PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 2
Liberty University
Question 1
Question
A steel rod of length 1.5 m and diameter 2 cm is initially at a temperature of
200
°
C. If the rod absorbs 1500 J of heat, calculate the final temperature of the
rod (specific heat capacity of steel = 490 J/kg
°
C and density of steel = 7850
kg/m3).
Solution
Step 1: Calculate the mass of the steel rod using its volume and density. The
volume of the rod can be calculated using the formula for the volume of a
cylinder:
V=πr2h
where ris the radius and his the height of the cylinder. Given that the diameter
is 2 cm, the radius ris 1 cm or 0.01 m and the height his 1.5 m.
Plugging in the values:
V=π×(0.01)2×1.5
V= 4.71 ×10−5m3
The mass mof the rod can be calculated using the formula:
m= density ×volume
Plugging in the values for density of steel:
m= 7850 ×4.71 ×10−5
m= 0.368 kg
Step 2: Calculate the change in temperature of the rod using the specific
heat capacity formula. The heat absorbed by the rod can be related to its mass,
specific heat capacity, and change in temperature using the formula:
Q=mc∆T
Given that the rod absorbs 1500 J of heat, the specific heat capacity of steel is
490 J/kg
°
C, and the mass of the rod is 0.368 kg, we have:
1500 = 0.368 ×490 ×∆T
∆T=1500
0.368 ×490
∆T≈7.06
°
C
Step 3: Calculate the final temperature of the rod. The final temperature
Tfof the rod can be calculated as:
Tf=Ti+ ∆T
Given that the initial temperature Tiis 200
°
C, we have:
Tf= 200 + 7.06
Tf≈207.06
°
C
Therefore, the final temperature of the rod is approximately 207.06
°
C.
Question 2
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
200◦C. It is then heated until the temperature reaches 400◦C. If the rod absorbs
500 J of heat while maintaining uniform temperature, find the change in length
of the rod (assuming coefficient of linear expansion of copper is 1.7×10−5K−1).
Solution
Step 1: Calculate the initial volume of the rod. The initial volume, Vi, of the
rod can be calculated as the volume of a cylinder:
Vi=πd
22
·L
where dis the diameter and Lis the length. Since the diameter is given in
centimeters, we need to convert it to meters:
Vi=π1 cm
2×1002
·2 m
2
Vi=π0.01 m
22
·2 m
Vi=π×0.01
2×0.01
2×2
Vi=π×0.012
4×2
Vi=π×0.012
2
Vi=π×0.0001
2
Vi= 0.000157 m3
Step 2: Calculate the final volume of the rod. As the rod heats up uniformly,
the change in temperature leads to the expansion of the rod. The change in
temperature is:
∆T= 400 −200 = 200◦C
The change in length, ∆L, is given by the expression:
∆L=Lα∆T
where: - Lis the original length, - αis the coefficient of linear expansion, -
∆Tis the change in temperature.
Given αand ∆T, we can find ∆L.
∆L= 2 m ×(1.7×10−5K−1)×200 K
∆L= 2 ×1.7×10−5×200 m
∆L= 6.8×10−3m
The final length, Lf, is:
Lf=L+ ∆L= 2 m + 6.8×10−3m
Lf= 2.0068 m
The final volume, Vf, of the rod can be calculated in the same way as the
initial volume:
Vf=πd
22
·Lf
Vf=π0.01 m
22
·2.0068 m
Vf=π0.01
22
×2.0068
3
Vf=π×0.012
4×2.0068
Vf=π×0.0001
4×2.0068
Vf=π×0.0001 ×2.0068
4
Vf=π×0.00020068
4
Vf=π×2.0068 ×10−4
4
Vf=2.0068π×10−4
4
Vf= 0.0001577 m3
Step 3: Calculate the change in volume. The change in volume, ∆V, is given
by:
∆V
Question 3
Question
A piece of aluminum with a mass of 250 g at a temperature of 100
°
C is placed
in 500 g of water at a temperature of 20
°
C. Assuming no heat is lost to the
surroundings, calculate the final temperature of the system when thermal equi-
librium is reached. (Specific heat capacity of aluminum = 0.902 J/g
°
C, specific
heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the aluminum. Given: - Mass
of aluminum, ma= 250 g - Initial temperature of aluminum, Ta,i = 100
°
C -
Specific heat capacity of aluminum, ca= 0.902 J/g
°
C - Final temperature of the
system, Tf
Using the formula for heat transfer:
Q=mc∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Since the aluminum is cooling down:
Qa=−mca∆Ta
Qa=−250 ×0.902 ×(Tf−100)
4
Step 2: Calculate the heat gained or lost by the water. Given: - Mass of
water, mw= 500 g - Initial temperature of water, Tw,i = 20
°
C - Specific heat
capacity of water, cw= 4.18 J/g
°
C
Since the water is getting warmer:
Qw=mcw∆Tw
Qw= 500 ×4.18 ×(Tf−20)
Step 3: Set up the equation for heat conservation. Since no heat is lost to
the surroundings, the heat lost by aluminum must be equal to the heat gained
by water:
−250 ×0.902 ×(Tf−100) = 500 ×4.18 ×(Tf−20)
Step 4: Solve for the final temperature of the system, Tf.
−225.5Tf+ 22550 = 2090Tf−41800
2315.5Tf= 64350
Tf=64350
2315.5
Tf≈27.8
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 27.8
°
C.
Question 4
Question
A 2 kg block of aluminum at an initial temperature of 100
°
C is dropped into a
large bath containing 20 kg of water at 20
°
C. Assuming no heat is lost to the sur-
roundings, calculate the final equilibrium temperature of the system. The spe-
cific heat capacity of aluminum is 900 J/kg◦C and that of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the aluminum block as it cools down from
100
°
C to the final equilibrium temperature. The change in temperature (∆T)
is the initial temperature (100
°
C) minus the final equilibrium temperature (T).
The heat lost (QAl) is given by QAl =mc∆T, where: - mis the mass of
aluminum (2 kg), - cis the specific heat capacity of aluminum (900 J/kg◦C).
Thus, QAl = 2 kg ×900 J/kg◦C×(100 −T).
Step 2: Calculate the heat gained by the water as it heats up from 20
°
C
to the final equilibrium temperature. The change in temperature (∆T) is the
final equilibrium temperature (T) minus the initial temperature (20
°
C). The
5
heat gained (Qw) is given by Qw=mc∆T, where: - mis the mass of water
(20 kg), - cis the specific heat capacity of water (4200 J/kg◦C). Thus, Qw=
20 kg ×4200 J/kg◦C×(T−20).
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block must equal the heat gained by the water. Thus, 2 kg×900 J/kg◦C×
(100 −T) = 20 kg ×4200 J/kg◦C×(T−20).
Step 4: Solve the equation from Step 3 to find the final equilibrium temper-
ature (T) of the system.
2×900 ×(100 −T) = 20 ×4200 ×(T−20)
Solving for T, we get:
1800(100 −T) = 84000(T−20)
180000 −1800T= 84000T−1680000
1800T+ 84000T= 180000 + 1680000
85800T= 1860000
T=1860000
85800 ≈21.67◦C
Therefore, the final equilibrium temperature of the system is approximately
21.67◦C.
Question 5
Question
A copper block weighing 500 grams is initially at a temperature of 100◦C. It is
then heated until its temperature rises to 200◦C. If the specific heat capacity of
copper is 0.385 J/g◦C, how much heat energy was added to the block?
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, T1= 100◦C Final temperature, T2= 200◦C
The change in temperature, ∆T=T2−T1= 200◦C - 100◦C = 100◦C
Step 2: Calculate the heat energy added to the copper block. The heat
energy added can be calculated using the formula:
q=mc∆T
where: q= heat energy added (in joules) m= mass of the copper block (in
grams) c= specific heat capacity of copper (in J/g◦C) ∆T= change in tem-
perature (in ◦C)
Given: m= 500 grams c= 0.385 J/g◦C ∆T= 100◦C
6
Substitute the values into the formula:
q= 500 ×0.385 ×100
q= 19250 joules
Therefore, the heat energy added to the copper block is 19250 joules.
Question 6
Question
A copper rod of length 2 meters and diameter 1 cm is initially at a temperature of
100◦C. It is placed in a water bath at 0◦C. Given that the thermal conductivity of
copper is 400 W/(m·K) and the specific heat capacity of copper is 390 J/(kg·K),
calculate the time it takes for the rod to reach thermal equilibrium with the
water bath.
Solution
Step 1: Calculate the surface area of the rod. The surface area of a cylinder is
given by the formula:
A= 2πrh
where ris the radius and his the height of the cylinder. Given the diameter of
the rod is 1 cm, the radius r=1
2cm. Converting the radius to meters: r=1
200
m. The height of the cylinder is 2 meters. Therefore, the surface area of the
rod is:
A= 2π1
200·2
Step 2: Calculate the volume of the rod. The volume of a cylinder is given
by the formula:
V=πr2h
The volume of the rod is:
V=π1
2002
·2
Step 3: Calculate the mass of the rod. The density of copper is approxi-
mately 8900 kg/m3. The mass of the rod is given by:
m= density ×volume
Step 4: Calculate the heat energy in the rod. The heat energy in the rod is
given by:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
7
Step 5: Calculate the rate of heat transfer from the rod to the water bath.
The rate of heat transfer is given by Fourier’s Law:
dQ
dt =kAdT
dx
where kis the thermal conductivity, Ais the surface area, dT
dx is the temperature
gradient, and dQ
dt is the rate of heat transfer.
Step 6: Equate the rates of heat transfer. At thermal equilibrium, the rate
of heat transfer from the rod to the water bath is equal to the rate of heat
transfer from the water bath to the rod.
kAdT
dx rod =kAdT
dx water bath
Step 7: Calculate the time it takes for the rod to reach thermal equilibrium.
Use the relation ∆T=Tfinal −Tinitial and the fact that the rod and water bath
will reach the same temperature at equilibrium.
Rate of heat transfer from rod to water bath = Rate of heat transfer from water bath to rod
Solve this equation to find the time it takes for the rod to reach thermal
equilibrium with the water bath.
Question 7
Question
A copper sphere of radius 5 cm is heated until its temperature reaches 100
°
C.
If the thermal conductivity of copper is 400 W/mK and the temperature of the
surrounding air is 25C, calculate the rate of heat loss from the sphere due to
conduction.
Solution
Step 1: First, calculate the surface area of the sphere using the formula A=
4πr2. Given that the radius r= 5 cm, we convert it to meters: r= 0.05 m.
A= 4π(0.05)2
= 4 ×π×0.0025
= 0.0314 m2
Step 2: Next, calculate the temperature difference between the sphere and
the surroundings.
∆T= 100C−25C= 75C= 75 K
Step 3: Now, use the formula for the rate of heat loss through conduction:
Rate of heat loss = k×A×∆T
d
8
where kis the thermal conductivity of copper, Ais the surface area, ∆Tis the
temperature difference, and dis the thickness of the material (which we assume
as negligible here since it is a sphere). Substitute the given values:
Rate of heat loss = 400 W/mK ×0.0314 m2×75 K
0
= 400 ×0.0314 ×75
= 942 W
Therefore, the rate of heat loss from the sphere due to conduction is 942 W .
Question 8
Question
A 2 kg block of metal at 80
°
C is dropped into 5 kg of water at 20
°
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 final temperature of the mixture assuming
no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the metal block as it cools down to the final
temperature. The heat lost by the metal block is given by the formula:
Qmetal =mmetal ×cmetal ×∆T
where: - mmetal = 2 kg (mass of the metal block), - cmetal = 450 J/kg
°
C (specific
heat capacity of the metal), - Tinitial = 80C(initial temperature of the metal),
and - Tfinal =Tfinal (final temperature of the mixture).
The change in temperature, ∆T, for the metal is given by:
∆Tmetal =Tfinal −Tinitial
Step 2: Calculate the heat gained by the water as it rises to the final tem-
perature. The heat gained by the water is given by the formula:
Qwater =mwater ×cwater ×∆T
where: - mwater = 5 kg (mass of the water), - cwater = 4186 J/kg
°
C (specific heat
capacity of water), and - Tinitial = 20C(initial temperature of the water).
Step 3: Set up the equation for heat transfer. Since in this case, no heat is
lost to the surroundings, the heat lost by the metal must be equal to the heat
gained by the water. Therefore:
Qmetal =Qwater
9
Step 4: Solve for the final temperature. Substitute the expressions for heat
lost by the metal and heat gained by the water into the equation from Step 3:
mmetal ×cmetal ×(Tfinal −Tinitial) = mwater ×cwater ×(Tfinal −Tinitial)
Solve for Tfinal to find the final temperature of the mixture.
Question 9
Question
A 0.5 kg block of copper is heated from 20
°
C to 80
°
C. If the specific heat capacity
of copper is 390 J/kg ·
°
C, how much heat energy is required?
Solution
Step 1: Calculate the change in temperature Given: Initial temperature, T1=
20CFinal temperature, T2= 80CChange in temperature, ∆T=T2−T1=
80C−20C= 60C
Step 2: Calculate the heat energy required using the formula:
Q=mc∆T
where: Q= heat energy m= mass of the block = 0.5 kg c= specific heat
capacity of copper = 390 J/kg ·
°
C ∆T= change in temperature = 60
°
C
Step 3: Substitute the given values into the formula and solve for Q:
Q= (0.5 kg)(390 J/kg ·
°
C)(60C)
Q= 0.5×390 ×60
Q= 11700 J
Answer: The heat energy required is 11700 J.
Question 10
Question
A solid copper sphere at an initial temperature of 100◦C is placed in a large
bath of water at a constant temperature of 0◦C. The sphere loses heat to the
water until it reaches thermal equilibrium. If the mass of the sphere is 2 kg and
its specific heat capacity is 0.385 J/g◦C, calculate the final temperature of the
copper sphere.
10
Solution
Step 1: Calculate the initial heat energy of the copper sphere using the formula:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the sphere, - cis the specific
heat capacity of copper (0.385 J/g◦C), - ∆Tis the change in temperature.
For the initial temperature of 100◦C and the final temperature Tf:
Q= 2 ×0.385 ×(100 −Tf)
Step 2: Calculate the heat energy transferred from the sphere to the water
using the same formula:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the sphere, - cis the specific
heat capacity of water (4.18 J/g◦C), - ∆Tis the change in temperature (from
100◦C to 0◦C).
The heat energy transferred is equal to the initial heat energy of the copper
sphere:
2×0.385 ×(100 −Tf)=2×4.18 ×100
Step 3: Solve the equation obtained in Step 2 to find the final temperature
Tf:
0.77 ×(100 −Tf) = 836
77 −0.77Tf= 836
0.77Tf=−759
Tf=−759
0.77 ≈ −985.71◦C
Since the final temperature must be between 0◦C and 100◦C, this negative
value indicates an error in the calculations. Double-check the calculations to
find the mistake.
Question 11
Question
A steel rod of length 2 m and diameter 1 cm is heated from 20
°
C to 200
°
C. If
the density of steel is 7850 kg/m3and its specific heat capacity is 460 J/kg
°
C,
calculate the heat energy required. Assume the rod is one-dimensional and
neglect heat losses to the surroundings.
11
Solution
Step 1: Calculate the mass of the steel rod. Given that the steel rod has a
length of 2 m and a diameter of 1 cm, we can first find the volume of the rod
and then calculate its mass.
The cross-sectional area of the rod is given by:
A=πd2
4=π×(0.01)2
4= 7.85398 ×10−5m2
The volume of the rod is:
V=A×length = 7.85398 ×10−5×2=1.5708 ×10−4m3
The mass of the rod can be found using the density formula:
mass = density ×volume = 7850 ×1.5708 ×10−4= 1.22887 kg
Therefore, the mass of the steel rod is 1.22887 kg.
Step 2: Calculate the heat energy required. The heat energy (Q) required
to change the temperature of an object can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, and - ∆Tis the change in temperature.
Given: - m= 1.22887 kg, - c= 460 J/kg
°
C, and - ∆T= 200 −20 = 180
°
C
Substitute these values into the formula to find the heat energy required:
Q= 1.22887 ×460 ×180 = 101238.642 J
Therefore, the heat energy required to heat the steel rod from 20
°
C to 200
°
C
is 101238.642 J.
Question 12
Question
A copper block with a mass of 500 g and at a temperature of 200
°
C is placed in
a container with 500 g of water at 20
°
C. If the final temperature of the system
is 40
°
C, calculate the specific heat capacity of copper. (Specific heat capacity
of water is 4186 J/kg
°
C.)
Solution
Step 1: Calculate the heat gained by the copper block. The heat absorbed by
the copper block is given by the formula:
Q=mc∆T
12
where: - Qis the heat absorbed, - mis the mass of the copper block, - cis the
specific heat capacity of copper, and - ∆Tis the change in temperature of the
copper block.
Given the mass of the copper block is 500 g (0.5 kg), the initial temperature
is 200
°
C, and the final temperature is 40
°
C, we can calculate ∆Tas:
∆T= 40C−200C=−160C=−160K
Note: The temperature needs to be converted to Kelvin for calculations involv-
ing temperature difference.
Substitute the values into the formula to find the heat absorbed by the
copper block:
Qcopper = (0.5kg)×c×(−160K)
Step 2: Calculate the heat lost by the water. The heat lost by the water is
given by the formula:
Q=mc∆T
where: - Qis the heat lost, - mis the mass of the water, - cis the specific heat
capacity of water, and - ∆Tis the change in temperature of the water.
Given the mass of the water is 500 g (0.5 kg) and the initial temperature is
20
°
C, we can calculate ∆Tas:
∆T= 40C−20C= 20K
Substitute the values into the formula to find the heat lost by the water:
Qwater = (0.5kg)×4186 J/kgC ×20K
Step 3: Since heat is conserved in the system, the heat lost by the water is
equal to the heat gained by the copper block. Therefore, equate Qcopper and
Qwater and solve for the specific heat capacity of copper. We have:
(0.5kg)×c×(−160K) = (0.5kg)×4186 J/kgC ×20K
c×(−160) = 4186 ×20
c=4186 ×20
−160
Thus, the specific heat capacity of copper is determined by solving the above
equation.
Question 13
Question
A certain metal has a specific heat capacity of 0.45 J/g
°
C. How much heat is
required to raise the temperature of a 250 g sample of this metal from 20
°
C to
80
°
C?
13
Solution
Step 1: Identify the given values and the formula for heat transfer: Given:
Specific heat capacity of the metal, c= 0.45 J/g
°
C Mass of the sample, m= 250
g Initial temperature, Ti= 20
°
C Final temperature, Tf= 80
°
C
The formula for heat transfer is:
Q=mc∆T
where: Q= heat transfer (in joules) m= mass of the sample (in grams) c=
specific heat capacity (in J/g
°
C) ∆T= change in temperature (in
°
C)
Step 2: Calculate the change in temperature, ∆T:
∆T=Tf−Ti= 80 −20 = 60
°
C
Step 3: Substitute the given values into the formula for heat transfer and
solve for Q:
Q= 250 g ×0.45 J/g
°
C×60
°
C
Step 4: Calculate the heat transfer Q:
Q= 250 ×0.45 ×60 = 6750 J
Therefore, the amount of heat required to raise the temperature of the 250
g sample of metal from 20
°
C to 80
°
C is 6750 J.
Question 14
Question
A steel rod of length 1 m and cross-sectional area 0.01 m2is initially at a
temperature of 300 K. It is heated until its temperature reaches 400 K. Given
that the linear expansion coefficient of steel is 1.2×10−5K−1and the Young’s
modulus of steel is 2 ×1011 Pa, calculate the change in length of the rod and
the amount of heat added to the rod.
Solution
Step 1: Calculate the change in length of the rod using the formula for linear
expansion:
∆L=αL∆T
where: ∆L= change in length of the rod, α= linear expansion coefficient
of steel, L= original length of the rod, ∆T= change in temperature.
Given: α= 1.2×10−5K−1,L= 1 m, Tinitial = 300 K, Tfinal = 400 K.
First, calculate ∆T:
∆T=Tfinal −Tinitial = 400 K −300 K = 100 K
14
Now, substitute the given values into the formula:
∆L= (1.2×10−5K−1)(1 m)(100 K) = 0.0012 m = 1.2 mm
Therefore, the change in length of the rod is 1.2 mm.
Step 2: Calculate the amount of heat added to the rod using the formula:
Q=F
A
where: Q= amount of heat added, F= force required to stretch the rod, A
= cross-sectional area of the rod.
Given: E= 2 ×1011 Pa, ∆L= 0.0012 m, A= 0.01 m2.
First, calculate the force using Hooke’s Law:
F=E∆L
L
Substitute the given values into the formula:
F= (2 ×1011 Pa)0.0012 m
1 m = 2.4×108N
Now, substitute the force and area values into the formula for heat:
Q=2.4×108N
0.01 m2= 2.4×1010 J
Therefore, the amount of heat added to the rod is 2.4×1010 J.
Question 15
Question
A 2 kg block of copper at 100
°
C is dropped into 5 kg of water at 20
°
C in a per-
fectly insulated container. Assuming no heat is lost to the surroundings, what
will be the final equilibrium temperature of the system? (Specific heat capacity
of copper = 390 J/kg◦C and specific heat capacity of water = 4184 J/kg◦C)
Solution
Step 1: Calculate the heat lost by the copper block The heat lost by the copper
block is given by the formula:
Qlost =mc∆T
where: - mis the mass of the copper block (2 kg) - cis the specific heat capacity
of copper (390 J/kg◦C) - ∆Tis the change in temperature of the copper block
The change in temperature of the copper block is:
∆T=Tfinal −Tinitial =Tfinal −100
15
Substitute this into the formula for heat lost and solve for Qlost.
Step 2: Calculate the heat gained by the water The heat gained by the water
is given by the formula:
Qgain =mc∆T
where: - mis the mass of the water (5 kg) - cis the specific heat capacity of
water (4184 J/kg◦C) - ∆Tis the change in temperature of the water
The change in temperature of the water is:
∆T=Tfinal −Tinitial =Tfinal −20
Substitute this into the formula for heat gained and solve for Qgain.
Step 3: Since there is no heat lost to the surroundings, the heat lost by the
copper block is equal to the heat gained by the water. Set up the equation:
Qlost =Qgain
Step 4: Solve for the final equilibrium temperature, Tfinal, by combining the
results from Step 1 and Step 2:
Qlost =Qgain
mcopperccopper(Tfinal −100) = mwatercwater(Tfinal −20)
2×390 ×(Tfinal −100) = 5 ×4184 ×(Tfinal −20)
Solve for Tfinal to find the final equilibrium temperature of the system.
Question 16
Question
A copper kettle is filled with 1.5 kg of water at 20
°
C. How much heat is required
to heat the water to its boiling point (100
°
C)? The specific heat capacity of
water is 4.18 J/g
°
C, and assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the initial temperature difference. The initial temperature of
the water is 20
°
C and the final temperature (boiling point) is 100
°
C. Therefore,
the initial temperature difference is:
∆T= 100C−20C= 80C
Step 2: Calculate the heat energy required to raise the temperature of the
water from 20
°
C to 100
°
C. The heat energy required is given by the formula:
Q=mc∆T
16
where: - mis the mass of water (1.5 kg = 1500 g), - cis the specific heat
capacity of water (4.18 J/g
°
C), and - ∆Tis the temperature difference (80
°
C).
Substituting the values:
Q= 1500g×4.18 J
gC ×80C
Q= 502800 J
Therefore, the heat energy required to heat the water to the boiling point
(100
°
C) is 502800 J.
Question 17
Question
A copper block with a mass of 2 kg is heated from an initial temperature of
20
°
C to a final temperature of 70
°
C. Calculate the amount of heat required
to raise the temperature of the block. The specific heat capacity of copper is
390 J/kg◦C.
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, Ti= 20◦C Final temperature, Tf= 70◦C
The change in temperature, ∆T=Tf−Ti∆T= 70◦C−20◦C = 50◦C
Step 2: Use the formula Q=mc∆Tto calculate the amount of heat required.
Given: Mass of the copper block, m= 2 kg Specific heat capacity of copper,
c= 390 J/kg◦C
Substitute the values into the formula: Q= (2 kg)(390 J/kg◦C)(50◦C)
Step 3: Calculate the amount of heat required. Q= 2 ×390 ×50 = 39000 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 70
°
C is 39000 J.
Question 18
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 45
°
C. If the rod is heated until its temperature reaches 85
°
C, calculate the
amount of heat energy absorbed by the rod. Assume the specific heat capacity
of copper is 390 J/kg ·K and the density of copper is 8,960 kg/m3.
17
Solution
Let’s denote: - L: length of the copper rod (1.5 m) - D: diameter of the
copper rod (2 cm) - Ti: initial temperature of the copper rod (45
°
C) - Tf: final
temperature of the copper rod (85
°
C) - C: specific heat capacity of copper
(390 J/kg ·K) - ρ: density of copper (8,960 kg/m3)
First, we need to find the mass of the copper rod:
Volume of the rod = πD
22
×L
Mass of the rod = Volume of the rod ×ρ
Step 1: Calculate the volume of the rod.
Volume of the rod = π0.02 m
22
×1.5 m
Step 2: Calculate the mass of the rod.
Mass of the rod = Volume of the rod ×ρ
Step 3: Calculate the heat energy absorbed by the rod using the formula:
Q=mc∆T
where - Q: heat energy absorbed by the rod - m: mass of the rod - c: specific
heat capacity of copper - ∆T: change in temperature
Step 4: Substitute the known values into the formula and calculate the heat
energy absorbed. Make sure to convert temperatures to Kelvin.
Q=mc(Tf−Ti)
Finally, calculate the heat energy absorbed by the copper rod.
Question 19
Question
A copper rod of length Land cross-sectional area Ais initially at a temperature
T1. It is then heated to a temperature T2. Calculate the change in length of the
rod. Assume that the coefficient of linear expansion of copper is α.
Solution
Step 1: Calculate the initial length of the copper rod using the formula for linear
expansion:
Linitial =L+ ∆Linitial =L(1 + α∆T1)
18
where ∆T1=T1−T0.
Step 2: Calculate the final length of the copper rod after heating to temper-
ature T2using the same formula:
Lfinal =L+ ∆Lfinal =L(1 + α∆T2)
where ∆T2=T2−T0.
Step 3: Calculate the change in length of the rod:
∆L=Lfinal −Linitial =L(α∆T2−α∆T1)
Therefore, the change in length of the copper rod when heated from T1to
T2is ∆L=Lα(∆T2−∆T1).
Question 20
Question
A metal rod of length L, cross-sectional area A, and thermal conductivity k
is initially at a temperature T0. One end of the rod is kept at a constant
temperature Th, while the other end is kept at a constant temperature Tcsuch
that Th> T0> Tc. Find the steady-state temperature distribution in the rod
and the rate at which heat is conducted through the rod.
Solution
Step 1: Let’s first define the heat conduction equation for steady-state condi-
tions:
−q=kAdT
dx
where qis the heat conducted through the rod per unit time, xis the distance
along the rod, and dT/dx is the temperature gradient.
Step 2: Since the rod is in a steady state, the temperature distribution within
the rod is linear:
T(x) = Tc+Th−Tc
Lx
Step 3: Now, let’s find the rate at which heat is conducted through the rod
by substituting the temperature distribution into the heat conduction equation:
−q=kA d
dx Tc+Th−Tc
Lx
Step 4: Taking the derivative with respect to xgives:
−q=kATh−Tc
L
19
Step 5: Therefore, the rate at which heat is conducted through the rod is
given by:
q=−kATh−Tc
L
Step 6: In conclusion, the steady-state temperature distribution in the rod
is linear with one end at temperature Tcand the other end at temperature Th.
The rate at which heat is conducted through the rod is q=−kA(Th−Tc)/L.
Question 21
Question
A copper rod of length 1 m and diameter 2 cm initially at 100
°
C is placed in
an ice-water mixture at 0
°
C. If the rod absorbs heat from the mixture until it
reaches thermal equilibrium, determine the final temperature of the rod. Given:
specific heat capacity of copper is 390 J/kg ·K, density of copper is 8930 kg/m3,
and latent heat of fusion of ice is 334 kJ/kg.
Solution
Step 1: Calculate the mass of the copper rod: The volume of the rod can be
calculated using V=πr2h, where ris the radius of the rod and his its length.
Substituting r= 0.01 m and h= 1 m, we get:
V=π×(0.01 m)2×1 m = 3.14 ×10−4m3
The mass of the rod can be calculated using m= density ×V. Substituting
the given density of copper, we get:
m= 8930 kg/m3×3.14 ×10−4m3= 2.805 kg
Step 2: Calculate the heat absorbed by the copper rod: The heat absorbed
by 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.
Let Tfbe the final temperature of the rod. The change in temperature is
then 100 −Tf(assuming Tf<100). Therefore, Q= 2.805 kg ×390 J/kg ·K×
(100 −Tf) = 280.215 (100 −Tf) J.
Step 3: Calculate the heat released by the ice-water mixture to cool down
the copper rod: The heat released by the ice-water mixture can be calculated
using the formula Q=mL, where mis the mass of melted ice and Lis the
latent heat of fusion of ice.
Given that the copper rod absorbs all the heat required to melt the ice, we
can equate the heat absorbed by the rod to the heat released by the ice-water
mixture:
280.215 (100 −Tf) J = m×334 ×103J
Step 4: Calculate the final temperature of the copper rod by solving the
equation from Step 3.
20
Question 22
Question
A copper rod of length 2 m and diameter 4 cm is initially at a temperature of
100
°
C. If the rod is heated until its temperature reaches 200
°
C, calculate the
amount of heat transferred to the rod. Assume the specific heat capacity of
copper is 390 J/kg◦C and the density of copper is 8900 kg/m3.
Solution
Step 1: First, calculate the mass of the copper rod using its dimensions and
density. The volume of the copper rod is given by V=π
4×(diameter)2×(length).
Substitute the values d= 4 cm = 0.04 m and l= 2 m:
V=π
4×(0.04)2×2
V=π
4×0.0016 ×2
V=π×0.0032
4
V= 0.00251327 m3
Given the density of copper, we can find the mass musing density = m
V.
m= density ×V= 8900 ×0.00251327 = 22.38315 kg
Step 2: Calculate the heat transferred using the formula Q=mc∆T. Given
that the specific heat capacity cof copper is 390 J/kg◦C, and the temperature
change ∆Tis 200◦C−100◦C = 100◦C. Substitute the values:
Q= 22.38315 ×390 ×100
Q= 873396 J
Therefore, the amount of heat transferred to the copper rod is 873396 J.
Question 23
Question
A copper rod of length 1.5 m has a cross-sectional area of 4 ×10−4m2. If a
temperature gradient of 20◦C is applied across the rod, determine the rate at
which heat flows through the rod assuming the thermal conductivity of copper
is 400 W/(m·K). Given that the thermal conductivity of copper is tempera-
ture dependent and can be approximated as k=k0(1 + αT ), where k0= 400
W/(m·K), α= 4 ×10−3K−1, and Tis the temperature in Kelvin.
21
Solution
Step 1: Calculate the temperature in Kelvin associated with the temperature
gradient. Given ∆T= 20◦C and T0= 20◦C, the temperature in Kelvin is
T= ∆T+T0= 20 + 20 = 40 K.
Step 2: Find the thermal conductivity of copper at the given temperature.
Given k=k0(1 + αT ), substitute k0= 400 W/(m·K), α= 4 ×10−3K−1,
and T= 40 K into the expression to find k:k= 400(1 + 4 ×10−3×40) =
400(1 + 0.16) = 400 ×1.16 = 464 W/(m·K).
Step 3: Calculate the rate at which heat flows through the rod. The rate
at which heat flows through the rod is given by Fourier’s Law: Q=−kAdT
dx ,
where Qis the heat flow rate, kis the thermal conductivity, Ais the cross-
sectional area of the rod, dT
dx is the temperature gradient, and the negative
sign indicates heat flow in the direction of decreasing temperature. Substitute
k= 464 W/(m·K), A= 4 ×10−4m2, and dT
dx = 20/1.5 K/m into the equation
to find Q:Q=−464 ×4×10−4×20
1.5=−464 ×4×2
15 =−464 ×8
15 =−247 W.
Hence, the rate at which heat flows through the rod is 247 W in the direction
of decreasing temperature.
Question 24
Question
A 1 kg block of copper at 100
°
C is placed in a 2 kg block of steel at 20
°
C.
Assuming no heat is lost to the surroundings, calculate the final temperature of
the two-block system. The specific heat capacity of copper is 385 J/(kg
°
C) and
the specific heat capacity of steel is 450 J/(kg
°
C).
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature. The heat lost is given by:
Qlost =mc∆T
where: m= mass of copper block = 1 kg, c= specific heat capacity of copper
= 385 J/(kg
°
C), ∆T= change in temperature for the copper block = (final
temperature - initial temperature).
Step 2: Calculate the heat gained by the steel block as it heats up to the
final temperature. The heat gained is given by:
Qgain =mc∆T
where: m= mass of steel block = 2 kg, c= specific heat capacity of steel = 450
J/(kg
°
C), ∆T= change in temperature for the steel block = (final temperature
- initial temperature).
22
Step 3: The heat lost by the copper block is equal to the heat gained by the
steel block (assuming no heat loss to the surroundings).
Qlost =Qgain
mc∆Tcopper =mc∆Tsteel
Step 4: Substitute the given values into the equation:
1×385 ×(final temperature −100) = 2 ×450 ×(final temperature −20)
Step 5: Solve for the final temperature:
385(final temperature −100) = 900(final temperature −20)
385final temperature −38500 = 900final temperature −18000
515final temperature = 20500
final temperature = 20500
515 ≈39.81C
So, the final temperature of the two-block system is approximately 39.81
°
C.
Question 25
Question
An aluminum rod of length 2.0 m and diameter 1.0 cm is initially at a tem-
perature of 20
°
C. It is immersed in a large water bath at 100
°
C. If the rod
is in perfect thermal contact with the water bath, calculate the time it takes
for the rod to reach a final temperature of 60
°
C. Assume the thermal conduc-
tivity of aluminum is 205 W/(m·K), the specific heat capacity of aluminum is
900 J/(kg·K), and the density of aluminum is 2700 kg/m3. (Neglect any heat
losses to the surroundings.)
Solution
Step 1: Calculate the volume and mass of the aluminum rod. The volume of the
aluminum rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where ris the radius of the rod and his the length of the rod. Given that
the diameter is 1.0 cm, the radius ris 0.5 cm or 0.005 m.
V=π×(0.005 m)2×2.0 m
V= 1.57 ×10−4m3
23
The mass of the aluminum rod can be calculated using the formula:
m= density ×volume
m= 2700 kg/m3×1.57 ×10−4m3
m= 0.425 kg
Step 2: Calculate the energy required to increase the temperature of the
aluminum rod from 20
°
C to 60
°
C. The energy required can be calculated using
the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity of aluminum,
and ∆Tis the change in temperature.
Given that ∆T= 60C−20C= 40C, we can substitute the values:
Q= 0.425 kg ×900 J/(kg·K) ×40C
Q= 15300 J
Step 3: Calculate the rate of heat transfer. The rate of heat transfer, dQ
dt , is
given by Fourier’s law of heat conduction:
dQ
dt =−kAdT
dx
where kis the thermal conductivity of aluminum, Ais the cross-sectional
area of the rod, and dT
dx is the temperature gradient.
The cross-sectional area of the rod can be calculated using the formula for
the area of a circle:
A=πr2
A=π×(0.005 m)2
A= 7.85 ×10−5m2
The temperature gradient, dT
dx , can be calculated as:
dT
dx =Tfinal −Tinitial
L
dT
dx =60C−20C
2.0 m
24
dT
dx = 20C/m
Now we can calculate the rate of heat transfer:
dQ
dt =−205 W/(m·K) ×7.85 ×10−5m2×20C/m
dQ
dt =−0.32 W
Step 4: Calculate the time taken for the rod to reach 60
°
C. The time taken
can be calculated using the formula:
Q=Zt
0
dQ
dt dt
15300 J = Zt
0
−0.32 Wdt
15300
Question 26
Question
A metal rod of length Land thermal conductivity kis initially at a uniform
temperature Tithroughout its length. The rod is then heated at one end,
causing a constant temperature gradient dT
dx to be established along the length
of the rod. If the temperature at the heated end is Thand at the other end
is Tc, derive an expression for the rate at which heat is conducted through the
rod.
Solution
Step 1: Determine the temperature distribution along the rod.
Let xbe the distance from the heated end. The temperature Tat a distance
xfrom the heated end satisfies the heat conduction equation:
d
dx kdT
dx = 0
Integrating twice:
kdT
dx =C1
dT
dx =C1
k
25
T=C1
kx+C2
Applying the boundary conditions T(0) = Thand T(L) = Tc:
Th=C2
Tc=C1
kL+C2=C1
kL+Th
C1=k(Tc−Th)/L
C2=Th
Thus, the temperature distribution is given by:
T(x) = Th+ (Tc−Th)x
L
Step 2: Calculate the rate of heat conduction.
The rate of heat conduction Qthrough the rod is given by Fourier’s Law:
Q=−kAdT
dx
where Ais the cross-sectional area of the rod.
The heat conducted through an element dx at a distance xfrom the heated
end is:
dQ =−kAdT
dx dx =−kA d
dx Th+ (Tc−Th)x
Ldx
=−kA(Tc−Th)dx
L
To find the total rate of heat conduction through the entire rod, integrate
dQ from 0 to L:
Q=ZL
0
−kA(Tc−Th)dx
L
Q=−kA(Tc−Th)L
L
Q=−kA(Tc−Th)
Question 27
Question
A copper rod of length 50 cm and diameter 2 cm is initially at a temperature
of 100
°
C. It is placed in a water bath at a temperature of 20
°
C. If the rod loses
heat to the water bath until it reaches thermal equilibrium, calculate the final
temperature of the rod. The specific heat capacity of copper is 0.385 J/g
°
C and
its density is 8.96 g/cm3. Assume the only heat lost by the rod is used to warm
the water bath.
26
Solution
Step 1: Calculate the mass of the copper rod. The volume of the copper rod is
given by V=πr2h, where ris the radius and his the height. Given that the
diameter is 2 cm, the radius is r= 1 cm. Thus, V=π(1 cm)2·50 cm = 50πcm3.
Since the density of copper is 8.96 g/cm3, the mass mof the copper rod is
m= density ×volume = 8.96 g/cm3×50πcm3= 448πg.
Step 2: Calculate the heat lost by the copper rod. The heat lost by the
copper rod can be calculated 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 copper rod is 100
°
C and the final temperature is
unknown, denoted as T. Therefore, the change in temperature ∆T= 100 −T.
Substitute m= 448πg, c= 0.385 J/g
°
C, and ∆T= 100 −Tinto the formula.
Step 3: Calculate the heat absorbed by the water bath. Since the only heat
lost by the copper rod is used to warm the water bath, the heat absorbed by
the water bath is equal to the heat lost by the copper rod. Therefore, the heat
absorbed by the water bath is mc∆T=Qas well. Substitute mand cinto the
formula and equate it to the heat lost by the copper rod.
Step 4: Solve for the final temperature. After equating the two heat quan-
tities, solve for the final temperature Tby setting up the equation and solving
for T.
448π×0.385 ×(100 −T) = 448π×(T−20)
Question 28
Question
A copper rod of length 2 m and cross-sectional area 5 cm2is initially at a
temperature of 100◦C. It is then placed in an ice bath at 0◦C. If the rod loses
heat to the ice at a constant rate of 200 J/s, how long will it take for the rod to
cool down to 10◦C? The specific heat capacity of copper is 390 J/kgK and its
density is 8900 kg/m3.
Solution
Step 1: Calculate the mass of the copper rod. Given: Length, L= 2 m Cross-
sectional area, A= 5 cm2= 5 ×10−4m2Density of copper, ρ= 8900 kg/m3
The volume of the rod is V=A·L= 5 ×10−4×2 = 0.001 m3. Therefore,
the mass of the rod is m=ρ·V= 8900 ×0.001 = 8.9 kg.
Step 2: Calculate the initial heat content of the rod. The initial temperature
of the rod, Ti= 100 ◦C = 100 + 273 = 373 K. The final temperature at which
we want to cool the rod, Tf= 10 ◦C = 10 + 273 = 283 K. The change in
temperature, ∆T=Ti−Tf= 373 −283 = 90 K.
The initial heat content of the rod is given by Qi=m·c·∆T, where c
is the specific heat capacity of copper. Substituting the values, we get Qi=
8.9·390 ·90 = 313650 J.
27
Step 3: Calculate the energy lost per degree drop in temperature. Given
that the rod loses heat at a constant rate of 200 J/s, the energy lost per second
per degree drop in temperature is 200 J/s.
Step 4: Calculate the time taken to cool the rod from an initial temperature
of 100◦C to 10◦C. The total energy required to cool the rod from 100◦C to
10◦C is Qi−(Qi−(90 ·200)) = Qi−90 ·200. Substitute the values to get
Qi−90 ·200 = 313650 −18000 = 295650 J.
The time taken to cool the rod to 10◦C is given by t=Qtotal
Rate of heat loss .
Substitute the values to get t=295650
200 = 1478.25 seconds.
Therefore, it will take approximately 1478 seconds for the rod to cool down
from 100◦C to 10◦C.
Question 29
Question
A metal rod of length 1 m is initially at a temperature of 500◦C. If one end
of the rod is submerged in an ice bath at 0◦C while the other end is heated to
1000◦C, what is the temperature at a point 30 cm from the colder end after the
rod reaches thermal equilibrium?
Solution
Given: Initial temperature of the rod, T1= 500◦C
Temperature at one end, T2= 1000◦C
Temperature at the other end, T3= 0◦C
Length of the rod, L= 1 m
Distance from the colder end, x= 30 cm = 0.3 m
We need to find the temperature at a point 30 cm from the colder end after
thermal equilibrium is reached.
Step 1: Firstly, we find the rate of heat flow per unit length of the rod when
the rod reaches thermal equilibrium. This is given by Fourier’s law:
dQ
dt =−kAdT
dx
where: dQ/dt = rate of heat flow per unit time, k= thermal conductivity
of the material, A= cross-sectional area of the rod, and dT/dx = temperature
gradient along the rod.
Since the rod has reached thermal equilibrium, the rate of heat flow at any
point along the rod is constant. Let this constant rate be denoted as q.
Step 2: Next, we express qin terms of the temperatures at the two ends of
the rod. We have:
q=kAT2−T3
L
Step 3: Substituting the values we have into the above formula, we find q:
28
q=kA1000 −0
1
q=kA(1000)
Step 4: Finally, we can find the temperature at a point 30 cm from the
colder end using the following relationship:
T(x) = T3+qL
kAx
Substitute the known values:
T(0.3) = 0 + (1000)(1)
kA (0.3)
T(0.3) = 300(0.3)
T(0.3) = 90◦C
Therefore, the temperature at a point 30 cm from the colder end after the
rod reaches thermal equilibrium is 90◦C.
Question 30
Question
A 200 g block of copper at 100
°
C is dropped into 400 g of water at 20
°
C in
a 200 g aluminum calorimeter cup. The final temperature of the mixture is
31
°
C. Assuming no heat is lost to the surroundings, calculate the specific heat
capacity of aluminum. The specific heat capacity of water is 4186 J/kg
°
C and
the specific heat capacity of copper is 387 J/kg
°
C.
Solution
Step 1: Calculate the total heat absorbed by copper block.
qcopper =mcopperccopper∆Tcopper
qcopper = (0.2 kg)(387 J/kg
°
C)(31 −100)
°
C
qcopper =−12900 J
Step 2: Calculate the total heat absorbed by water.
qwater =mwatercwater∆Twater
qwater = (0.4 kg)(4186 J/kg
°
C)(31 −20)
°
C
qwater = 14692 J
29
Step 3: Calculate the change in heat for the aluminum calorimeter cup.
qaluminum =−(qcopper +qwater)
qaluminum =−(−12900 + 14692) J
qaluminum = 1792 J
Step 4: Calculate the specific heat capacity of aluminum.
qaluminum =maluminumcaluminum∆Taluminum
1792 J = (0.2 kg)caluminum(31 −20)
°
C
caluminum =1792
0.2×11
caluminum = 814.54 J/kg
°
C
Therefore, the specific heat capacity of aluminum is 814.54 J/kg
°
C.
Question 31
Question
A copper rod of length 1 m and cross-sectional area 0.01 m2is heated to a
temperature of 100◦C. The rod loses heat energy to the surroundings at a con-
stant rate of 500 W. If the thermal conductivity of copper is 400 W/(m·K) and
its specific heat capacity is 400 J/(kg·K), calculate the time it takes for the
temperature of the rod to drop to 50◦C. Assume the density of copper is 8900
kg/m3.
Solution
Step 1: Calculate the mass of the copper rod using its volume and density.
The volume of the rod is given by V=A·L= 0.01 m2·1 m = 0.01 m3.
The mass can be calculated as m=ρ·V= 8900 kg/m3·0.01 m3= 89 kg.
Step 2: Calculate the initial heat content of the rod.
The specific heat capacity of copper is 400 J/(kg·K), so the heat content of the
rod at 100◦C is Qinitial =m·c·∆T= 89 kg ·400 J/(kg·K) ·100 K = 3.56 ×106J.
Step 3: Determine the rate of temperature decrease in the rod.
The rate at which the rod loses heat is given as 500 W. By the formula P=
kA∆T
L, we can solve for the rate of temperature decrease ∆T
∆t.
Substitute the known values: 500 = 400 W/(m·K) ·0.01 m2·∆T
1 m .
Solving for ∆T, we get ∆T= 5 K/s.
Step 4: Determine the time it takes for the rod to cool to 50◦C.
The final temperature is 50◦C, so ∆T= 100 −50 = 50 K. The time it takes for
the rod to cool to this temperature can be found using ∆t=∆T
∆T
∆t
=50 K
5 K/s = 10 s.
Therefore, it takes 10 seconds for the temperature of the rod to drop to
50◦C.
30
Question 32
Question
A sample of copper with mass 500 g is heated from 20
°
C to 90
°
C. Calculate the
amount of heat required to bring about this temperature change. The specific
heat capacity of copper is 0.385 J/g
°
C.
Solution
Step 1: Determine the change in temperature. Step 2: Calculate the amount of
heat transferred.
Step 1: The change in temperature can be calculated using the formula:
∆T=Tf−Ti= 90C−20C= 70C
Step 2: The amount of heat transferred can be calculated using the formula:
Q=mc∆T
where: - Qis the amount of heat transferred, - mis the mass of the sample (500
g), - cis the specific heat capacity of copper (0.385 J/g
°
C), - ∆Tis the change
in temperature (70
°
C).
Substitute the values into the formula:
Q= (500 g)(0.385 J/g
°
C)(70
°
C)
Q= 500 ×0.385 ×70
Q= 19250 J
Therefore, the amount of heat required to bring about this temperature
change is 19250 J.
Question 33
Question
An aluminum rod of length 2.0 m and diameter 0.02 m is heated from an initial
temperature of 20
°
C to a final temperature of 80
°
C. The rod has a thermal
conductivity of 237 W/(m*K), a specific heat capacity of 900 J/(kg*K), and a
density of 2700 kg/m
³
. Calculate the heat energy transferred to the rod during
this heating process.
31
Solution
Step 1: Calculate the mass of the aluminum rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder: V=πr2h, where ris
the radius and his the height (length) of the cylinder. Given that the diameter
is 0.02 m, the radius r= 0.01 m. The volume Vof the rod is:
V=π(0.01)2×2.0=0.000628 m3
The mass mof the rod can be calculated using the formula m= density×volume.
Substitute the density of aluminum, ρ= 2700 kg/m
³
:
m= 2700 ×0.000628 = 1.6764 kg
Step 2: Calculate the heat energy transferred. The heat energy transferred
to the rod can be calculated using the formula Q=mc∆T, where mis the mass,
cis the specific heat capacity, and ∆Tis the change in temperature. Given that
the initial temperature Ti= 20
°
C and the final temperature Tf= 80
°
C, the
change in temperature is ∆T=Tf−Ti= 80 −20 = 60
°
C (note that the change
in temperature must be in Kelvin, so ∆T= 60 K). Now, substitute the values
into the formula:
Q= 1.6764 ×900 ×60 = 90679.2 J
Hence, the heat energy transferred to the aluminum rod during this heating
process is 90679.2 J.
Question 34
Question
A 200 g piece of copper is heated until its temperature reaches 65
°
C. The piece
is then placed in 500 g of water at 20
°
C. What will be the final temperature
when thermal equilibrium is reached? (Specific heat capacity of copper is 0.385
J/g
°
C and of water is 4.18 J/g
°
C.)
Solution
Step 1: Calculate the heat gained by the copper piece: The formula for heat
transfer is: Q=mc∆T, where Qis heat energy, mis mass, cis specific heat
capacity, and ∆Tis the change in temperature. Given: m= 200 g, cCu = 0.385
J/g
°
C, ∆TCu = 65 −Tf, where Tfis the final temperature. So, QCu = 200 ×
0.385 ×(65 −Tf).
Step 2: Calculate the heat lost by the copper to the water: The heat lost
by the copper will be equal to the heat gained by the water, as no heat is lost
to the surroundings. Given: QCu =QH2O, where QH2O =mc∆T. Using the
mass of water m= 500 g, cH2O = 4.18 J/g
°
C, and ∆TH2O =Tf−20, we have
QH2O = 500 ×4.18 ×(Tf−20).
32
Step 3: Set up the equation and solve for the final temperature: Since
QCu =QH2O, we can set the two expressions for heat equal to each other and
solve for Tf: 200 ×0.385 ×(65 −Tf) = 500 ×4.18 ×(Tf−20). Solving this
equation will give us the final temperature Tf.
Question 35
Question
A copper rod of length 1.5 m and diameter 2 cm is heated from 20
°
C to 100
°
C. If
the thermal conductivity of copper is 400 W/mK and its density is 8960 kg/m3,
calculate the total heat energy transferred to the rod during this process.
Solution
Step 1: Find the cross-sectional area of the copper 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. Given that the diameter of the rod is 2 cm, the radius
ris half of the diameter, so r= 1 cm = 0.01 m. Therefore, the cross-sectional
area A=π(0.01)2= 3.14 ×10−4m2.
Step 2: Find the volume of the copper rod. The volume of the rod can be
calculated as V=A×L, where Lis the length of the rod. Given that the length
of the rod is 1.5 m, the volume V= 3.14 ×10−4m2×1.5 m = 4.71 ×10−4m3.
Step 3: Find the mass of the copper rod. The mass of the rod can be
calculated using the formula m= density ×volume. Given that the density of
copper is 8960 kg/m3, the mass m= 8960 kg/m3×4.71 ×10−4m3= 4.22 kg.
Step 4: Calculate the heat energy transferred. The heat energy transferred
to the rod can be calculated using the formula Q=mc∆T, where mis the mass,
cis the specific heat capacity (which for copper is approximately 390 J/kgK),
and ∆Tis the change in temperature. Given that the initial temperature T1=
20Cand the final temperature T2= 100C, the change in temperature ∆T=
T2−T1= 100C−20C= 80C. Plugging in the values, we get Q= 4.22 kg ×
390 J/kgK ×80 K = 132768 J. Therefore, the total heat energy transferred to
the rod during this process is 132768 J.
33
Step 2: Calculate the change in temperature of the rod using the specific
heat capacity formula. The heat absorbed by the rod can be related to its mass,
specific heat capacity, and change in temperature using the formula:
Q=mc∆T
Given that the rod absorbs 1500 J of heat, the specific heat capacity of steel is
490 J/kg
°
C, and the mass of the rod is 0.368 kg, we have:
1500 = 0.368 ×490 ×∆T
∆T=1500
0.368 ×490
∆T≈7.06
°
C
Step 3: Calculate the final temperature of the rod. The final temperature
Tfof the rod can be calculated as:
Tf=Ti+ ∆T
Given that the initial temperature Tiis 200
°
C, we have:
Tf= 200 + 7.06
Tf≈207.06
°
C
Therefore, the final temperature of the rod is approximately 207.06
°
C.
Question 2
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
200◦C. It is then heated until the temperature reaches 400◦C. If the rod absorbs
500 J of heat while maintaining uniform temperature, find the change in length
of the rod (assuming coefficient of linear expansion of copper is 1.7×10−5K−1).
Solution
Step 1: Calculate the initial volume of the rod. The initial volume, Vi, of the
rod can be calculated as the volume of a cylinder:
Vi=πd
22
·L
where dis the diameter and Lis the length. Since the diameter is given in
centimeters, we need to convert it to meters:
Vi=π1 cm
2×1002
·2 m
2
Vi=π0.01 m
22
·2 m
Vi=π×0.01
2×0.01
2×2
Vi=π×0.012
4×2
Vi=π×0.012
2
Vi=π×0.0001
2
Vi= 0.000157 m3
Step 2: Calculate the final volume of the rod. As the rod heats up uniformly,
the change in temperature leads to the expansion of the rod. The change in
temperature is:
∆T= 400 −200 = 200◦C
The change in length, ∆L, is given by the expression:
∆L=Lα∆T
where: - Lis the original length, - αis the coefficient of linear expansion, -
∆Tis the change in temperature.
Given αand ∆T, we can find ∆L.
∆L= 2 m ×(1.7×10−5K−1)×200 K
∆L= 2 ×1.7×10−5×200 m
∆L= 6.8×10−3m
The final length, Lf, is:
Lf=L+ ∆L= 2 m + 6.8×10−3m
Lf= 2.0068 m
The final volume, Vf, of the rod can be calculated in the same way as the
initial volume:
Vf=πd
22
·Lf
Vf=π0.01 m
22
·2.0068 m
Vf=π0.01
22
×2.0068
3
Vf=π×0.012
4×2.0068
Vf=π×0.0001
4×2.0068
Vf=π×0.0001 ×2.0068
4
Vf=π×0.00020068
4
Vf=π×2.0068 ×10−4
4
Vf=2.0068π×10−4
4
Vf= 0.0001577 m3
Step 3: Calculate the change in volume. The change in volume, ∆V, is given
by:
∆V
Question 3
Question
A piece of aluminum with a mass of 250 g at a temperature of 100
°
C is placed
in 500 g of water at a temperature of 20
°
C. Assuming no heat is lost to the
surroundings, calculate the final temperature of the system when thermal equi-
librium is reached. (Specific heat capacity of aluminum = 0.902 J/g
°
C, specific
heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the aluminum. Given: - Mass
of aluminum, ma= 250 g - Initial temperature of aluminum, Ta,i = 100
°
C -
Specific heat capacity of aluminum, ca= 0.902 J/g
°
C - Final temperature of the
system, Tf
Using the formula for heat transfer:
Q=mc∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Since the aluminum is cooling down:
Qa=−mca∆Ta
Qa=−250 ×0.902 ×(Tf−100)
4
Step 2: Calculate the heat gained or lost by the water. Given: - Mass of
water, mw= 500 g - Initial temperature of water, Tw,i = 20
°
C - Specific heat
capacity of water, cw= 4.18 J/g
°
C
Since the water is getting warmer:
Qw=mcw∆Tw
Qw= 500 ×4.18 ×(Tf−20)
Step 3: Set up the equation for heat conservation. Since no heat is lost to
the surroundings, the heat lost by aluminum must be equal to the heat gained
by water:
−250 ×0.902 ×(Tf−100) = 500 ×4.18 ×(Tf−20)
Step 4: Solve for the final temperature of the system, Tf.
−225.5Tf+ 22550 = 2090Tf−41800
2315.5Tf= 64350
Tf=64350
2315.5
Tf≈27.8
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 27.8
°
C.
Question 4
Question
A 2 kg block of aluminum at an initial temperature of 100
°
C is dropped into a
large bath containing 20 kg of water at 20
°
C. Assuming no heat is lost to the sur-
roundings, calculate the final equilibrium temperature of the system. The spe-
cific heat capacity of aluminum is 900 J/kg◦C and that of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the aluminum block as it cools down from
100
°
C to the final equilibrium temperature. The change in temperature (∆T)
is the initial temperature (100
°
C) minus the final equilibrium temperature (T).
The heat lost (QAl) is given by QAl =mc∆T, where: - mis the mass of
aluminum (2 kg), - cis the specific heat capacity of aluminum (900 J/kg◦C).
Thus, QAl = 2 kg ×900 J/kg◦C×(100 −T).
Step 2: Calculate the heat gained by the water as it heats up from 20
°
C
to the final equilibrium temperature. The change in temperature (∆T) is the
final equilibrium temperature (T) minus the initial temperature (20
°
C). The
5
heat gained (Qw) is given by Qw=mc∆T, where: - mis the mass of water
(20 kg), - cis the specific heat capacity of water (4200 J/kg◦C). Thus, Qw=
20 kg ×4200 J/kg◦C×(T−20).
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block must equal the heat gained by the water. Thus, 2 kg×900 J/kg◦C×
(100 −T) = 20 kg ×4200 J/kg◦C×(T−20).
Step 4: Solve the equation from Step 3 to find the final equilibrium temper-
ature (T) of the system.
2×900 ×(100 −T) = 20 ×4200 ×(T−20)
Solving for T, we get:
1800(100 −T) = 84000(T−20)
180000 −1800T= 84000T−1680000
1800T+ 84000T= 180000 + 1680000
85800T= 1860000
T=1860000
85800 ≈21.67◦C
Therefore, the final equilibrium temperature of the system is approximately
21.67◦C.
Question 5
Question
A copper block weighing 500 grams is initially at a temperature of 100◦C. It is
then heated until its temperature rises to 200◦C. If the specific heat capacity of
copper is 0.385 J/g◦C, how much heat energy was added to the block?
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, T1= 100◦C Final temperature, T2= 200◦C
The change in temperature, ∆T=T2−T1= 200◦C - 100◦C = 100◦C
Step 2: Calculate the heat energy added to the copper block. The heat
energy added can be calculated using the formula:
q=mc∆T
where: q= heat energy added (in joules) m= mass of the copper block (in
grams) c= specific heat capacity of copper (in J/g◦C) ∆T= change in tem-
perature (in ◦C)
Given: m= 500 grams c= 0.385 J/g◦C ∆T= 100◦C
6
Substitute the values into the formula:
q= 500 ×0.385 ×100
q= 19250 joules
Therefore, the heat energy added to the copper block is 19250 joules.
Question 6
Question
A copper rod of length 2 meters and diameter 1 cm is initially at a temperature of
100◦C. It is placed in a water bath at 0◦C. Given that the thermal conductivity of
copper is 400 W/(m·K) and the specific heat capacity of copper is 390 J/(kg·K),
calculate the time it takes for the rod to reach thermal equilibrium with the
water bath.
Solution
Step 1: Calculate the surface area of the rod. The surface area of a cylinder is
given by the formula:
A= 2πrh
where ris the radius and his the height of the cylinder. Given the diameter of
the rod is 1 cm, the radius r=1
2cm. Converting the radius to meters: r=1
200
m. The height of the cylinder is 2 meters. Therefore, the surface area of the
rod is:
A= 2π1
200·2
Step 2: Calculate the volume of the rod. The volume of a cylinder is given
by the formula:
V=πr2h
The volume of the rod is:
V=π1
2002
·2
Step 3: Calculate the mass of the rod. The density of copper is approxi-
mately 8900 kg/m3. The mass of the rod is given by:
m= density ×volume
Step 4: Calculate the heat energy in the rod. The heat energy in the rod is
given by:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
7
Step 5: Calculate the rate of heat transfer from the rod to the water bath.
The rate of heat transfer is given by Fourier’s Law:
dQ
dt =kAdT
dx
where kis the thermal conductivity, Ais the surface area, dT
dx is the temperature
gradient, and dQ
dt is the rate of heat transfer.
Step 6: Equate the rates of heat transfer. At thermal equilibrium, the rate
of heat transfer from the rod to the water bath is equal to the rate of heat
transfer from the water bath to the rod.
kAdT
dx rod =kAdT
dx water bath
Step 7: Calculate the time it takes for the rod to reach thermal equilibrium.
Use the relation ∆T=Tfinal −Tinitial and the fact that the rod and water bath
will reach the same temperature at equilibrium.
Rate of heat transfer from rod to water bath = Rate of heat transfer from water bath to rod
Solve this equation to find the time it takes for the rod to reach thermal
equilibrium with the water bath.
Question 7
Question
A copper sphere of radius 5 cm is heated until its temperature reaches 100
°
C.
If the thermal conductivity of copper is 400 W/mK and the temperature of the
surrounding air is 25C, calculate the rate of heat loss from the sphere due to
conduction.
Solution
Step 1: First, calculate the surface area of the sphere using the formula A=
4πr2. Given that the radius r= 5 cm, we convert it to meters: r= 0.05 m.
A= 4π(0.05)2
= 4 ×π×0.0025
= 0.0314 m2
Step 2: Next, calculate the temperature difference between the sphere and
the surroundings.
∆T= 100C−25C= 75C= 75 K
Step 3: Now, use the formula for the rate of heat loss through conduction:
Rate of heat loss = k×A×∆T
d
8
where kis the thermal conductivity of copper, Ais the surface area, ∆Tis the
temperature difference, and dis the thickness of the material (which we assume
as negligible here since it is a sphere). Substitute the given values:
Rate of heat loss = 400 W/mK ×0.0314 m2×75 K
0
= 400 ×0.0314 ×75
= 942 W
Therefore, the rate of heat loss from the sphere due to conduction is 942 W .
Question 8
Question
A 2 kg block of metal at 80
°
C is dropped into 5 kg of water at 20
°
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 final temperature of the mixture assuming
no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the metal block as it cools down to the final
temperature. The heat lost by the metal block is given by the formula:
Qmetal =mmetal ×cmetal ×∆T
where: - mmetal = 2 kg (mass of the metal block), - cmetal = 450 J/kg
°
C (specific
heat capacity of the metal), - Tinitial = 80C(initial temperature of the metal),
and - Tfinal =Tfinal (final temperature of the mixture).
The change in temperature, ∆T, for the metal is given by:
∆Tmetal =Tfinal −Tinitial
Step 2: Calculate the heat gained by the water as it rises to the final tem-
perature. The heat gained by the water is given by the formula:
Qwater =mwater ×cwater ×∆T
where: - mwater = 5 kg (mass of the water), - cwater = 4186 J/kg
°
C (specific heat
capacity of water), and - Tinitial = 20C(initial temperature of the water).
Step 3: Set up the equation for heat transfer. Since in this case, no heat is
lost to the surroundings, the heat lost by the metal must be equal to the heat
gained by the water. Therefore:
Qmetal =Qwater
9
Step 4: Solve for the final temperature. Substitute the expressions for heat
lost by the metal and heat gained by the water into the equation from Step 3:
mmetal ×cmetal ×(Tfinal −Tinitial) = mwater ×cwater ×(Tfinal −Tinitial)
Solve for Tfinal to find the final temperature of the mixture.
Question 9
Question
A 0.5 kg block of copper is heated from 20
°
C to 80
°
C. If the specific heat capacity
of copper is 390 J/kg ·
°
C, how much heat energy is required?
Solution
Step 1: Calculate the change in temperature Given: Initial temperature, T1=
20CFinal temperature, T2= 80CChange in temperature, ∆T=T2−T1=
80C−20C= 60C
Step 2: Calculate the heat energy required using the formula:
Q=mc∆T
where: Q= heat energy m= mass of the block = 0.5 kg c= specific heat
capacity of copper = 390 J/kg ·
°
C ∆T= change in temperature = 60
°
C
Step 3: Substitute the given values into the formula and solve for Q:
Q= (0.5 kg)(390 J/kg ·
°
C)(60C)
Q= 0.5×390 ×60
Q= 11700 J
Answer: The heat energy required is 11700 J.
Question 10
Question
A solid copper sphere at an initial temperature of 100◦C is placed in a large
bath of water at a constant temperature of 0◦C. The sphere loses heat to the
water until it reaches thermal equilibrium. If the mass of the sphere is 2 kg and
its specific heat capacity is 0.385 J/g◦C, calculate the final temperature of the
copper sphere.
10
Solution
Step 1: Calculate the initial heat energy of the copper sphere using the formula:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the sphere, - cis the specific
heat capacity of copper (0.385 J/g◦C), - ∆Tis the change in temperature.
For the initial temperature of 100◦C and the final temperature Tf:
Q= 2 ×0.385 ×(100 −Tf)
Step 2: Calculate the heat energy transferred from the sphere to the water
using the same formula:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the sphere, - cis the specific
heat capacity of water (4.18 J/g◦C), - ∆Tis the change in temperature (from
100◦C to 0◦C).
The heat energy transferred is equal to the initial heat energy of the copper
sphere:
2×0.385 ×(100 −Tf)=2×4.18 ×100
Step 3: Solve the equation obtained in Step 2 to find the final temperature
Tf:
0.77 ×(100 −Tf) = 836
77 −0.77Tf= 836
0.77Tf=−759
Tf=−759
0.77 ≈ −985.71◦C
Since the final temperature must be between 0◦C and 100◦C, this negative
value indicates an error in the calculations. Double-check the calculations to
find the mistake.
Question 11
Question
A steel rod of length 2 m and diameter 1 cm is heated from 20
°
C to 200
°
C. If
the density of steel is 7850 kg/m3and its specific heat capacity is 460 J/kg
°
C,
calculate the heat energy required. Assume the rod is one-dimensional and
neglect heat losses to the surroundings.
11
Solution
Step 1: Calculate the mass of the steel rod. Given that the steel rod has a
length of 2 m and a diameter of 1 cm, we can first find the volume of the rod
and then calculate its mass.
The cross-sectional area of the rod is given by:
A=πd2
4=π×(0.01)2
4= 7.85398 ×10−5m2
The volume of the rod is:
V=A×length = 7.85398 ×10−5×2=1.5708 ×10−4m3
The mass of the rod can be found using the density formula:
mass = density ×volume = 7850 ×1.5708 ×10−4= 1.22887 kg
Therefore, the mass of the steel rod is 1.22887 kg.
Step 2: Calculate the heat energy required. The heat energy (Q) required
to change the temperature of an object can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the object, - cis the specific heat capacity of the
material, and - ∆Tis the change in temperature.
Given: - m= 1.22887 kg, - c= 460 J/kg
°
C, and - ∆T= 200 −20 = 180
°
C
Substitute these values into the formula to find the heat energy required:
Q= 1.22887 ×460 ×180 = 101238.642 J
Therefore, the heat energy required to heat the steel rod from 20
°
C to 200
°
C
is 101238.642 J.
Question 12
Question
A copper block with a mass of 500 g and at a temperature of 200
°
C is placed in
a container with 500 g of water at 20
°
C. If the final temperature of the system
is 40
°
C, calculate the specific heat capacity of copper. (Specific heat capacity
of water is 4186 J/kg
°
C.)
Solution
Step 1: Calculate the heat gained by the copper block. The heat absorbed by
the copper block is given by the formula:
Q=mc∆T
12
where: - Qis the heat absorbed, - mis the mass of the copper block, - cis the
specific heat capacity of copper, and - ∆Tis the change in temperature of the
copper block.
Given the mass of the copper block is 500 g (0.5 kg), the initial temperature
is 200
°
C, and the final temperature is 40
°
C, we can calculate ∆Tas:
∆T= 40C−200C=−160C=−160K
Note: The temperature needs to be converted to Kelvin for calculations involv-
ing temperature difference.
Substitute the values into the formula to find the heat absorbed by the
copper block:
Qcopper = (0.5kg)×c×(−160K)
Step 2: Calculate the heat lost by the water. The heat lost by the water is
given by the formula:
Q=mc∆T
where: - Qis the heat lost, - mis the mass of the water, - cis the specific heat
capacity of water, and - ∆Tis the change in temperature of the water.
Given the mass of the water is 500 g (0.5 kg) and the initial temperature is
20
°
C, we can calculate ∆Tas:
∆T= 40C−20C= 20K
Substitute the values into the formula to find the heat lost by the water:
Qwater = (0.5kg)×4186 J/kgC ×20K
Step 3: Since heat is conserved in the system, the heat lost by the water is
equal to the heat gained by the copper block. Therefore, equate Qcopper and
Qwater and solve for the specific heat capacity of copper. We have:
(0.5kg)×c×(−160K) = (0.5kg)×4186 J/kgC ×20K
c×(−160) = 4186 ×20
c=4186 ×20
−160
Thus, the specific heat capacity of copper is determined by solving the above
equation.
Question 13
Question
A certain metal has a specific heat capacity of 0.45 J/g
°
C. How much heat is
required to raise the temperature of a 250 g sample of this metal from 20
°
C to
80
°
C?
13
Solution
Step 1: Identify the given values and the formula for heat transfer: Given:
Specific heat capacity of the metal, c= 0.45 J/g
°
C Mass of the sample, m= 250
g Initial temperature, Ti= 20
°
C Final temperature, Tf= 80
°
C
The formula for heat transfer is:
Q=mc∆T
where: Q= heat transfer (in joules) m= mass of the sample (in grams) c=
specific heat capacity (in J/g
°
C) ∆T= change in temperature (in
°
C)
Step 2: Calculate the change in temperature, ∆T:
∆T=Tf−Ti= 80 −20 = 60
°
C
Step 3: Substitute the given values into the formula for heat transfer and
solve for Q:
Q= 250 g ×0.45 J/g
°
C×60
°
C
Step 4: Calculate the heat transfer Q:
Q= 250 ×0.45 ×60 = 6750 J
Therefore, the amount of heat required to raise the temperature of the 250
g sample of metal from 20
°
C to 80
°
C is 6750 J.
Question 14
Question
A steel rod of length 1 m and cross-sectional area 0.01 m2is initially at a
temperature of 300 K. It is heated until its temperature reaches 400 K. Given
that the linear expansion coefficient of steel is 1.2×10−5K−1and the Young’s
modulus of steel is 2 ×1011 Pa, calculate the change in length of the rod and
the amount of heat added to the rod.
Solution
Step 1: Calculate the change in length of the rod using the formula for linear
expansion:
∆L=αL∆T
where: ∆L= change in length of the rod, α= linear expansion coefficient
of steel, L= original length of the rod, ∆T= change in temperature.
Given: α= 1.2×10−5K−1,L= 1 m, Tinitial = 300 K, Tfinal = 400 K.
First, calculate ∆T:
∆T=Tfinal −Tinitial = 400 K −300 K = 100 K
14
Now, substitute the given values into the formula:
∆L= (1.2×10−5K−1)(1 m)(100 K) = 0.0012 m = 1.2 mm
Therefore, the change in length of the rod is 1.2 mm.
Step 2: Calculate the amount of heat added to the rod using the formula:
Q=F
A
where: Q= amount of heat added, F= force required to stretch the rod, A
= cross-sectional area of the rod.
Given: E= 2 ×1011 Pa, ∆L= 0.0012 m, A= 0.01 m2.
First, calculate the force using Hooke’s Law:
F=E∆L
L
Substitute the given values into the formula:
F= (2 ×1011 Pa)0.0012 m
1 m = 2.4×108N
Now, substitute the force and area values into the formula for heat:
Q=2.4×108N
0.01 m2= 2.4×1010 J
Therefore, the amount of heat added to the rod is 2.4×1010 J.
Question 15
Question
A 2 kg block of copper at 100
°
C is dropped into 5 kg of water at 20
°
C in a per-
fectly insulated container. Assuming no heat is lost to the surroundings, what
will be the final equilibrium temperature of the system? (Specific heat capacity
of copper = 390 J/kg◦C and specific heat capacity of water = 4184 J/kg◦C)
Solution
Step 1: Calculate the heat lost by the copper block The heat lost by the copper
block is given by the formula:
Qlost =mc∆T
where: - mis the mass of the copper block (2 kg) - cis the specific heat capacity
of copper (390 J/kg◦C) - ∆Tis the change in temperature of the copper block
The change in temperature of the copper block is:
∆T=Tfinal −Tinitial =Tfinal −100
15
Substitute this into the formula for heat lost and solve for Qlost.
Step 2: Calculate the heat gained by the water The heat gained by the water
is given by the formula:
Qgain =mc∆T
where: - mis the mass of the water (5 kg) - cis the specific heat capacity of
water (4184 J/kg◦C) - ∆Tis the change in temperature of the water
The change in temperature of the water is:
∆T=Tfinal −Tinitial =Tfinal −20
Substitute this into the formula for heat gained and solve for Qgain.
Step 3: Since there is no heat lost to the surroundings, the heat lost by the
copper block is equal to the heat gained by the water. Set up the equation:
Qlost =Qgain
Step 4: Solve for the final equilibrium temperature, Tfinal, by combining the
results from Step 1 and Step 2:
Qlost =Qgain
mcopperccopper(Tfinal −100) = mwatercwater(Tfinal −20)
2×390 ×(Tfinal −100) = 5 ×4184 ×(Tfinal −20)
Solve for Tfinal to find the final equilibrium temperature of the system.
Question 16
Question
A copper kettle is filled with 1.5 kg of water at 20
°
C. How much heat is required
to heat the water to its boiling point (100
°
C)? The specific heat capacity of
water is 4.18 J/g
°
C, and assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the initial temperature difference. The initial temperature of
the water is 20
°
C and the final temperature (boiling point) is 100
°
C. Therefore,
the initial temperature difference is:
∆T= 100C−20C= 80C
Step 2: Calculate the heat energy required to raise the temperature of the
water from 20
°
C to 100
°
C. The heat energy required is given by the formula:
Q=mc∆T
16
where: - mis the mass of water (1.5 kg = 1500 g), - cis the specific heat
capacity of water (4.18 J/g
°
C), and - ∆Tis the temperature difference (80
°
C).
Substituting the values:
Q= 1500g×4.18 J
gC ×80C
Q= 502800 J
Therefore, the heat energy required to heat the water to the boiling point
(100
°
C) is 502800 J.
Question 17
Question
A copper block with a mass of 2 kg is heated from an initial temperature of
20
°
C to a final temperature of 70
°
C. Calculate the amount of heat required
to raise the temperature of the block. The specific heat capacity of copper is
390 J/kg◦C.
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, Ti= 20◦C Final temperature, Tf= 70◦C
The change in temperature, ∆T=Tf−Ti∆T= 70◦C−20◦C = 50◦C
Step 2: Use the formula Q=mc∆Tto calculate the amount of heat required.
Given: Mass of the copper block, m= 2 kg Specific heat capacity of copper,
c= 390 J/kg◦C
Substitute the values into the formula: Q= (2 kg)(390 J/kg◦C)(50◦C)
Step 3: Calculate the amount of heat required. Q= 2 ×390 ×50 = 39000 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 70
°
C is 39000 J.
Question 18
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 45
°
C. If the rod is heated until its temperature reaches 85
°
C, calculate the
amount of heat energy absorbed by the rod. Assume the specific heat capacity
of copper is 390 J/kg ·K and the density of copper is 8,960 kg/m3.
17
Solution
Let’s denote: - L: length of the copper rod (1.5 m) - D: diameter of the
copper rod (2 cm) - Ti: initial temperature of the copper rod (45
°
C) - Tf: final
temperature of the copper rod (85
°
C) - C: specific heat capacity of copper
(390 J/kg ·K) - ρ: density of copper (8,960 kg/m3)
First, we need to find the mass of the copper rod:
Volume of the rod = πD
22
×L
Mass of the rod = Volume of the rod ×ρ
Step 1: Calculate the volume of the rod.
Volume of the rod = π0.02 m
22
×1.5 m
Step 2: Calculate the mass of the rod.
Mass of the rod = Volume of the rod ×ρ
Step 3: Calculate the heat energy absorbed by the rod using the formula:
Q=mc∆T
where - Q: heat energy absorbed by the rod - m: mass of the rod - c: specific
heat capacity of copper - ∆T: change in temperature
Step 4: Substitute the known values into the formula and calculate the heat
energy absorbed. Make sure to convert temperatures to Kelvin.
Q=mc(Tf−Ti)
Finally, calculate the heat energy absorbed by the copper rod.
Question 19
Question
A copper rod of length Land cross-sectional area Ais initially at a temperature
T1. It is then heated to a temperature T2. Calculate the change in length of the
rod. Assume that the coefficient of linear expansion of copper is α.
Solution
Step 1: Calculate the initial length of the copper rod using the formula for linear
expansion:
Linitial =L+ ∆Linitial =L(1 + α∆T1)
18
where ∆T1=T1−T0.
Step 2: Calculate the final length of the copper rod after heating to temper-
ature T2using the same formula:
Lfinal =L+ ∆Lfinal =L(1 + α∆T2)
where ∆T2=T2−T0.
Step 3: Calculate the change in length of the rod:
∆L=Lfinal −Linitial =L(α∆T2−α∆T1)
Therefore, the change in length of the copper rod when heated from T1to
T2is ∆L=Lα(∆T2−∆T1).
Question 20
Question
A metal rod of length L, cross-sectional area A, and thermal conductivity k
is initially at a temperature T0. One end of the rod is kept at a constant
temperature Th, while the other end is kept at a constant temperature Tcsuch
that Th> T0> Tc. Find the steady-state temperature distribution in the rod
and the rate at which heat is conducted through the rod.
Solution
Step 1: Let’s first define the heat conduction equation for steady-state condi-
tions:
−q=kAdT
dx
where qis the heat conducted through the rod per unit time, xis the distance
along the rod, and dT/dx is the temperature gradient.
Step 2: Since the rod is in a steady state, the temperature distribution within
the rod is linear:
T(x) = Tc+Th−Tc
Lx
Step 3: Now, let’s find the rate at which heat is conducted through the rod
by substituting the temperature distribution into the heat conduction equation:
−q=kA d
dx Tc+Th−Tc
Lx
Step 4: Taking the derivative with respect to xgives:
−q=kATh−Tc
L
19
Step 5: Therefore, the rate at which heat is conducted through the rod is
given by:
q=−kATh−Tc
L
Step 6: In conclusion, the steady-state temperature distribution in the rod
is linear with one end at temperature Tcand the other end at temperature Th.
The rate at which heat is conducted through the rod is q=−kA(Th−Tc)/L.
Question 21
Question
A copper rod of length 1 m and diameter 2 cm initially at 100
°
C is placed in
an ice-water mixture at 0
°
C. If the rod absorbs heat from the mixture until it
reaches thermal equilibrium, determine the final temperature of the rod. Given:
specific heat capacity of copper is 390 J/kg ·K, density of copper is 8930 kg/m3,
and latent heat of fusion of ice is 334 kJ/kg.
Solution
Step 1: Calculate the mass of the copper rod: The volume of the rod can be
calculated using V=πr2h, where ris the radius of the rod and his its length.
Substituting r= 0.01 m and h= 1 m, we get:
V=π×(0.01 m)2×1 m = 3.14 ×10−4m3
The mass of the rod can be calculated using m= density ×V. Substituting
the given density of copper, we get:
m= 8930 kg/m3×3.14 ×10−4m3= 2.805 kg
Step 2: Calculate the heat absorbed by the copper rod: The heat absorbed
by 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.
Let Tfbe the final temperature of the rod. The change in temperature is
then 100 −Tf(assuming Tf<100). Therefore, Q= 2.805 kg ×390 J/kg ·K×
(100 −Tf) = 280.215 (100 −Tf) J.
Step 3: Calculate the heat released by the ice-water mixture to cool down
the copper rod: The heat released by the ice-water mixture can be calculated
using the formula Q=mL, where mis the mass of melted ice and Lis the
latent heat of fusion of ice.
Given that the copper rod absorbs all the heat required to melt the ice, we
can equate the heat absorbed by the rod to the heat released by the ice-water
mixture:
280.215 (100 −Tf) J = m×334 ×103J
Step 4: Calculate the final temperature of the copper rod by solving the
equation from Step 3.
20
Question 22
Question
A copper rod of length 2 m and diameter 4 cm is initially at a temperature of
100
°
C. If the rod is heated until its temperature reaches 200
°
C, calculate the
amount of heat transferred to the rod. Assume the specific heat capacity of
copper is 390 J/kg◦C and the density of copper is 8900 kg/m3.
Solution
Step 1: First, calculate the mass of the copper rod using its dimensions and
density. The volume of the copper rod is given by V=π
4×(diameter)2×(length).
Substitute the values d= 4 cm = 0.04 m and l= 2 m:
V=π
4×(0.04)2×2
V=π
4×0.0016 ×2
V=π×0.0032
4
V= 0.00251327 m3
Given the density of copper, we can find the mass musing density = m
V.
m= density ×V= 8900 ×0.00251327 = 22.38315 kg
Step 2: Calculate the heat transferred using the formula Q=mc∆T. Given
that the specific heat capacity cof copper is 390 J/kg◦C, and the temperature
change ∆Tis 200◦C−100◦C = 100◦C. Substitute the values:
Q= 22.38315 ×390 ×100
Q= 873396 J
Therefore, the amount of heat transferred to the copper rod is 873396 J.
Question 23
Question
A copper rod of length 1.5 m has a cross-sectional area of 4 ×10−4m2. If a
temperature gradient of 20◦C is applied across the rod, determine the rate at
which heat flows through the rod assuming the thermal conductivity of copper
is 400 W/(m·K). Given that the thermal conductivity of copper is tempera-
ture dependent and can be approximated as k=k0(1 + αT ), where k0= 400
W/(m·K), α= 4 ×10−3K−1, and Tis the temperature in Kelvin.
21
Solution
Step 1: Calculate the temperature in Kelvin associated with the temperature
gradient. Given ∆T= 20◦C and T0= 20◦C, the temperature in Kelvin is
T= ∆T+T0= 20 + 20 = 40 K.
Step 2: Find the thermal conductivity of copper at the given temperature.
Given k=k0(1 + αT ), substitute k0= 400 W/(m·K), α= 4 ×10−3K−1,
and T= 40 K into the expression to find k:k= 400(1 + 4 ×10−3×40) =
400(1 + 0.16) = 400 ×1.16 = 464 W/(m·K).
Step 3: Calculate the rate at which heat flows through the rod. The rate
at which heat flows through the rod is given by Fourier’s Law: Q=−kAdT
dx ,
where Qis the heat flow rate, kis the thermal conductivity, Ais the cross-
sectional area of the rod, dT
dx is the temperature gradient, and the negative
sign indicates heat flow in the direction of decreasing temperature. Substitute
k= 464 W/(m·K), A= 4 ×10−4m2, and dT
dx = 20/1.5 K/m into the equation
to find Q:Q=−464 ×4×10−4×20
1.5=−464 ×4×2
15 =−464 ×8
15 =−247 W.
Hence, the rate at which heat flows through the rod is 247 W in the direction
of decreasing temperature.
Question 24
Question
A 1 kg block of copper at 100
°
C is placed in a 2 kg block of steel at 20
°
C.
Assuming no heat is lost to the surroundings, calculate the final temperature of
the two-block system. The specific heat capacity of copper is 385 J/(kg
°
C) and
the specific heat capacity of steel is 450 J/(kg
°
C).
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature. The heat lost is given by:
Qlost =mc∆T
where: m= mass of copper block = 1 kg, c= specific heat capacity of copper
= 385 J/(kg
°
C), ∆T= change in temperature for the copper block = (final
temperature - initial temperature).
Step 2: Calculate the heat gained by the steel block as it heats up to the
final temperature. The heat gained is given by:
Qgain =mc∆T
where: m= mass of steel block = 2 kg, c= specific heat capacity of steel = 450
J/(kg
°
C), ∆T= change in temperature for the steel block = (final temperature
- initial temperature).
22
Step 3: The heat lost by the copper block is equal to the heat gained by the
steel block (assuming no heat loss to the surroundings).
Qlost =Qgain
mc∆Tcopper =mc∆Tsteel
Step 4: Substitute the given values into the equation:
1×385 ×(final temperature −100) = 2 ×450 ×(final temperature −20)
Step 5: Solve for the final temperature:
385(final temperature −100) = 900(final temperature −20)
385final temperature −38500 = 900final temperature −18000
515final temperature = 20500
final temperature = 20500
515 ≈39.81C
So, the final temperature of the two-block system is approximately 39.81
°
C.
Question 25
Question
An aluminum rod of length 2.0 m and diameter 1.0 cm is initially at a tem-
perature of 20
°
C. It is immersed in a large water bath at 100
°
C. If the rod
is in perfect thermal contact with the water bath, calculate the time it takes
for the rod to reach a final temperature of 60
°
C. Assume the thermal conduc-
tivity of aluminum is 205 W/(m·K), the specific heat capacity of aluminum is
900 J/(kg·K), and the density of aluminum is 2700 kg/m3. (Neglect any heat
losses to the surroundings.)
Solution
Step 1: Calculate the volume and mass of the aluminum rod. The volume of the
aluminum rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where ris the radius of the rod and his the length of the rod. Given that
the diameter is 1.0 cm, the radius ris 0.5 cm or 0.005 m.
V=π×(0.005 m)2×2.0 m
V= 1.57 ×10−4m3
23
The mass of the aluminum rod can be calculated using the formula:
m= density ×volume
m= 2700 kg/m3×1.57 ×10−4m3
m= 0.425 kg
Step 2: Calculate the energy required to increase the temperature of the
aluminum rod from 20
°
C to 60
°
C. The energy required can be calculated using
the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity of aluminum,
and ∆Tis the change in temperature.
Given that ∆T= 60C−20C= 40C, we can substitute the values:
Q= 0.425 kg ×900 J/(kg·K) ×40C
Q= 15300 J
Step 3: Calculate the rate of heat transfer. The rate of heat transfer, dQ
dt , is
given by Fourier’s law of heat conduction:
dQ
dt =−kAdT
dx
where kis the thermal conductivity of aluminum, Ais the cross-sectional
area of the rod, and dT
dx is the temperature gradient.
The cross-sectional area of the rod can be calculated using the formula for
the area of a circle:
A=πr2
A=π×(0.005 m)2
A= 7.85 ×10−5m2
The temperature gradient, dT
dx , can be calculated as:
dT
dx =Tfinal −Tinitial
L
dT
dx =60C−20C
2.0 m
24
dT
dx = 20C/m
Now we can calculate the rate of heat transfer:
dQ
dt =−205 W/(m·K) ×7.85 ×10−5m2×20C/m
dQ
dt =−0.32 W
Step 4: Calculate the time taken for the rod to reach 60
°
C. The time taken
can be calculated using the formula:
Q=Zt
0
dQ
dt dt
15300 J = Zt
0
−0.32 Wdt
15300
Question 26
Question
A metal rod of length Land thermal conductivity kis initially at a uniform
temperature Tithroughout its length. The rod is then heated at one end,
causing a constant temperature gradient dT
dx to be established along the length
of the rod. If the temperature at the heated end is Thand at the other end
is Tc, derive an expression for the rate at which heat is conducted through the
rod.
Solution
Step 1: Determine the temperature distribution along the rod.
Let xbe the distance from the heated end. The temperature Tat a distance
xfrom the heated end satisfies the heat conduction equation:
d
dx kdT
dx = 0
Integrating twice:
kdT
dx =C1
dT
dx =C1
k
25
T=C1
kx+C2
Applying the boundary conditions T(0) = Thand T(L) = Tc:
Th=C2
Tc=C1
kL+C2=C1
kL+Th
C1=k(Tc−Th)/L
C2=Th
Thus, the temperature distribution is given by:
T(x) = Th+ (Tc−Th)x
L
Step 2: Calculate the rate of heat conduction.
The rate of heat conduction Qthrough the rod is given by Fourier’s Law:
Q=−kAdT
dx
where Ais the cross-sectional area of the rod.
The heat conducted through an element dx at a distance xfrom the heated
end is:
dQ =−kAdT
dx dx =−kA d
dx Th+ (Tc−Th)x
Ldx
=−kA(Tc−Th)dx
L
To find the total rate of heat conduction through the entire rod, integrate
dQ from 0 to L:
Q=ZL
0
−kA(Tc−Th)dx
L
Q=−kA(Tc−Th)L
L
Q=−kA(Tc−Th)
Question 27
Question
A copper rod of length 50 cm and diameter 2 cm is initially at a temperature
of 100
°
C. It is placed in a water bath at a temperature of 20
°
C. If the rod loses
heat to the water bath until it reaches thermal equilibrium, calculate the final
temperature of the rod. The specific heat capacity of copper is 0.385 J/g
°
C and
its density is 8.96 g/cm3. Assume the only heat lost by the rod is used to warm
the water bath.
26
Solution
Step 1: Calculate the mass of the copper rod. The volume of the copper rod is
given by V=πr2h, where ris the radius and his the height. Given that the
diameter is 2 cm, the radius is r= 1 cm. Thus, V=π(1 cm)2·50 cm = 50πcm3.
Since the density of copper is 8.96 g/cm3, the mass mof the copper rod is
m= density ×volume = 8.96 g/cm3×50πcm3= 448πg.
Step 2: Calculate the heat lost by the copper rod. The heat lost by the
copper rod can be calculated 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 copper rod is 100
°
C and the final temperature is
unknown, denoted as T. Therefore, the change in temperature ∆T= 100 −T.
Substitute m= 448πg, c= 0.385 J/g
°
C, and ∆T= 100 −Tinto the formula.
Step 3: Calculate the heat absorbed by the water bath. Since the only heat
lost by the copper rod is used to warm the water bath, the heat absorbed by
the water bath is equal to the heat lost by the copper rod. Therefore, the heat
absorbed by the water bath is mc∆T=Qas well. Substitute mand cinto the
formula and equate it to the heat lost by the copper rod.
Step 4: Solve for the final temperature. After equating the two heat quan-
tities, solve for the final temperature Tby setting up the equation and solving
for T.
448π×0.385 ×(100 −T) = 448π×(T−20)
Question 28
Question
A copper rod of length 2 m and cross-sectional area 5 cm2is initially at a
temperature of 100◦C. It is then placed in an ice bath at 0◦C. If the rod loses
heat to the ice at a constant rate of 200 J/s, how long will it take for the rod to
cool down to 10◦C? The specific heat capacity of copper is 390 J/kgK and its
density is 8900 kg/m3.
Solution
Step 1: Calculate the mass of the copper rod. Given: Length, L= 2 m Cross-
sectional area, A= 5 cm2= 5 ×10−4m2Density of copper, ρ= 8900 kg/m3
The volume of the rod is V=A·L= 5 ×10−4×2 = 0.001 m3. Therefore,
the mass of the rod is m=ρ·V= 8900 ×0.001 = 8.9 kg.
Step 2: Calculate the initial heat content of the rod. The initial temperature
of the rod, Ti= 100 ◦C = 100 + 273 = 373 K. The final temperature at which
we want to cool the rod, Tf= 10 ◦C = 10 + 273 = 283 K. The change in
temperature, ∆T=Ti−Tf= 373 −283 = 90 K.
The initial heat content of the rod is given by Qi=m·c·∆T, where c
is the specific heat capacity of copper. Substituting the values, we get Qi=
8.9·390 ·90 = 313650 J.
27
Step 3: Calculate the energy lost per degree drop in temperature. Given
that the rod loses heat at a constant rate of 200 J/s, the energy lost per second
per degree drop in temperature is 200 J/s.
Step 4: Calculate the time taken to cool the rod from an initial temperature
of 100◦C to 10◦C. The total energy required to cool the rod from 100◦C to
10◦C is Qi−(Qi−(90 ·200)) = Qi−90 ·200. Substitute the values to get
Qi−90 ·200 = 313650 −18000 = 295650 J.
The time taken to cool the rod to 10◦C is given by t=Qtotal
Rate of heat loss .
Substitute the values to get t=295650
200 = 1478.25 seconds.
Therefore, it will take approximately 1478 seconds for the rod to cool down
from 100◦C to 10◦C.
Question 29
Question
A metal rod of length 1 m is initially at a temperature of 500◦C. If one end
of the rod is submerged in an ice bath at 0◦C while the other end is heated to
1000◦C, what is the temperature at a point 30 cm from the colder end after the
rod reaches thermal equilibrium?
Solution
Given: Initial temperature of the rod, T1= 500◦C
Temperature at one end, T2= 1000◦C
Temperature at the other end, T3= 0◦C
Length of the rod, L= 1 m
Distance from the colder end, x= 30 cm = 0.3 m
We need to find the temperature at a point 30 cm from the colder end after
thermal equilibrium is reached.
Step 1: Firstly, we find the rate of heat flow per unit length of the rod when
the rod reaches thermal equilibrium. This is given by Fourier’s law:
dQ
dt =−kAdT
dx
where: dQ/dt = rate of heat flow per unit time, k= thermal conductivity
of the material, A= cross-sectional area of the rod, and dT/dx = temperature
gradient along the rod.
Since the rod has reached thermal equilibrium, the rate of heat flow at any
point along the rod is constant. Let this constant rate be denoted as q.
Step 2: Next, we express qin terms of the temperatures at the two ends of
the rod. We have:
q=kAT2−T3
L
Step 3: Substituting the values we have into the above formula, we find q:
28
q=kA1000 −0
1
q=kA(1000)
Step 4: Finally, we can find the temperature at a point 30 cm from the
colder end using the following relationship:
T(x) = T3+qL
kAx
Substitute the known values:
T(0.3) = 0 + (1000)(1)
kA (0.3)
T(0.3) = 300(0.3)
T(0.3) = 90◦C
Therefore, the temperature at a point 30 cm from the colder end after the
rod reaches thermal equilibrium is 90◦C.
Question 30
Question
A 200 g block of copper at 100
°
C is dropped into 400 g of water at 20
°
C in
a 200 g aluminum calorimeter cup. The final temperature of the mixture is
31
°
C. Assuming no heat is lost to the surroundings, calculate the specific heat
capacity of aluminum. The specific heat capacity of water is 4186 J/kg
°
C and
the specific heat capacity of copper is 387 J/kg
°
C.
Solution
Step 1: Calculate the total heat absorbed by copper block.
qcopper =mcopperccopper∆Tcopper
qcopper = (0.2 kg)(387 J/kg
°
C)(31 −100)
°
C
qcopper =−12900 J
Step 2: Calculate the total heat absorbed by water.
qwater =mwatercwater∆Twater
qwater = (0.4 kg)(4186 J/kg
°
C)(31 −20)
°
C
qwater = 14692 J
29
Step 3: Calculate the change in heat for the aluminum calorimeter cup.
qaluminum =−(qcopper +qwater)
qaluminum =−(−12900 + 14692) J
qaluminum = 1792 J
Step 4: Calculate the specific heat capacity of aluminum.
qaluminum =maluminumcaluminum∆Taluminum
1792 J = (0.2 kg)caluminum(31 −20)
°
C
caluminum =1792
0.2×11
caluminum = 814.54 J/kg
°
C
Therefore, the specific heat capacity of aluminum is 814.54 J/kg
°
C.
Question 31
Question
A copper rod of length 1 m and cross-sectional area 0.01 m2is heated to a
temperature of 100◦C. The rod loses heat energy to the surroundings at a con-
stant rate of 500 W. If the thermal conductivity of copper is 400 W/(m·K) and
its specific heat capacity is 400 J/(kg·K), calculate the time it takes for the
temperature of the rod to drop to 50◦C. Assume the density of copper is 8900
kg/m3.
Solution
Step 1: Calculate the mass of the copper rod using its volume and density.
The volume of the rod is given by V=A·L= 0.01 m2·1 m = 0.01 m3.
The mass can be calculated as m=ρ·V= 8900 kg/m3·0.01 m3= 89 kg.
Step 2: Calculate the initial heat content of the rod.
The specific heat capacity of copper is 400 J/(kg·K), so the heat content of the
rod at 100◦C is Qinitial =m·c·∆T= 89 kg ·400 J/(kg·K) ·100 K = 3.56 ×106J.
Step 3: Determine the rate of temperature decrease in the rod.
The rate at which the rod loses heat is given as 500 W. By the formula P=
kA∆T
L, we can solve for the rate of temperature decrease ∆T
∆t.
Substitute the known values: 500 = 400 W/(m·K) ·0.01 m2·∆T
1 m .
Solving for ∆T, we get ∆T= 5 K/s.
Step 4: Determine the time it takes for the rod to cool to 50◦C.
The final temperature is 50◦C, so ∆T= 100 −50 = 50 K. The time it takes for
the rod to cool to this temperature can be found using ∆t=∆T
∆T
∆t
=50 K
5 K/s = 10 s.
Therefore, it takes 10 seconds for the temperature of the rod to drop to
50◦C.
30
Question 32
Question
A sample of copper with mass 500 g is heated from 20
°
C to 90
°
C. Calculate the
amount of heat required to bring about this temperature change. The specific
heat capacity of copper is 0.385 J/g
°
C.
Solution
Step 1: Determine the change in temperature. Step 2: Calculate the amount of
heat transferred.
Step 1: The change in temperature can be calculated using the formula:
∆T=Tf−Ti= 90C−20C= 70C
Step 2: The amount of heat transferred can be calculated using the formula:
Q=mc∆T
where: - Qis the amount of heat transferred, - mis the mass of the sample (500
g), - cis the specific heat capacity of copper (0.385 J/g
°
C), - ∆Tis the change
in temperature (70
°
C).
Substitute the values into the formula:
Q= (500 g)(0.385 J/g
°
C)(70
°
C)
Q= 500 ×0.385 ×70
Q= 19250 J
Therefore, the amount of heat required to bring about this temperature
change is 19250 J.
Question 33
Question
An aluminum rod of length 2.0 m and diameter 0.02 m is heated from an initial
temperature of 20
°
C to a final temperature of 80
°
C. The rod has a thermal
conductivity of 237 W/(m*K), a specific heat capacity of 900 J/(kg*K), and a
density of 2700 kg/m
³
. Calculate the heat energy transferred to the rod during
this heating process.
31
Solution
Step 1: Calculate the mass of the aluminum rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder: V=πr2h, where ris
the radius and his the height (length) of the cylinder. Given that the diameter
is 0.02 m, the radius r= 0.01 m. The volume Vof the rod is:
V=π(0.01)2×2.0=0.000628 m3
The mass mof the rod can be calculated using the formula m= density×volume.
Substitute the density of aluminum, ρ= 2700 kg/m
³
:
m= 2700 ×0.000628 = 1.6764 kg
Step 2: Calculate the heat energy transferred. The heat energy transferred
to the rod can be calculated using the formula Q=mc∆T, where mis the mass,
cis the specific heat capacity, and ∆Tis the change in temperature. Given that
the initial temperature Ti= 20
°
C and the final temperature Tf= 80
°
C, the
change in temperature is ∆T=Tf−Ti= 80 −20 = 60
°
C (note that the change
in temperature must be in Kelvin, so ∆T= 60 K). Now, substitute the values
into the formula:
Q= 1.6764 ×900 ×60 = 90679.2 J
Hence, the heat energy transferred to the aluminum rod during this heating
process is 90679.2 J.
Question 34
Question
A 200 g piece of copper is heated until its temperature reaches 65
°
C. The piece
is then placed in 500 g of water at 20
°
C. What will be the final temperature
when thermal equilibrium is reached? (Specific heat capacity of copper is 0.385
J/g
°
C and of water is 4.18 J/g
°
C.)
Solution
Step 1: Calculate the heat gained by the copper piece: The formula for heat
transfer is: Q=mc∆T, where Qis heat energy, mis mass, cis specific heat
capacity, and ∆Tis the change in temperature. Given: m= 200 g, cCu = 0.385
J/g
°
C, ∆TCu = 65 −Tf, where Tfis the final temperature. So, QCu = 200 ×
0.385 ×(65 −Tf).
Step 2: Calculate the heat lost by the copper to the water: The heat lost
by the copper will be equal to the heat gained by the water, as no heat is lost
to the surroundings. Given: QCu =QH2O, where QH2O =mc∆T. Using the
mass of water m= 500 g, cH2O = 4.18 J/g
°
C, and ∆TH2O =Tf−20, we have
QH2O = 500 ×4.18 ×(Tf−20).
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Step 3: Set up the equation and solve for the final temperature: Since
QCu =QH2O, we can set the two expressions for heat equal to each other and
solve for Tf: 200 ×0.385 ×(65 −Tf) = 500 ×4.18 ×(Tf−20). Solving this
equation will give us the final temperature Tf.
Question 35
Question
A copper rod of length 1.5 m and diameter 2 cm is heated from 20
°
C to 100
°
C. If
the thermal conductivity of copper is 400 W/mK and its density is 8960 kg/m3,
calculate the total heat energy transferred to the rod during this process.
Solution
Step 1: Find the cross-sectional area of the copper 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. Given that the diameter of the rod is 2 cm, the radius
ris half of the diameter, so r= 1 cm = 0.01 m. Therefore, the cross-sectional
area A=π(0.01)2= 3.14 ×10−4m2.
Step 2: Find the volume of the copper rod. The volume of the rod can be
calculated as V=A×L, where Lis the length of the rod. Given that the length
of the rod is 1.5 m, the volume V= 3.14 ×10−4m2×1.5 m = 4.71 ×10−4m3.
Step 3: Find the mass of the copper rod. The mass of the rod can be
calculated using the formula m= density ×volume. Given that the density of
copper is 8960 kg/m3, the mass m= 8960 kg/m3×4.71 ×10−4m3= 4.22 kg.
Step 4: Calculate the heat energy transferred. The heat energy transferred
to the rod can be calculated using the formula Q=mc∆T, where mis the mass,
cis the specific heat capacity (which for copper is approximately 390 J/kgK),
and ∆Tis the change in temperature. Given that the initial temperature T1=
20Cand the final temperature T2= 100C, the change in temperature ∆T=
T2−T1= 100C−20C= 80C. Plugging in the values, we get Q= 4.22 kg ×
390 J/kgK ×80 K = 132768 J. Therefore, the total heat energy transferred to
the rod during this process is 132768 J.
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