature
of 20◦C, how long will it take for the rod to reach a temperature of 30◦C? The
thermal conductivity of steel is 50 W/(m ·K), the heat capacity of steel is 500
J/kg·K, and the density of steel is 8000 kg/m3.
Solution
Step 1: Calculate the cross-sectional area of the steel rod. The cross-sectional
area of the rod can be calculated using the formula for the area of a circle:
A=πr2, where ris the radius of the rod. Given that the diameter of the rod
is 1 cm, the radius rcan be calculated as 0.5 cm or 0.005 m. Therefore, the
cross-sectional area of the rod is:
A=π(0.005)2≈7.85 ×10−5m2
Step 2: Calculate the volume of the steel rod. The volume of the rod can
be calculated using the formula for the volume of a cylinder: V=A×L, where
2
Lis the length of the rod. Given that the length of the rod is 2 meters, the
volume of the rod is:
V= 7.85 ×10−5×2=1.57 ×10−4m3
Step 3: Calculate the mass of the steel rod. The mass of the rod can be
calculated using the formula m= density ×volume. Given that the density of
steel is 8000 kg/m3, the mass of the rod is:
m= 8000 ×1.57 ×10−4= 1.256 kg
Step 4: Calculate the thermal energy required to heat the rod from 20◦C
to 30◦C. The thermal energy Qrequired can be calculated using the formula
Q=mc∆T, where mis the mass of the rod, cis the specific heat capacity
of steel, and ∆Tis the change in temperature. Given that the specific heat
capacity of steel is 500 J/kg·K, the change in temperature is 10◦C, and the
mass of the rod is 1.256 kg, the thermal energy required is:
Q= 1.256 ×500 ×10 = 6280 J
Step 5: Calculate the rate of heat transfer through the rod. The rate of
heat transfer Pcan be calculated using the formula P=kA∆T
L, where kis the
thermal conductivity of steel, Ais the cross-sectional area of the rod, ∆Tis the
temperature difference between the rod and the water bath, and Lis the length
of the rod. Given that the thermal conductivity of steel is 50 W/(m·K), the
temperature difference is 80◦C, the cross-sectional area is 7.85 ×10−5m2, and
the length of the rod is 2 meters, the rate of heat transfer is:
P= 50 ×7.85 ×10−5×80
2= 0.314 W
Step 6: Calculate the time taken for the rod to reach 30◦C. The time taken
tcan be calculated using the formula Q=P t, where Qis the thermal energy
required and Pis the rate of heat transfer. Substitute the values of Qand P
into the formula:
6280 = 0.314t
Solving for t:
t≈6280
0.314 ≈20000 s
Therefore, it will take approximately 20000 seconds for the rod to reach a tem-
perature of 30◦C.
Question 4
Question
A copper bar with a length of 2 meters and a cross-sectional area of 10 cm2is
initially at a temperature of 100◦C. If 5000 J of heat energy is added to the bar,
calculate the final temperature of the bar given that the specific heat capacity
of copper is 390 J/kg◦C and its density is 8900 kg/m3.
3
Solution
Step 1: Calculate the mass of the copper bar. The formula to calculate mass is
given by:
mass = density ×volume
The volume of the copper bar can be calculated using its length and cross-
sectional area:
volume = length ×cross-sectional area
Substitute the given values:
volume = 2 m ×10 ×10−4m2= 2 ×10−3m3
mass = 8900 kg/m3×2×10−3m3= 17.8 kg
Step 2: Calculate the change in temperature. The heat energy added to
the copper bar causes a temperature rise. The change in temperature can be
calculated using the formula:
Q=mc∆T
where: - Q= 5000 J (heat energy added) - m= 17.8 kg (mass of the copper
bar) - c= 390 J/kg◦C (specific heat capacity of copper) - ∆Tis the change in
temperature (unknown)
Solve for ∆T:
5000 = 17.8×390 ×∆T
∆T=5000
17.8×390 ≈7.95◦C
Step 3: Calculate the final temperature. The final temperature of the copper
bar is the initial temperature plus the change in temperature:
Final temperature = 100 + 7.95 = 107.95◦C
Therefore, the final temperature of the copper bar is 107.95◦C after 5000 J
of heat energy is added.
Question 5
Question
A metal rod of length 50 cm and thermal conductivity 200 W/(m2K) is being
heated at one end with a constant heat flux of 500 W/m2. The ambient tem-
perature surrounding the rod is 20
°
C. If the rod has a uniform cross-sectional
area of 2 cm2, calculate the temperature at a distance of 20 cm from the heated
end after 5 minutes.
Given: Thermal conductivity k= 200 W/(m2K), Heat flux q= 500 W/m2,
Rod length L= 0.5 m, Cross-sectional area A= 2 ×10−4m2, Time t= 5
minutes, Ambient temperature Ta= 20
°
C.
4
Solution
Step 1: Calculate the heat transfer rate through the metal rod using Fourier’s
law:
q=−k·AdT
dx
Where: q= Heat flux = 500 W/m2,k= Thermal conductivity = 200
W/(m2K), A= Cross-sectional area = 2×10−4m2,dT
dx = Temperature gradient.
Given that the heat flux q= 500 W/m2is constant, we can rewrite the
equation as:
−500 = −200 ·2×10−4·dT
dx
Solving for dT
dx gives:
dT
dx =500
200 ·2×10−4
dT
dx = 125,000 K/m
Step 2: Integrate to find the temperature distribution in the rod:
ZdT =Z125,000dx
T= 125,000x+C
Where Cis the integration constant.
Step 3: Apply the boundary condition that the temperature at the heated
end (x= 0) is Th(unknown) and at the ambient end (x= 0.5 m) is Ta= 20
°
C:
Th= 125,000 ×0 + C
Ta= 125,000 ×0.5 + C
20 = 62,500 + C
C=−62,480
Step 4: Substitute back into the temperature distribution equation to find
the temperature profile:
T= 125,000x−62,480
Step 5: Calculate the temperature at x= 0.2 m:
5
T(0.2) = 125,000 ×0.2−62,480
T(0.2) = 25,000 −62,480 = −37,480 K
Therefore, the temperature at a distance of 20 cm from the heated end after
5 minutes is −37,480 K.
Question 6
Question
A copper block at 100
°
C is dropped into a calorimeter containing 200 g of water
at 20
°
C. The final temperature of the system is found to be 30
°
C. If the specific
heat capacity of water is 4.18 J/g
°
C, determine the mass of the copper block.
Assume no heat is lost to the surroundings.
Solution
Step 1: Determine the heat gained by the water and calorimeter. The heat
gained by the water and calorimeter is equal to the heat lost by the copper block.
The formula for heat transfer is Q=mc∆T, where Qis the heat transferred, m
is the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Let mcbe the mass of the copper block. The heat lost by the copper block
is equal to the heat gained by the water and calorimeter:
mc·cc·∆T= (mw+mcal)·cw·∆T
Step 2: Substitute the known values into the equation. Substitute the given
mass and specific heat capacity values into the equation:
mc·cc·(100 −30) = (200 + 200) ·4.18 ·(30 −20)
Step 3: Solve for the mass of the copper block. Simplify the equation and
solve for mc:
mc·0.387 = 800 ·4.18
mc=800 ·4.18
0.387
mc≈8678.27 g
Therefore, the mass of the copper block is approximately 8678.27 g.
6
Question 7
Question
A steel rod of length 2.0 m and diameter 1.0 cm is initially at a temperature
of 30
°
C. If the rod is heated until its temperature reaches 100
°
C, determine the
amount of heat transferred to the rod. Assume the linear expansion coefficient
of steel is 1.2×10−5K−1and the specific heat capacity of steel is 450 J/kg ·K.
Solution
Step 1: Calculate the initial volume of the steel rod. The initial volume of
the steel rod can be calculated using the formula for the volume of a cylinder:
V=πr2h, where ris the radius and his the length. Given that r= 0.5 cm
(since the diameter is 1 cm), we have r= 0.005 m and h= 2.0 m. Plugging
these values into the formula, we get:
V=π(0.005 m)2·2.0 m
Step 2: Calculate the final volume of the steel rod after heating. When
heated, the steel rod will expand due to the increase in temperature. The
change in length ∆Lof the rod can be calculated using the formula:
∆L=αL0∆T
where αis the linear expansion coefficient, L0is the initial length (2.0 m), and
∆Tis the change in temperature (100
°
C - 30
°
C). Using this information, we can
calculate the final length Lfas:
Lf=L0+ ∆L
Step 3: Calculate the final volume of the steel rod. Once we have the final
length of the rod, we can calculate the final volume using the formula for the
volume of a cylinder:
Vf=πr2Lf
Step 4: Calculate the mass of the steel rod. The mass of the steel rod can
be calculated using the formula:
m= density ×V
Given that the density of steel is approximately 8000 kg/m3, we can calculate
the mass of the rod.
Step 5: Calculate the heat transferred to the rod. The amount of heat
transferred to the rod can be calculated using the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity of steel, and ∆T
is the change in temperature. Substitute the values into the formula to find the
heat transferred.
7
Question 8
Question
A copper rod of length Land cross-sectional area Ais used to connect two
large metal blocks at different temperatures. The ends of the rod are in thermal
contact with the blocks. If the rod is initially at a uniform temperature T0and
the steady-state temperature distribution is linear, find the rate at which heat
flows through the rod. Assume that the thermal conductivity of copper is k.
Solution
The rate at which heat flows through the rod can be found using Fourier’s Law of
Heat Conduction, which states that the rate of heat transfer through a material
is directly proportional to the negative gradient of temperature in the direction
of heat flow.
Step 1: Determine the temperature distribution along the rod.
Let T(x) be the temperature at a distance xfrom the end in contact with
the hotter block. Since the steady-state temperature distribution is linear, we
have
T(x) = Th−Th−T0
Lx,
where This the temperature of the hotter block.
Step 2: Calculate the temperature gradient.
The temperature gradient ∆T
∆xcan be found as
dT
dx =−Th−T0
L.
Step 3: Apply Fourier’s Law of Heat Conduction.
According to Fourier’s Law, the rate at which heat flows through the rod
(˙
Q) is given by
˙
Q=−kAdT
dx .
Substitute the expression for dT
dx into the equation above to obtain
˙
Q=kATh−T0
L.
Step 4: Finalize the answer.
Therefore, the rate at which heat flows through the rod is
˙
Q=kATh−T0
L.
8
Question 9
Question
A copper block of mass 0.2 kg is heated to a temperature of 100◦C. The block
is then placed in 0.5 kg of water at 20◦C. Assuming no heat is lost to the
surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of copper is 390 J/kg◦Cand specific heat capacity of
water is 4186 J/kg◦C)
Solution
Step 1: Calculate the heat gained by the copper block. The formula to calculate
heat is:
Q=mc∆T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, - ∆Tis the change in temperature.
Given: - m= 0.2 kg, - c= 390 J/kg◦C, - Initial temperature of the copper
block = 100◦C, - Final temperature to be determined.
The heat gained by the copper block is:
Qcopper = 0.2×390 ×(Tf−100)
Step 2: Calculate the heat lost by the water. Given that the heat lost by the
copper block is equal to the heat gained by the water (assuming no heat loss to
the surroundings), the formula is:
Qcopper =Qwater
Substitute in the known values:
0.2×390 ×(Tf−100) = 0.5×4186 ×(Tf−20)
Step 3: Solve for the final equilibrium temperature, Tf.
78(Tf−100) = 2093(Tf−20)
78Tf−7800 = 2093Tf−41860
2015Tf= 34060
Tf= 16.91◦C
Therefore, the final equilibrium temperature of the system will be 16.91◦C.
9
Question 10
Question
A 0.5 kg aluminum block at 30
°
C is submerged in 2 kg of water at 20
°
C. As-
suming no heat is lost to the surroundings, what will be the final temperature
of the system when thermal equilibrium is reached? (Specific heat capacity of
aluminum = 900 J/kg
°
C, specific heat capacity of water = 4186 J/kg
°
C, and
latent heat of fusion of water = 334 kJ/kg)
Solution
Step 1: Calculate the heat gained or lost by the aluminum block using the
formula:
QAl =mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Given: mAl = 0.5 kg cAl = 900 J/kg
°
CTinitial, Al = 30C Tfinal =Tfinal, system
Calculating the heat lost by the aluminum block:
QAl = 0.5×900 ×(Tfinal −30)
Step 2: Calculate the heat gained or lost by the water through temperature
change using the formula:
Qw=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Given: mw= 2 kg cw= 4186 J/kg
°
CTinitial, w = 20C Tfinal =Tfinal, system
Calculating the heat gained by the water:
Qw= 2 ×4186 ×(Tfinal −20)
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block is equal to the heat gained by the water:
0.5×900 ×(Tfinal −30) = 2 ×4186 ×(Tfinal −20)
Solving this equation will give us the final temperature of the system.
Question 11
Question
A cylinder filled with an ideal gas is placed on a stove. The gas occupies a
volume of 2.0 m3at a pressure of 1.0 atm and a temperature of 20◦C. The
stove is turned on and the gas is rapidly heated, causing the volume of the gas
to increase to 4.0 m3. If the pressure remains constant during this process,
calculate the final temperature of the gas in degrees Celsius.
10
Solution
Step 1: Convert the initial temperature to Kelvin. Since the temperature is
given in degrees Celsius, we need to convert it to Kelvin using the formula:
T(K) = T(◦C) + 273.15
Plugging in the values:
Tinitial = 20◦C+ 273.15 = 293.15K
Step 2: Use the ideal gas law to calculate the initial number of moles of gas.
The ideal gas law is given by:
P V =nRT
where: P= 1.0 atm = 1.013 ×105Pa (since 1 atm = 1.013 ×105Pa)
V= 2.0 m3R= 8.314 J/(mol K) (universal gas constant)
Substitute the values into the equation to solve for the initial number of
moles (n):
n=P V
RT
n=(1.013 ×105Pa)(2.0 m3)
(8.314 J/(mol K))(293.15 K)
n≈817.73 mol
Step 3: Calculate the final temperature of the gas in Kelvin. Since the
pressure remains constant, we can use the formula:
P1V1
T1
=P2V2
T2
Substitute the initial and final conditions into the equation:
(1.013 ×105Pa)(2.0 m3)
293.15 K =(1.013 ×105Pa)(4.0 m3)
T2
Solve for T2:
T2=(1.013 ×105Pa)(4.0 m3)
(1.013 ×105Pa)(2.0 m3)(293.15 K)
T2= 586.3 K
Step 4: Convert the final temperature back to degrees Celsius.
Tfinal = 586.3 K −273.15 = 313.15◦C
Therefore, the final temperature of the gas is 313.15◦C.
11
Question 12
Question
A piece of copper of mass 250 g at a temperature of 150◦C is placed in a
calorimeter containing 500 g of water at 20◦C. If the final temperature of the
mixture 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 copper and the heat gained by water. The
heat lost by copper is equal to the heat gained by water. Let cCu be the specific
heat capacity of copper. Using the formula:
Qcopper =mcCu∆T
Qwater =mcwater∆T
where mis the mass, and ∆Tis the change in temperature.
Step 2: Calculate the heat lost by copper. Substitute the values into the
formula:
Qcopper = (0.25 kg)(cCu)(150 −25)
Step 3: Calculate the heat gained by water. Substitute the values into the
formula:
Qwater = (0.5 kg)(4.18 J/g◦C)(25 −20)
Step 4: Set the heat lost by copper equal to the heat gained by water and
solve for cCu.
(0.25)(cCu)(125) = (0.5)(4.18)(5)
31.25cCu = 10.45
cCu =10.45
31.25
cCu = 0.3344 J/g◦C
Therefore, the specific heat capacity of copper is 0.3344 J/g◦C.
Question 13
Question
A copper kettle with a mass of 500g contains 1 kg of water at 100
°
C. If the
specific heat capacity of copper is 390 J/kg·
°
C, and that of water is 4186 J/kg·
°
C,
what will be the final temperature of the system once the kettle and water come
to thermal equilibrium?
12
Solution
Step 1: Let the final temperature of the system be T
°
C. The heat lost by the
water will be equal to the heat gained by the copper kettle. We can set up the
following equation:
mwcw(T−100) = mccc(T−100)
where mw= 1 kg (mass of water), cw= 4186 J/kg ·
°
C (specific heat capacity
of water), mc= 0.5 kg (mass of copper), and cc= 390 J/kg ·
°
C (specific heat
capacity of copper).
Step 2: Substituting the values into the equation, we get:
1×4186 ×(T−100) = 0.5×390 ×(T−100)
Step 3: Simplifying the equation, we have:
4186T−418600 = 195T−19500
Step 4: Rearranging the equation, we get:
4186T−195T= 418600 −19500
3991T= 399100
Step 5: Solving for T, we find:
T=399100
3991 ≈100
°
C
Therefore, the final temperature of the system once the kettle and water
come to thermal equilibrium is approximately 100
°
C.
Question 14
Question
A brass bar of length Land cross-sectional area Ais initially at a temperature
T1. If heat is added to the bar at a constant rate P, what is the final temperature
of the bar in terms of its initial temperature, the rate of heat supplied, and its
thermal conductivity?
Solution
Step 1: Determine the rate of heat conduction. Given that the rate of heat
supplied is P, the rate at which heat is conducted through the brass bar is also
P.
13
Step 2: Apply the formula for rate of heat conduction. The rate of heat
conduction through a material is given by
P=kA∆T
L,
where kis the thermal conductivity of the material, Ais the cross-sectional area
of the bar, ∆Tis the temperature difference across the bar, and Lis the length
of the bar.
Step 3: Evaluate the temperature difference. Since the initial temperature
is T1, the final temperature can be denoted as Tf. Therefore, the temperature
difference is ∆T=Tf−T1.
Step 4: Rewrite the formula in terms of the quantities given. Based on Step
2, we can rewrite the formula as
P=kATf−T1
L.
Step 5: Solve for the final temperature, Tf. Rearranging the equation from
Step 4 to solve for Tf, we have
Tf=T1+P L
kA .
Therefore, the final temperature of the brass bar in terms of its initial tem-
perature, the rate of heat supplied, and its thermal conductivity is Tf=T1+P L
kA .
Question 15
Question
A copper container with a mass of 1.5 kg contains 2.0 kg of water at a tem-
perature of 80◦C. The container and water are cooled to a final temperature of
20◦C. If the specific heat capacity of water is 4186 J/kg◦C and the specific heat
capacity of copper is 390 J/kg◦C, calculate the heat lost by the water and the
copper container during this process.
Solution
Step 1: Calculate the heat lost by the water.
The formula for calculating heat lost or gained is Q=mc∆T, where: - Qis
the heat lost or gained, - mis the mass of the object, - cis the specific heat
capacity of the object, - ∆Tis the change in temperature.
Given that m= 2.0 kg, c= 4186 J/kg◦C, and ∆T= 80◦C−20◦C = 60◦C,
we can calculate the heat lost by the water:
Q= (2.0 kg)(4186 J/kg◦C)(60◦C)
Q= 502320 J
14
Step 2: Calculate the heat lost by the copper container.
Using the same formula, Q=mc∆T, for the copper container: - m= 1.5 kg, -
c= 390 J/kg◦C, - ∆T= 60◦C.
Therefore, the heat lost by the copper container is:
Q= (1.5 kg)(390 J/kg◦C)(60◦C)
Q= 35100 J
Thus, the heat lost by the water is 502320 J and the heat lost by the copper
container is 35100 J.
Question 16
Question
A 5 kg block of copper is initially at a temperature of 80
°
C. How much heat
would need to be added to raise the temperature of the block to 200
°
C? (Specific
heat capacity of copper is 386 J/kg◦C)
Solution
Step 1: Find the change in temperature: The change in temperature (∆T) is
given by:
∆T=Tf−Ti
where Tiis the initial temperature and Tfis the final temperature. Given
Ti= 80Cand Tf= 200C,
∆T= 200 −80 = 120C
Step 2: Calculate the heat energy required: The heat energy required to
change the temperature of an object is given by the formula:
Q=mc∆T
where Qis the heat energy, mis the mass of the object, cis the specific heat
capacity of the material, and ∆Tis the change in temperature. Given m= 5 kg,
c= 386 J/kg◦C, and ∆T= 120C,
Q= 5 ×386 ×120 = 231600 J
Therefore, the amount of heat that needs to be added to raise the tempera-
ture of the copper block to 200
°
C is 231600 J.
15
Question 17
Question
A 2 kg block of iron at 200
°
C is dropped into a container of water at 20
°
C. If the
final temperature of the system is 50
°
C, how much heat was transferred from
the iron to the water? (Specific heat capacity of iron = 450 J/kg
°
C, specific heat
capacity of water = 4186 J/kg
°
C)
Solution
Step 1: Calculate the heat lost by the iron block to reach the final temperature.
The heat lost by the iron block can be calculated using the formula:
Q=m·c·∆T
Where: - Qis the heat lost, - mis the mass of the iron block (2 kg), - c
is the specific heat capacity of iron (450 J/kg
°
C), and - ∆Tis the change in
temperature of the iron (200
°
C - 50
°
C).
Plugging in the values:
Qiron = 2 kg ×450 J/kg
°
C×(200C−50C)
Qiron = 2 kg ×450 J/kg
°
C×150C
Qiron = 2 ×450 ×150 J
Qiron = 135000 J
Step 2: Calculate the heat gained by the water to reach the final temperature.
The heat gained by the water can be calculated using the same formula:
Q=m·c·∆T
Where: - Qis the heat gained, - mis the mass of the water (let’s assume it’s the
same as the mass of the iron block for simplicity, i.e., 2 kg), - cis the specific
heat capacity of water (4186 J/kg
°
C), and - ∆Tis the change in temperature
of the water (50
°
C - 20
°
C).
Plugging in the values:
Qwater = 2 kg ×4186 J/kg
°
C×(50C−20C)
Qwater = 2 kg ×4186 J/kg
°
C×30C
Qwater = 2 ×4186 ×30 J
Qwater = 251160 J
Step 3: The heat transferred from the iron to the water is the same as the
heat lost by the iron (since energy is conserved). Hence, the amount of heat
transferred from the iron to the water is:
Qtransfer =Qiron = 135000 J
Therefore, the amount of heat transferred from the iron to the water is
135,000 J.
16
Question 18
Question
A block of iron initially at a temperature of 100
°
C is dropped into a container of
water at 20
°
C. The mass of the iron block is 2 kg and the mass of the water in
the container is 5 kg. If the specific heat capacity of iron is 450 J/kg ·K and the
specific heat capacity of water is 4186 J/kg ·K, calculate the final temperature
of the system when thermal equilibrium is reached. Assume no heat is lost to
the surroundings.
Solution
Step 1: Calculate the heat lost by the iron block using the formula Q=mc∆T,
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature. The initial temperature of the iron block is 100
°
C and the final
temperature is the same as the final temperature of the system.
Qlost =mironciron∆Tiron
Qlost = 2 kg ×450 J/kg ·K×(Tfinal −100)
Step 2: Calculate the heat gained by the water using the same formula.
Qgain =mwatercwater∆Twater
Qgain = 5 kg ×4186 J/kg ·K×(Tfinal −20)
Step 3: Since no heat is lost to the surroundings, the heat lost by the iron
block is equal to the heat gained by the water.
2 kg ×450 J/kg ·K×(Tfinal −100) = 5 kg ×4186 J/kg ·K×(Tfinal −20)
Step 4: Solve the equation to find the final temperature Tfinal.
900 J/K ×(Tfinal −100) = 20930 J/K ×(Tfinal −20)
900Tfinal −90000 = 20930Tfinal −418600
20930Tfinal −900Tfinal = 418600 −90000
20030Tfinal = 328600
Tfinal = 16.4◦C
Therefore, the final temperature of the system when thermal equilibrium is
reached is 16.4◦C.
17
Question 19
Question
A block of copper with a mass of 0.5 kg is heated from 20
°
C to 90
°
C. Calcu-
late the amount of heat required to raise the temperature of the copper block.
(Specific heat of copper = 385 J/kg
°
C)
Solution
Step 1: Determine the change in temperature Given: Initial temperature, T1=
20CFinal temperature, T2= 90CChange in temperature, ∆T=T2−T1
∆T= 90C−20C
∆T= 70C
Step 2: Calculate the amount of heat using the formula Q = mcT Given:
Mass of copper block, m= 0.5 kg Specific heat of copper, c= 385 J/kg
°
C
Change in temperature, ∆T= 70C
Q=mc∆T
Q= 0.5 kg ×385 J/kg
°
C×70C
Q= 0.5×385 ×70 J
Q= 13,475 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 90
°
C is 13,475 J.
Question 20
Question
A 2 kg block of metal is heated to a temperature of 300
°
C. If the specific heat
of the metal is 450 J/kg
°
C, how much heat energy is required to raise the
temperature of the block by 100
°
C?
Solution
Step 1: Identify the given values.
The mass of the metal block, m= 2 kg
The initial temperature of the block, T1= 300◦C
The specific heat of the metal, c= 450 J/kg
°
C
The change in temperature, ∆T= 100◦C
Step 2: Calculate the heat energy using the formula
Q=mc∆T
18
where: Qis the heat energy, mis the mass of the block, cis the specific heat of
the metal, and ∆Tis the change in temperature.
Step 3: Substitute the given values into the formula
Q= (2 kg)(450 J/kg
°
C)(100◦C)
Q= 90000 J
Therefore, the amount of heat energy required to raise the temperature of
the block by 100
°
C is 90,000 Joules.
Question 21
Question
Two identical blocks of aluminum are heated separately to different tempera-
tures. Block A is heated to 200◦C, while block B is heated to 300◦C. If both
blocks are placed in contact with each other and allowed to reach thermal equi-
librium, what will be the final temperature of the system?
Solution
Step 1: Calculate the specific heat capacity of aluminum. The specific heat
capacity of aluminum is approximately 900 J/kg◦C.
Step 2: Determine the mass of the aluminum blocks. Let’s assume the mass
of each block is mkg.
Step 3: Calculate the heat absorbed by each block using the formula: Q=
mc∆T, where mis the mass, cis the specific heat capacity, and ∆Tis the
change in temperature.
For block A: QA=m·900 ·(300 −200) QA= 100m·900 QA= 90000m
Joules
For block B: QB=m·900 ·(300 −200) QB= 100m·900 QB= 90000m
Joules
Step 4: Since no heat is lost to the surroundings, the total heat gained
by block A must be equal to the total heat lost by block B in order to reach
thermal equilibrium. Therefore, we can set QA=QBand solve for the final
temperature.
90000m= 90000m200 ·c·∆TA= 200 ·c·∆TB200 ·∆TA= 300 ·∆TB
∆TB=2
3·∆TA
The final equation we will solve is: 200 + ∆TA= 300 −∆TB.
200 + ∆TA= 300 −2
3·∆TA5
3·∆TA= 100 ∆TA=3
5·100 ∆TA= 60
Therefore, the final temperature of the system is 300 −60 = 240◦C.
19
Question 22
Question
A copper bar of length 2 m and cross-sectional area 0.01 m2is heated from 20
°
C
to 80
°
C. If the thermal conductivity of copper is 400 W/(m K), calculate the
rate at which heat flows through the bar.
Solution
Step 1: Calculate the temperature difference Given that the initial temperature,
T1, is 20
°
C and the final temperature, T2, is 80
°
C, the temperature difference,
∆T, is:
∆T=T2−T1= 80
°
C−20
°
C = 60
°
C
Step 2: Find the rate at which heat flows through the bar The rate at which
heat flows through the bar can be calculated using the formula:
P=kA∆T
L
where: P= rate of heat transfer through the bar, k= thermal conductivity of
copper (400 W/(m K)), A= cross-sectional area of the bar (0.01 m2), ∆T=
temperature difference (60
°
C), and L= length of the bar (2 m).
Plugging in the values:
P=400 ×0.01 ×60
2
P=240
2
P= 120 W
Therefore, the rate at which heat flows through the copper bar is 120 W.
Question 23
Question
A 2 kg block of copper initially at 300 K is heated until it reaches a final
temperature of 600 K. If the specific heat capacity of copper is 386 J/kg-K, how
much heat is added to the block during this process?
Solution
Step 1: Calculate the heat added using the equation Q=mc∆T, where Qis
the heat added, mis the mass of the block, cis the specific heat capacity of
copper, and ∆Tis the change in temperature.
Given: m= 2 kg, c = 386 J/kg-K,∆T= 600 K −300 K = 300 K
20
Q= (2 kg)(386 J/kg-K)(300 K)
Q= 2 ×386 ×300 J
Q= 231600 J
Therefore, 231600 J of heat is added to the block during this process.
Question 24
Question
A block of ice at -10
°
C is initially on a scale and its melting produces water
at 0
°
C. If the block’s mass is 500 grams and no heat is exchanged with the
surroundings, determine:
1. the amount of heat involved in the process,
2. the entropy change for the ice-water process, and
3. the entropy change for the water increase in temperature to 10
°
C.
(The specific heat capacity of ice is cice = 0.5 cal/g
°
C, the specific heat capacity
of water is cwater = 1 cal/g
°
C, and the heat of fusion for ice is 80 cal/g.)
Solution
1. Let’s start by calculating the heat involved in the process.
Step 1: The heat required to raise the temperature of the ice from -10
°
C to
0
°
C is given by the formula:
Q1=mcice∆T
where mis the mass of the ice, cice is the specific heat capacity of ice, and ∆T
is the temperature change. Substitute the given values to obtain:
Q1= 500 ×0.5×(0 −(−10))
Q1= 500 ×0.5×10 = 2500 cal
Step 2: The heat of fusion required to melt the ice into water at 0
°
C is
given by:
Q2=m·heat of fusion
Substitute the values to find:
Q2= 500 ×80 = 40,000 cal
Therefore, the total heat involved in the process is:
Qtotal =Q1+Q2= 2500 + 40,000 = 42,500 cal
21
2. Next, let’s calculate the entropy change for the ice-water process.
Step 3: The entropy change for the ice-water process is given by:
∆S1=Qtotal
T
where Qtotal is the total heat involved in the process and Tis the temperature
of the process. Substitute the values to find:
∆S1=42,500
273
∆S1= 156.02 cal/K
3. Finally, let’s calculate the entropy change for the water increasing in
temperature from 0
°
C to 10
°
C.
Step 4: The entropy change for the water heating process is given by:
∆S2=Q3
T
where Q3is the heat required for the water to increase in temperature and Tis
the temperature of the process.
Q3=mcwater∆T= 500 ×1×(10 −0) = 5000 cal
Therefore,
∆S2=5000
283 = 18.01 cal/K
Question 25
Question
A copper rod of length 2 m and cross-sectional area 2 cm2is initially at a
temperature of 200◦C. It is placed in an ice-water mixture at 0◦C. If the rod
absorbs 5000 J of heat from the surrounding, calculate the final temperature of
the rod. Assume the specific heat capacity of copper is 400 J/(kg·K), density
of copper is 8.96 g/cm3, and latent heat of fusion of ice is 334 kJ/kg.
Solution
Step 1: Calculate the mass of the copper rod.
Given the density of copper is 8.96 g/cm3, we can convert the volume of the rod
(2 m ×2 cm2) into grams:
Volume = length ×cross-sectional area = 2 m ×2 cm2= 4 ×104cm3
Mass = Density×Volume = 8.96 g/cm3×4×104cm3= 3.584 ×105g = 358.4 kg
22
Step 2: Calculate the initial heat content of the copper rod.
The initial temperature of the copper rod is 200◦C. The specific heat capacity
of copper is 400 J/(kg·K). The formula to calculate heat content is:
Q=mc∆T
where Qis the heat content, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature. Substitute the given values:
Q= 358.4 kg ×400 J/(kg ·K) ×(200 −T)
Step 3: Calculate the heat required to change the temperature of the copper
rod to 0◦C.
Using the same formula as in Step 2, but with the final temperature as 0◦C:
Q2= 358.4 kg ×400 J/(kg ·K) ×(T−0)
Step 4: Calculate the heat required to freeze the water in the ice-water
mixture.
Since the rod absorbs 5000 J of heat, the heat required to freeze the water can
be calculated as:
Q3= 5000 J + Q2+Q
Step 5: Calculate the final temperature of the rod.
Substitute Q3into the equation from Step 4 and solve for T:
5000 J = Q3
Question 26
Question
A copper rod of length 2 meters is initially at a temperature of 100
°
C. If one
end of the rod is placed in a furnace at 300
°
C and the other end is placed in an
ice bath at 0
°
C, what is the temperature at a point 0.5 meters from the end in
the furnace after 10 minutes? Assume that the rod is well insulated and can be
approximated as one-dimensional. The thermal conductivity of copper is 401
W/(m
·
K) and its density is 8.96 ×103kg/m
³
. Specific heat capacity of copper
is 385 J/(kg
·
K).
Solution
Step 1: We first need to calculate the thermal diffusivity of copper, given by
the formula
α=κ
ρcp
,
23
where κis the thermal conductivity, ρis the density, and cpis the specific heat
capacity. Substituting the given values, we have
α=401
8.96 ×103×385 ≈1.41 ×10−4m2/s.
Step 2: Next, we can use the solution to the heat equation in one dimension
with a fixed end to find the temperature distribution along the rod. The general
form of the temperature distribution is given by
T(x, t) = T1+T2−T1
Lx+
∞
X
n=1
Bnsin nπx
Le−α(nπ/L)2t,
where T1= 100◦C, T2= 300◦C, and L= 2 m.
Step 3: To find the coefficients Bn, we use the initial condition T(x, 0) =
100◦C for 0 < x < L. The uniform temperature distribution at t= 0 implies
that Bn= 0 for all nexcept for n= 0. Thus, the temperature distribution
simplifies to
T(x, t) = 100 + 200
2x−400
π
∞
X
n=1
(−1)n
nsin nπx
2e−α(nπ/2)2t.
Step 4: Now, we need to find the temperature at x= 0.5 m and t= 10
minutes. Substituting x= 0.5 m, t= 600 s into the equation above and
evaluating the series numerically, we can find the temperature.
This solution involves solving a partial differential equation and evaluating
an infinite series, which may require numerical techniques.
Question 27
Question
A metal block of mass 0.5 kg is initially at a temperature of 200
°
C. It is then
heated until it reaches a temperature of 400
°
C. If the specific heat capacity
of the metal is 500 J/kg
°
C, calculate the heat energy transferred to the block
during this process.
Solution
Step 1: Identify the given values and the formula for calculating heat energy
transferred. Given: Mass of the metal block, m= 0.5 kg Initial temperature,
Ti= 200CFinal temperature, Tf= 400CSpecific heat capacity, c= 500 J/kg
°
C
The formula for calculating heat energy transferred is:
Q=mc∆T
24
where Qis the heat energy transferred, mis the mass of the object, cis the
specific heat capacity of the material, and ∆Tis the change in temperature
(Tf−Ti).
Step 2: Calculate the change in temperature (∆T).
∆T=Tf−Ti= 400C−200C= 200C
Step 3: Substitute the given values into the formula to find the heat energy
transferred.
Q= (0.5 kg) ×(500 J/kg
°
C) ×200C
Q= 100 J/
°
C×200C= 20000 J
Therefore, the heat energy transferred to the metal block during this process
is 20,000 J.
Question 28
Question
A 500 g block of copper at 100
°
C is dropped into a container of water at 20
°
C.
If the final temperature of the system is 30
°
C, determine the mass of water in
the container. (Specific heat capacity of copper: cCu = 0.385 J/g
°
C, specific
heat capacity of water: cH2O = 4.18 J/g
°
C)
Solution
Step 1: Identify the heat lost by the copper block and the heat gained by the
water. The heat lost by the copper block can be calculated as:
Qlost =mCu ×cCu ×(Tf−TCu)
Where: - mCu is the mass of the copper block (500 g), - cCu is the specific heat
capacity of copper (0.385 J/g
°
C), - Tfis the final temperature of the system
(30
°
C), - TCu is the initial temperature of the copper block (100
°
C).
The heat gained by the water can be calculated as:
Qgain =mH2O ×cH2O ×(Tf−TH2O)
Where: - mH2O is the mass of water in the container, - cH2O is the specific heat
capacity of water (4.18 J/g
°
C), - TH2O is the initial temperature of the water
(20
°
C).
Step 2: Set up the equation with the heat lost equal to the heat gained.
Since the system is isolated, the heat lost by the copper block must equal the
heat gained by the water.
mCu ×cCu ×(Tf−TCu) = mH2O ×cH2O ×(Tf−TH2O)
25
Step 3: Solve for the mass of water in the container. Substitute the given
values into the equation:
500 ×0.385 ×(30 −100) = mH2O ×4.18 ×(30 −20)
−500 ×0.385 ×70 = mH2O ×4.18 ×10
−133.75 = mH2O ×41.8
mH2O =−133.75
41.8
mH2O ≈ −3.194 g
The negative mass indicates there was an error in the calculations or as-
sumptions made. Please review the calculations and assumptions to find the
mistake.
Question 29
Question
A block of copper of mass 0.5 kg at a temperature of 200
°
C is dropped into 1 kg
of water at 20
°
C. If the specific heat capacity of copper is 0.39 J/g
°
C and the
specific heat capacity of water is 4.18 J/g
°
C, determine the final equilibrium
temperature of the system. Assume no heat is lost to the surroundings and
neglect any heat capacity of the container.
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
equilibrium temperature. The formula to calculate heat exchanged is given by:
Q=mc∆T
where: - Qis the heat exchanged, - mis the mass of the substance, - cis the
specific heat capacity, - ∆Tis the change in temperature.
For the copper block: m= 0.5 kg = 500 g c= 0.39 J/g
°
C Initial tempera-
ture, Tcopper = 200CFinal temperature, Tf
The heat lost by the copper block is equal to the heat gained by the water:
500 ×0.39 ×(200 −Tf) = 1000 ×4.18 ×(Tf−20)
Step 2: Solve for the final equilibrium temperature. Solving the equation
from Step 1 will give us the value of Tf:
500 ×0.39 ×(200 −Tf) = 1000 ×4.18 ×(Tf−20)
195 ×(200 −Tf) = 4180 ×(Tf−20)
26
39000 −195Tf= 4180Tf−83600
43780 = 4375Tf
Tf=43780
4375 ≈10C
Therefore, the final equilibrium temperature of the system is approximately
10C.
Question 30
Question
A solid metal cube with a side length of 10 cm is heated from an initial temper-
ature of 20
°
C to a final temperature of 80
°
C. The specific heat capacity of the
metal is 0.4 J/g
°
C. If 1000 J of heat is added to the cube, calculate the mass of
the metal cube.
Solution
Step 1: Calculate the change in temperature using the formula:
∆T=Tf−Ti= 80◦C−20◦C= 60◦C
Step 2: Next, calculate the mass of the metal cube using the formula:
Q=mc∆T
Step 3: Substitute the given values into the formula:
1000 = m×0.4×60
Step 4: Solve for the mass m:
m=1000
0.4×60 =1000
24 = 41.67 g
Therefore, the mass of the metal cube is 41.67 g.
Question 31
Question
A copper rod of length 1 m and cross-sectional area 1 cm2is used to transfer
heat from one end to the other. The temperature at one end is 200
°
C, while the
temperature at the other end is 100
°
C. If the thermal conductivity of copper is
400 W/(m·K), calculate the rate of heat transfer through the rod.
27
Solution
Step 1: Calculate the temperature difference across the rod. Given that the
temperature at one end is 200
°
C and the temperature at the other end is 100
°
C,
the temperature difference across the rod is:
∆T= 200
°
C−100
°
C = 100
°
C
Step 2: Convert the temperature to Kelvin. To convert temperature in
degrees Celsius to Kelvin, we use the formula T(K) = T(
°
C) + 273.15. Thus,
the temperature difference in Kelvin is:
∆T= 100
°
C + 273.15 = 373.15 K
Step 3: Calculate the rate of heat transfer using Fourier’s Law. The rate of
heat transfer through the rod is given by Fourier’s Law:
Q=kA∆T
L
where Qis the rate of heat transfer, kis the thermal conductivity of copper, A
is the cross-sectional area of the rod, ∆Tis the temperature difference across
the rod, and Lis the length of the rod.
Step 4: Substitute the known values into Fourier’s Law. Substitute k= 400
W/(m·K), A= 1 cm2= 0.0001 m2, ∆T= 373.15 K, and L= 1 m into Fourier’s
Law:
Q= (400 W/(m·K)) ×(0.0001 m2)×373.15 K
1 m
Step 5: Calculate the rate of heat transfer.
Q= 400 ×0.0001 ×373.15 W = 14.926 W
Therefore, the rate of heat transfer through the rod is 14.926 W.
Question 32
Question
A copper rod of length 1 m and cross-sectional area 1 ×10−4m2is heated from
20
°
C to 60
°
C. If the thermal conductivity of copper is 400 W/(m·K) and the
density of copper is 8.96 ×103kg/m3, calculate the amount of heat transferred
to the rod during this process.
Solution
Step 1: Calculate the volume of the copper rod. The volume of the copper rod
can be calculated using the formula:
Volume = Length ×Area
28
Volume = 1 m ×1×10−4m2
Volume = 1 ×10−4m3
Step 2: Calculate the mass of the copper rod. The mass of the copper rod
can be calculated using the formula:
Mass = Volume ×Density
Mass = 1 ×10−4m3×8.96 ×103kg/m3
Mass = 8.96 ×10−1kg
Step 3: Calculate the initial temperature of the copper rod in Kelvin.
T1= 20 + 273.15
T1= 293.15 K
Step 4: Calculate the final temperature of the copper rod in Kelvin.
T2= 60 + 273.15
T2= 333.15 K
Step 5: Calculate the change in temperature.
∆T=T2−T1
∆T= 333.15 −293.15
∆T= 40 K
Step 6: Calculate the amount of heat transferred. The amount of heat
transferred can be calculated using the formula:
Heat = Thermal Conductivity ×Area ×∆T×Time
Heat = 400 W/(m ·K) ×1×10−4m2×40 K
Heat = 160 W
Therefore, the amount of heat transferred to the copper rod during this
process is 160 J.
Question 33
Question
A metal object of mass 0.5 kg at a temperature of 100
°
C is placed in a container
of water at 20
°
C. The final temperature of the metal object and water is 30
°
C.
If the specific heat capacity of the metal is 0.5 J/g
°
C, determine the mass of
water in the container. Assume no heat is lost to the surroundings.
29
Solution
Step 1: Calculate the heat absorbed by the metal object. The heat absorbed
by the metal object can be calculated using the formula:
Qmetal =mc∆T
where: - m= 0.5 kg is the mass of the metal object, - c= 0.5 J/g
°
C is the specific
heat capacity of the metal, and - ∆T=Tfinal −Tinitial = 30 −100 = −70
°
C is
the change in temperature of the metal object.
Plugging in the values, we get:
Qmetal = 0.5 kg ×0.5 J/g
°
C× −70
°
C
Step 2: Calculate the heat released by the metal object to the water. Since
the heat lost by the metal object is equal to the heat gained by the water
(assuming no heat is lost to the surroundings), we have:
Qmetal =Qwater
Step 3: Calculate the heat absorbed by the water. The heat absorbed by
the water can be calculated using the formula:
Qwater =mc∆T
where: - cwater = 4.18 J/g
°
C is the specific heat capacity of water, and - ∆T=
Tfinal −Tinitial = 30 −20 = 10
°
C is the change in temperature of the water.
Plugging in the values, we get:
Qwater =mwater ×4.18 J/g
°
C×10
°
C
Step 4: Solve for the mass of water. Since Qmetal =Qwater, we can set the
two expressions equal to each other:
0.5 kg ×0.5 J/g
°
C× −70
°
C = mwater ×4.18 J/g
°
C×10
°
C
Solving for mwater, we find:
mwater =0.5×0.5× −70
4.18 ×10
Therefore, the mass of water in the container is mwater kg.
Question 34
Question
A copper block with a mass of 500 g is initially at a temperature of 100◦C. The
block is placed in 200 g of water at 20◦C. If all the heat lost by the copper block
is absorbed by the water, what will be the final equilibrium temperature of the
system? Assume specific heat capacity of copper is 0.385 J/g◦C and specific
heat capacity of water is 4.18 J/g◦C.
30
Solution
Step 1: Calculate the heat lost by the copper block: The heat lost by the copper
block can be calculated using the formula:
Q=mc∆T
where - mis the mass of the copper block (500 g), - cis the specific heat capacity
of copper (0.385 J/g◦C), - ∆Tis the change in temperature of the copper block.
The change in temperature (∆T) of the copper block is:
∆T= (final temperature) −(initial temperature)
∆T=Tf−100
Step 2: Calculate the heat gained by the water: The heat gained by the
water can also be calculated using the formula:
Q=mc∆T
where - mis the mass of the water (200 g), - cis the specific heat capacity of
water (4.18 J/g◦C), - ∆Tis the change in temperature of the water.
The change in temperature (∆T) of the water is:
∆T= (final temperature) −20
Step 3: Since the heat lost by the copper block is equal to the heat gained
by the water (assuming no heat losses to the surroundings), we can set the two
equations equal to each other and solve for the final temperature (Tf):
mcopperccopper∆Tcopper =mwatercwater∆Twater
Substitute the given values and expressions for ∆Tcopper and ∆Twater into
the equation and solve for Tf.
Question 35
Question
A copper rod of length 1 m and diameter 2 cm is initially at a temperature
of 100
°
C. The rod is then submerged in a water bath at 0
°
C. Given that the
thermal conductivity of copper is 390 W/(m·K), determine how long it will take
for the rod to cool down to 10
°
C.
Solution
Step 1: Calculate the cross-sectional area of the rod. The cross-sectional area
of the rod can be calculated using the formula for the area of a circle: A=πd2
4,
31
where dis the diameter of the rod. Substituting d= 2 cm into the formula, we
have:
A=π·(0.02 m)2
4=π·0.0004 m2
4= 0.00003142 m2
Step 2: Calculate the volume of the rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder: V=A·L, where L
is the length of the rod. Thus,
V= 0.00003142 m2·1 m = 0.00003142 m3
Step 3: Calculate the thermal energy of the rod. The thermal energy of
the rod can be calculated using the formula: Q=mc∆T, where mis the mass
of the rod, cis the specific heat capacity of copper, and ∆Tis the change in
temperature. Since we are given the temperature in Celsius, we should convert
it to Kelvin: ∆T= 10 −100 = −90
°
C = -90 K. The specific heat capacity of
copper is 385 J/(kg
K), and the density of copper is 8900 kg/m
³
. Thus,
m= Density ×Volume = 8900 kg/m3×0.00003142 m3= 0.280 kg
Therefore, Q= 0.280 kg ×385
32
