PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 9
Liberty University
Question 1
Question
A 500 g aluminum block at a temperature of 100
°
C is dropped into a calorimeter
containing 500 g of water at 20
°
C. The final temperature of the mixture is 22
°
C.
Assuming no heat is lost to the surroundings, what is the specific heat capacity
of the aluminum block? (Specific heat capacity of water is 4.18 J/g
°
C)
Solution
Step 1: First, calculate the heat lost by the aluminum block as it cools down
from 100
°
C to the final temperature. The heat lost by the aluminum block is
given by:
Qaluminum =m·c·∆T
where: m= mass of the aluminum block = 500 g, c= specific heat capacity
of aluminum (unknown), ∆T= change in temperature of the aluminum block
= 100C−22C= 78C.
Substitute the given values into the equation:
Qaluminum = 500 g ×cJ/g
°
C×78C
Step 2: Next, calculate the heat gained by the water as it warms up from
20
°
C to the final temperature. The heat gained by the water is given by:
Qwater =m·c·∆T
where: m= mass of the water = 500 g, c= specific heat capacity of water
= 4.18 J/g
°
C, ∆T= change in temperature of the water = 22C−20C= 2C.
Substitute the given values into the equation:
Qwater = 500 g ×4.18 J/g
°
C×2C
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block is equal to the heat gained by the water. Set both heat equations
equal to each other:
500 g ×cJ/g
°
C×78C= 500 g ×4.18 J/g
°
C×2C
Solve for the specific heat capacity cof aluminum.
Question 2
Question
A copper container holds 0.5 kg of water at 20◦C. How much heat is required
to raise the temperature of the water to 90◦C? Given: Specific heat capacity
of water = 4200 J/kg◦C, Specific heat capacity of copper = 380 J/kg◦C, and
neglect any heat loss to the surroundings.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
20◦C to 90◦C. Step 2: Calculate the heat required to raise the temperature of
the copper container from 20◦C to 90◦C. Step 3: Add the heat required for
the water and the heat required for the copper container to find the total heat
required.
Step 1: The heat required to raise the temperature of the water is given by
the formula:
Qwater =mc∆T
where: m= 0.5 kg (mass of water), c= 4200 J/kg◦C (specific heat capacity of
water), ∆T= 90◦C−20◦C = 70◦C (change in temperature).
Substitute the values into the formula:
Qwater = 0.5×4200 ×70
Qwater = 147000 J
So, the heat required to raise the temperature of the water is 147000 J.
Step 2: The heat required to raise the temperature of the copper container
is given by the formula:
Qcopper =mc∆T
where: m= mass of the copper container, c= 380 J/kg◦C (specific heat capacity
of copper), ∆T= 70◦C (change in temperature).
As the mass of the copper container is not given, we will assume it is very
large so that it effectively absorbs all the heat energy from the water. Therefore,
the heat required to raise the temperature of the copper container will be equal
to the heat lost by the water, which is 147000 J.
2
Step 3: The total heat required is the sum of the heat required for the
water and the copper container:
Qtotal =Qwater +Qcopper
Qtotal = 147000 + 147000
Qtotal = 294000 J
Therefore, the total heat required to raise the temperature of the water to
90◦C is 294000 J.
Question 3
Question
A metal bar of length Land thermal conductivity kis initially at a uniform
temperature of T0throughout. One end of the bar is then placed in an ice bath
at 0◦C and the other end is placed in a boiling water at 100◦C. The bar reaches
a steady state where the temperature at xmeters from the ice bath, T(x), is
given by the equation T(x) = (T0−0◦C) 1−x
L+ 0◦C. Calculate the rate of
heat transfer through the bar in terms of the given variables.
Solution
Step 1: To find the rate of heat transfer, we need to calculate the heat flux
density qat any point xalong the metal bar. The heat flux density is given by
Fourier’s law:
q=−kAdT
dx ,
where Ais the cross-sectional area of the metal bar.
Step 2: We can differentiate the equation for temperature T(x) to find dT
dx :
dT
dx =−T0−0◦C
L.
Step 3: Substituting the expression for dT
dx into Fourier’s law, we get
q=kAT0−0◦C
L.
Step 4: The heat transfer rate Qthrough the bar is given by
Q=qA =kA2T0−0◦C
L.
Therefore, the rate of heat transfer through the bar in terms of the given
variables is
Q=kA2T0−0◦C
L.
3
Question 4
Question
A copper rod of length 1 m has a temperature of 100◦C at one end and 0◦C
at the other end. If the rod is insulated along its length, what is the average
temperature of the rod?
Solution
To find the average temperature of the rod, we need to consider the temperature
distribution along the length of the rod.
Step 1: Calculate the change in temperature along the length of the rod.
The change in temperature along the length of the rod is given by the temper-
ature difference between the two ends: ∆T= 100◦C - 0◦C = 100◦C.
Step 2: Find the average temperature. The average temperature of the rod
can be calculated using the formula: Average temperature = Initial temperature
+∆T
2. Substitute the values: Average temperature = 0◦C + 100
2= 50◦C.
Therefore, the average temperature of the rod is 50◦C.
Question 5
Question
A copper cylinder of mass 500 g at a temperature of 100
°
C is dropped into 400
g of water at 20
°
C. If the final temperature of the mixture is 30
°
C, determine
the specific heat capacity of copper. Assume that all the heat lost by the copper
is gained by the water and that there are no heat losses to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper cylinder. Let the specific heat
capacity of copper be denoted by ccand the initial temperature of the copper
be denoted by Tc1. The heat lost by the copper cylinder can be calculated using
the formula:
Qlost =mc·cc·(Tfinal −Tc1)
where mcis the mass of the copper cylinder and Tfinal is the final temperature of
the mixture. Substitute the given values: mc= 500 g = 0.5 kg, Tc1= 100◦C =
373 K, and Tfinal = 30◦C = 303 K.
Qlost = 0.5·cc·(303 −373)
Step 2: Calculate the heat gained by the water. Let the specific heat capacity
of water be denoted by cwand the initial temperature of the water be denoted
by Tw1= 20◦C = 293 K. The heat gained by the water can be calculated using
the formula:
Qgain =mw·cw·(Tfinal −Tw1)
4
where mwis the mass of the water. Given that the mass of water, mw= 400 g =
0.4 kg, and the final temperature, Tfinal = 30◦C = 303 K.
Qgain = 0.4·cw·(303 −293)
Step 3: Since there is no heat loss to the surroundings, the heat lost by
the copper must be equal to the heat gained by the water. Equating the two
equations from Step 1 and Step 2:
0.5·cc·(303 −373) = 0.4·cw·(303 −293)
Step 4: Solve for cc(specific heat capacity of copper). Simplify and solve
the equation from Step 3 to find the specific heat capacity of copper, cc.
Question 6
Question
A piece of aluminum initially at 100◦C is dropped into a container of water at
20◦C. The mass of the aluminum is 500 g and the mass of the water is 2 kg.
Assuming no heat is lost to the surroundings, calculate the final equilibrium
temperature of the system. The specific heat capacity of aluminum is 0.9 J/g◦C
and that of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat gained or lost by the aluminum. The heat lost by
the aluminum is equal to the heat gained by the water, assuming no heat is lost
to the surroundings. Let Tfbe the final equilibrium temperature of the system.
The heat lost by the aluminum is given by:
Qalu =malu ·calu ·(Tf−Tinitial, alu)
where: malu = mass of aluminum = 0.5 kg = 500 g, calu = specific heat capacity
of aluminum = 0.9 J/g◦C, Tinitial, alu = initial temperature of aluminum =
100◦C, Tf= final equilibrium temperature of the system, Qalu = heat lost by
aluminum.
Substitute the given values into the formula:
Qalu = 500 g ·0.9 J/g◦C·(Tf−100◦C)
Qalu = 450 ·(Tf−100)
Step 2: Calculate the heat gained by the water. The heat gained by the
water is given by:
Qwater =mwater ·cwater ·(Tf−Tinitial, water)
5
where: mwater = mass of water = 2 kg = 2000 g, cwater = specific heat capacity
of water = 4.18 J/g◦C, Tinitial, water = initial temperature of water = 20◦C, Tf
= final equilibrium temperature of the system, Qwater = heat gained by water.
Substitute the given values into the formula:
Qwater = 2000 g ·4.18 J/g◦C·(Tf−20◦C)
Qwater = 8360 ·(Tf−20)
Step 3: Set up the equation. Since no heat is lost to the surroundings,
Qalu =Qwater.
450 ·(Tf−100) = 8360 ·(Tf−20)
Step 4: Solve for Tf.
450Tf−45000 = 8360Tf−167200
450Tf−8360Tf=−167200 + 45000
7910Tf=−122200
Tf=−122200
7910
Tf≈15.44◦C
Therefore, the final equilibrium temperature of the system is approximately
15.44◦C.
Question 7
Question
A block of copper with a mass of 500 g undergoes a temperature change of 50
°
C.
How much heat energy is required to cause this temperature change? (Specific
heat capacity of copper is 0.385 J/g
°
C)
Solution
Step 1: First, calculate the amount of heat energy using the formula:
Q=mc∆T
where: Q= heat energy (in Joules), m= mass (in grams) = 500 g, c=
specific heat capacity of copper = 0.385 J/g
°
C, ∆T= temperature change =
50
°
C.
Step 2: Substitute the values into the formula:
Q= (500 g)(0.385 J/g
°
C)(50C)
Step 3: Calculate the heat energy:
6
Q= 500 ×0.385 ×50
Q= 9625 J
Therefore, the amount of heat energy required to cause a temperature change
of 50
°
C in a 500 g block of copper is 9625 J.
Question 8
Question
A gas at pressure P1and volume V1is heated from an initial temperature T1to
a final temperature T2. The gas follows the equation of state P V =nRT where
nis the number of moles and Ris the ideal gas constant. If the final volume of
the gas is V2, find the work done on the gas during the process.
Solution
Step 1: First, let’s find the final pressure P2of the gas using the ideal gas law:
P1V1=nRT1and P2V2=nRT2
Since the number of moles nand the ideal gas constant Rare constant, we have:
P1V1
T1
=P2V2
T2
=⇒P2=P1V1T2
V2T1
Step 2: The work done on the gas during the process can be calculated using
the formula:
W=ZV2
V1
P dV
Substitute the expression for Pin terms of Vinto the integral:
W=ZV2
V1
P1V1T2
V2T1
dV =P1V1T2
V2T1ZV2
V1
dV
W=P1V1T2
T1
[ln(V2)−ln(V1)] = P1V1T2ln V2
V1
Therefore, the work done on the gas during the process is P1V1T2ln V2
V1.
Question 9
Question
A copper cylinder of length 50 cm and diameter 10 cm is heated from 20
°
C to
80
°
C. If the coefficient of linear expansion for copper is 1.7×10−5per degree
Celsius, determine the change in volume of the cylinder.
7
Solution
Step 1: Find the change in length of the cylinder. Step 2: Use the change in
length to find the change in volume.
Step 1: The formula for linear expansion is given by:
∆L=α·L·∆T
where: ∆L= change in length, α= coefficient of linear expansion, L= original
length, and ∆T= change in temperature.
Substitute the given values:
∆L= (1.7×10−5)·(50 cm) ·(80 −20)
°
C
∆L= 0.000085 cm ×60
°
C
∆L= 0.0051 cm
Step 2: The formula for volume expansion is given by:
∆V= 3α·V·∆T
where: ∆V= change in volume, α= coefficient of linear expansion, V= original
volume, and ∆T= change in temperature.
First, calculate the original volume of the cylinder:
V=πD2L
4=π(10 cm)2·50 cm
4
V= 1250πcm3
Now, substitute the values to find the change in volume:
∆V= 3(1.7×10−5)·(1250π)·60
∆V= 0.000051 ·7500π
∆V= 0.38πcm3
Therefore, the change in volume of the copper cylinder is 0.38πcm3.
Question 10
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 100
°
C. It is then placed in an ice bath at 0
°
C. If the specific heat capacity
of copper is 390 J/kg ·K and its density is 8900 kg/m3, calculate the amount of
heat transferred to or from the rod to bring it to thermal equilibrium with the
ice bath.
8
Solution
Step 1: Calculate the mass of the copper rod.
Given that the rod is cylindrical, we can find its volume using the formula for
the volume of a cylinder: V=πr2h, where ris the radius and his the height.
The mass of the rod can be calculated using the density formula: m=ρV ,
where ρis the density of copper.
Given: Diameter = 2 cm = 0.02 m (radius, r= 0.01 m), Height, h= 1.5 m,
Density, ρ= 8900 kg/m3.
Plugging in these values, we get:
V=π(0.01 m)2×1.5 m
V=π×10−4×1.5
V= 1.5π×10−4m3
m= 8900 kg/m3×1.5π×10−4m3
m= 1.3345 kg
Therefore, the mass of the copper rod is 1.3345 kg.
Step 2: Calculate the heat lost by the rod.
The heat lost by 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: Initial temperature, Tinitial = 100
°
C, Final temperature, Tfinal = 0
°
C,
Specific heat capacity, c= 390 J/kg ·K.
Plugging in these values, we get:
∆T=Tfinal −Tinitial = 0 −100 = −100 K
Q= 1.3345 ×390 ×(−100)
Q=−519465 J
Therefore, the heat lost by the rod is 519465 J.
Question 11
Question
A copper sphere with a radius of 10 cm is initially at a temperature of 100◦C. It
is then placed in a large vat of water at 20◦C. Assuming the only heat transfer is
between the sphere and the water, calculate the time it takes for the temperature
of the sphere to reach 25◦C. The specific heat capacity of copper is 0.385 J/g◦C,
and the density of copper is 8.96 g/cm3. Assume the sphere loses heat primarily
through radiation with a constant loss rate of 5W.
9
Solution
Step 1: Determine the mass of the copper sphere. Given that density of copper
is 8.96 g/cm3and the sphere has a radius of 10 cm, the volume of the sphere
can be calculated using the formula for the volume of a sphere:
V=4
3πr3
Substitute the given radius:
V=4
3π(10 cm)3=4000
3πcm3
Then, the mass of the sphere can be calculated using the density:
Mass = Density ×Volume = 8.96 ×4000
3πg≈37642 g
Step 2: Calculate the energy needed to change the temperature of the sphere
from 100◦C to 25◦C. The heat energy required to change the temperature can
be calculated as:
Q=mc∆T
Where: - m= 37642 g is the mass of the copper sphere, - c= 0.385 J/g◦C is
the specific heat capacity of copper, and - ∆T= 100 −25 = 75 ◦C is the change
in temperature. Substitute the values:
Q= 37642 ×0.385 ×75 = 1099957 J
Step 3: Calculate the time it takes for the temperature to decrease by 5◦C.
The rate at which the sphere loses heat energy through radiation is given as 5
W. This means:
P=∆Q
∆t
Where ∆Q= 1099957 J is the heat energy to be lost, and P= 5 W is the
power. Solve for ∆t:
5 = 1099957
∆t
∆t=1099957
5≈219991.4 seconds
Therefore, it will take approximately 219991.4 seconds for the temperature
of the copper sphere to decrease from 100◦C to 25◦C.
Question 12
Question
A 5 kg iron bar at 100◦C is dropped into 10 liters of water at 20◦C. Assuming
no heat is lost to the surroundings, calculate the final temperature of the iron
and water system. (Specific heat capacity of iron is 450 J/kg◦C, specific heat
capacity of water is 4186 J/kg◦C, and the density of water is 1000 kg/m3.)
10
Solution
Step 1: Find the heat gained by the iron bar as it cools down from 100◦C to
the final temperature. The formula for heat transfer is given by:
Q=mc∆T
where: Q= heat transferred, m= mass, c= specific heat capacity, and ∆T=
change in temperature.
For the iron bar: m= 5 kg, ciron = 450 J/kg◦C, Initial temperature Ti= 100
◦C, Final temperature Tf=T◦C, and ∆T=T−Ti.
The heat lost by the iron bar is equal to the heat gained by the water,
therefore:
mciron(T−Ti) = mcwater(T−Twater)
Substitute in the given values:
5×450(T−100) = 10 ×4186(T−20)
Step 2: Solve for Tto find the final temperature of the iron and water
system. Expand and solve the equation:
2250T−225000 = 41860T−837200
39610T= 612200
T≈15.47◦C
Therefore, the final temperature of the iron and water system is approxi-
mately 15.47◦C.
Question 13
Question
A metal rod of length Land thermal conductivity kis heated at one end to a
temperature T1and kept at a constant temperature of T1while the other end
is kept at a constant temperature of T2such that (T1> T2). The rod loses heat
to the surroundings at a rate of Qper unit time. If the rod is in a steady state,
determine the temperature distribution along the rod and express the heat flux
qin terms of T1,T2,L, and Q.
Solution
Step 1: Set up the heat conduction equation
d2T
dx2= 0
11
where T(x) is the temperature distribution along the rod and xis the distance
from the end at temperature T1.
Step 2: Integrate the heat conduction equation twice Integrating once, we
get dT
dx =c1
where c1is an arbitrary constant.
Integrating again, we get
T(x) = c1x+c2
where c2is another arbitrary constant.
Step 3: Apply boundary conditions At x= 0, T=T1. So,
T(0) = c2=T1
At x=L,T=T2. So,
T(L) = c1L+T1=T2
c1=T2−T1
L
Step 4: Express the heat flux The heat flux qis given by Fourier’s law as
q=−kdT
dx
Plugging in our expression for dT
dx ,
q=−kc1=kT1−T2
L
Therefore, the temperature distribution along the rod is given by
T(x) = T2−T1
Lx+T1
and the heat flux qis given by
q=kT1−T2
L
Question 14
Question
A metal cylinder of mass 2 kg and specific heat capacity 420 J/(kg·K) is heated
to a temperature of 100◦C. It is then placed in a calorimeter containing 1 kg of
water initially at 20◦C. If the final temperature of the system is 30◦C and no
heat is lost to the surroundings, find the specific heat capacity of the calorimeter.
12
Solution
Step 1: Calculate the heat absorbed by the metal cylinder.
Qmetal =mc∆T
= 2 kg ×420 J/(kg·K) ×(100 −30) K
= 2 ×420 ×70
= 58800 J
Step 2: Calculate the heat lost by the metal cylinder in order to raise the
temperature of the water and calorimeter to the final temperature of 30◦C.
Qmetal =Qwater +Qcalorimeter
58800 = (1 kg ×cwater ×(30 −20) K) + (mcalorimeter ×ccalorimeter ×(30 −20) K)
= 10cwater + 10ccalorimeter
Step 3: From the information given, we know that the specific heat capacity
of water is 4186 J/(kg·K). Substituting this into the equation above, we can
solve for the specific heat capacity of the calorimeter.
58800 = 10(4186) + 10ccalorimeter
10ccalorimeter = 58800 −10(4186)
10ccalorimeter = 50540
ccalorimeter =50540
10
ccalorimeter = 5054 J/(kg·K)
Therefore, the specific heat capacity of the calorimeter is 5054 J/(kg·K).
Question 15
Question
A metal rod of length 2 m and diameter 5 cm is heated from 20 ◦C to 80 ◦C.
If the coefficient of linear expansion for the metal is 1.8×10−5◦C−1and the
specific heat capacity is 450 J/kg ·◦C, calculate the amount of heat transferred
to the rod.
Solution
Step 1: Determine the change in length of the metal rod due to heating. Given:
Initial temperature, Ti= 20 ◦C Final temperature, Tf= 80 ◦C Coefficient of
linear expansion, α= 1.8×10−5◦C−1Original length, L0= 2 m
The change in length, ∆L, of the rod can be calculated using the formula:
∆L=L0·α·(Tf−Ti)
13
Substitute the given values:
∆L= 2 ·1.8×10−5·(80 −20) = 0.000 12 m
Step 2: Calculate the volume of the metal rod. The radius, r, of the rod is
half of the diameter:
r=diameter
2=5
2cm = 0.025 m
Volume of the rod, V=πr2L
V=π(0.025)2·2 = π·0.00125 ·2 = 0.0025 m3
Step 3: Calculate the mass of the metal rod. Given: Density of the metal
= 8000 kg/m3Mass, m= density ×volume
m= 8000 ×0.0025 = 20 kg
Step 4: Determine the amount of heat transferred to the rod. The heat
transferred, Q, is given by the formula:
Q=mc∆T
where cis the specific heat capacity and ∆Tis the temperature change.
Q= 20 ×450 ×(80 −20) = 720 000 J
Therefore, the amount of heat transferred to the rod is 720 000 J.
Question 16
Question
A copper rod of length 2 m and diameter 1 cm is heated from 20◦C to 100◦C.
Given that the coefficient of linear expansion of copper is 1.7×10−5C−1, deter-
mine the change in the volume of the rod.
Solution
Step 1: Find the initial volume of the rod. The initial volume of the copper rod
can be calculated as the volume of a cylinder:
Vinitial =πr2h
where ris the initial radius, his the length of the rod, and Vinitial is the initial
volume.
Given that the initial diameter is 1 cm, the initial radius ris 0.5 cm =
0.005 m, and the length his 2 m, we can calculate:
Vinitial =π(0.005 m)2×2 m
14
Vinitial =π×0.000025 ×2
Vinitial = 0.00005πm3
Step 2: Find the final volume of the rod. The final volume of the rod is
obtained by considering the increase in length of the rod due to heating. The
change in length can be calculated using the formula:
∆L=αL0∆T
where αis the coefficient of linear expansion, L0is the initial length, and ∆T
is the change in temperature.
Given that α= 1.7×10−5C−1,L0= 2 m, and ∆T= 80◦C (as the rod is
heated from 20◦C to 100◦C), we can find:
∆L= 1.7×10−5×2×80
∆L= 0.00272 m
The final volume can now be calculated using the change in length:
Vfinal =π(0.005 m)2×(2 + 0.00272) m
Vfinal =π×0.000025 ×2.00272
Vfinal ≈0.0000501πm3
Step 3: Determine the change in volume. The change in volume (∆V) is
calculated by finding the difference between the final volume and the initial
volume:
∆V=Vfinal −Vinitial
∆V= 0.0000501π−0.00005π
∆V≈0.0000001πm3
Therefore, the change in volume of the copper rod when heated from 20◦C
to 100◦C is approximately 0.0000001πm3.
Question 17
Question
A copper block with a mass of 500 g is heated from 25
°
C to 75
°
C. The specific
heat capacity of copper is 0.386 J/g
°
C. How much heat energy was supplied to
the copper block?
15
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, T1= 25CFinal temperature, T2= 75C
The change in temperature, ∆T=T2−T1= 75C−25C= 50C.
Step 2: Calculate the heat energy supplied to the copper block. The formula
for calculating heat energy is:
Q=mc∆T
where: Q= heat energy supplied (in joules) m= mass of the object (in grams)
c= specific heat capacity (in J/g
°
C) ∆T= change in temperature (in
°
C)
Given: m= 500 g c= 0.386 J/g
°
C ∆T= 50
°
C
Substitute the values into the formula:
Q= 500 g ×0.386 J/g
°
C×50
°
C
Q= 500 ×0.386 ×50
Q= 9650 joules
Therefore, the amount of heat energy supplied to the copper block is 9650
joules.
Question 18
Question
A metal rod of length 1 m has one end kept at a temperature of 100
°
C and the
other end at 0
°
C. The thermal conductivity of the rod is 200 W/mK and its
area of cross-section is 10−4m
²
. Calculate the rate of transfer of heat along the
rod.
Solution
Step 1: Calculate the temperature gradient along the rod. Given: - Length of
the rod, L= 1 m - Area of cross-section, A= 10−4m
²
- Temperature at one
end, T1= 100C= 373 K - Temperature at the other end, T2= 0C= 273 K -
Thermal conductivity, k= 200 W/mK
The temperature gradient dT
dx along the rod is given by:
dT
dx =T1−T2
L
dT
dx =373 −273
1
16
dT
dx = 100K/m
Step 2: Calculate the rate of transfer of heat along the rod. The rate of
transfer of heat, dQ
dt , is given by Fourier’s Law:
dQ
dt =−kAdT
dx
dQ
dt =−200 ×10−4×100
dQ
dt =−2 W
Therefore, the rate of transfer of heat along the rod is 2 W.
Question 19
Question
A copper cylinder of mass 0.5 kg is initially at a temperature of 200
°
C. It is
placed in a container of water at 20
°
C, and after some time, the temperature
of the copper cylinder and water reach a common temperature of 30
°
C. If the
specific heat capacity of copper is 390 J/kg·
°
C and that of water is 4200 J/kg·
°
C,
determine the mass of water in the container.
Solution
Step 1: Calculate the heat lost by the copper cylinder and the heat gained by
the water. The heat lost by the copper cylinder is equal to the heat gained by
the water:
mcCu∆TCu =mcwater∆Twater
where: mCu = 0.5 kg (mass of copper), cCu = 390 J/kg·
°
C (specific heat capacity
of copper), ∆TCu = 200C−30C= 170C(temperature change of copper),
cwater = 4200 J/kg·
°
C (specific heat capacity of water), ∆Twater = 30C−20C=
10C(temperature change of water).
Step 2: Substitute the values into the equation and solve for the mass of
water.
0.5×390 ×170 = m×4200 ×10
85,500 = 42,000m
m=85,500
42,000 ≈2.02 kg
Therefore, the mass of water in the container is approximately 2.02 kg.
17
Question 20
Question
A copper bar of length 2 m and cross-sectional area 0.004 m2is heated from
20
°
C to 75
°
C. If the thermal conductivity of copper is 400 W/(m·K) and the
specific heat capacity of copper is 390 J/(kg ·K), calculate the heat transfer in
the bar.
Solution
Step 1: Calculate the volume of the copper bar. The volume of the bar is given
by:
V= length ×cross-sectional area = 2 m ×0.004 m2= 0.008 m3
Step 2: Calculate the mass of the copper bar. The density of copper is
approximately 8900 kg/m3. Therefore, the mass of the copper bar can be found
as:
m= density ×volume = 8900 kg/m3×0.008 m3= 71.2kg
Step 3: Calculate the change in temperature. The change in temperature is
given by:
∆T=Tf−Ti= 75◦C−20◦C= 55◦C
Step 4: Calculate the heat transfer using the formula:
Q=mc∆T
where: - mis the mass of the copper bar, - cis the specific heat capacity of
copper, and - ∆Tis the change in temperature.
Plugging in the values:
Q= 71.2kg ×390 J/(kg ·K)×55 K= 1,394,200 J
Step 5: Convert the heat transfer to kilowatt-hours (kWh). Since 1 J=
2.78 ×10−7kW h, we have:
Q= 1,394,200 J×2.78 ×10−7kW h/J ≈0.387 kW h
Therefore, the heat transfer in the bar is approximately 0.387 kWh.
Question 21
Question
A copper vessel of mass 0.5 kg contains 2 kg of water at 20
°
C. If 0.5 kg of ice at
-10
°
C is added to the vessel, calculate the final temperature of the system when
thermal equilibrium is reached. Assume all heat losses to the surroundings are
negligible.
Given: Specific heat capacity of copper, cCu = 390 J/kg◦C Specific heat
capacity of water, cw= 4200 J/kg◦C Heat of fusion of ice, L= 3.36 ×105J/kg
18
Solution
Let the final temperature of the system be T
°
C.
Step 1: Calculate the heat lost by the copper vessel to reach the
final temperature T
°
CThe heat lost by the copper vessel is given by the
formula:
QCu =mCu ·cCu ·(T−20)
Substitute the given values:
QCu = 0.5 kg ×390 J/kg◦C×(T−20)
Step 2: Calculate the heat gained by the water The heat gained by
the water is given by the formula:
Qw=mw·cw·(T−20)
Where mw= 2 kg (mass of water).
Substitute the given values:
Qw= 2 kg ×4200 J/kg◦C×(T−20)
Step 3: Calculate the heat gained by the ice to reach 0
°
CThe heat
gained by the ice to reach 0
°
C is given by the formula:
Qice =mice ·L
Where mice = 0.5 kg (mass of ice).
Substitute the given values:
Qice = 0.5 kg ×3.36 ×105J/kg = 1.68 ×105J
Step 4: Write the heat balance equation At thermal equilibrium, the
total heat lost and gained must be equal.
QCu =Qw+Qice
Substitute the calculated values:
0.5·390 ·(T−20) = 2 ·4200 ·(T−20) + 1.68 ×105
Step 5: Solve for the final temperature TSolve the equation obtained
in Step 4 to find the final temperature T.
Question 22
Question
A steel rod of length 2.5 m and diameter 2 cm is initially at a temperature of
100◦C. It is then placed in a large body of ice at 0◦C. Assuming the specific heat
capacity of steel is 450 J/kg·K and its density is 7850 kg/m3, calculate the heat
transferred from the rod to the ice as the rod cools down to the temperature of
the ice.
19
Solution
Step 1: First, we need to find the mass of the steel rod. The volume of the steel
rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where r=d
2is the radius and his the length of the rod. Thus,
V=π2 cm
22
×2.5 m
V=π×1 cm2×2.5 m
V= 2.5πcm3≈7.85 cm3
Converting the volume to m3:
V= 7.85 cm3×1 m
100 cm3
V= 7.85 ×10−6m3
The mass of the steel rod can be calculated using the formula:
mass = density ×volume
mass = 7850 kg/m3×7.85 ×10−6m3
mass ≈0.0617 kg
Step 2: Next, we calculate the heat transferred from the rod to the ice using
the formula:
Q=mc∆T
where: - Qis the heat transferred, - mis the mass of the steel rod, - cis the
specific heat capacity of steel, and - ∆Tis the change in temperature.
Given that the final temperature of the rod is 0◦C and the initial temperature
is 100◦C, ∆T= 100 K.
Substitute the values into the formula:
Q= 0.0617 kg ×450 J/kg ·K×100 K
Q= 2781.75 J
Therefore, the heat transferred from the rod to the ice as it cools down to
the temperature of the ice is 2781.75 Joules.
20
Question 23
Question
A metal rod of length 50 cm and diameter 2 cm is initially at a temperature of
100
°
C. If the rod is placed in a room at 20
°
C, what is the rate at which heat is
transferred from the rod to the room? Assume the thermal conductivity of the
metal is 200 W/(m*K) and its specific heat capacity is 0.5 J/(g*K). (Hint: The
surface area of the rod can be calculated using the formula for the surface area
of a cylinder.)
Solution
Step 1: Calculate the surface area of the rod. The surface area Aof a cylinder
can be calculated using the formula:
A= 2πrh + 2πr2
where ris the radius and his the height of the cylinder (in this case, the length
of the rod). Given that the diameter is 2 cm, the radius ris 1 cm or 0.01 m and
the height his 50 cm or 0.5 m. Plugging in these values, we get:
A= 2π×0.01 ×0.5+2π×0.012
A= 0.02π+ 0.0002π
A= 0.0202π≈0.0635 m2
Step 2: Calculate the rate of heat transfer. The rate of heat transfer can be
calculated using the formula:
Q=kA∆T
L
where Qis the rate of heat transfer, kis the thermal conductivity, Ais the
surface area, ∆Tis the temperature difference, and Lis the thickness of the
material. Given that the metal rod is uniform in temperature, we can consider
Lto be negligible. The temperature difference ∆Tis 100−20 = 80
°
C. Plugging
in the values, we get:
Q= 200 ×0.0635 ×80
0.5
Q= 200 ×0.0635 ×160
Q= 2048 W
Therefore, the rate at which heat is transferred from the rod to the room is
2048 W.
21
Question 24
Question
A 2 kg block of copper at 200◦C is dropped into a 1 kg block of aluminum at
0◦C. If the specific heat capacities of copper and aluminum are 390 J/(kg·K)
and 900 J/(kg·K) respectively, determine the final temperature of the two-block
system when thermal equilibrium is reached. Assume no heat is lost to the
surroundings.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
aluminum block.
The heat lost by the copper block is given by
Qcopper =mcopper ·ccopper ·(Tf−Tinitial,copper)
where: - mcopper = 2 kg is the mass of copper, - ccopper = 390 J/(kg·K) is the
specific heat capacity of copper, - Tfis the final temperature of the two-block
system, - Tinitial,copper = 200 ◦C is the initial temperature of copper.
The heat gained by the aluminum block is given by
Qaluminum =maluminum ·caluminum ·(Tf−Tinitial,aluminum)
where: - maluminum = 1 kg is the mass of aluminum, - caluminum = 900 J/(kg·K)
is the specific heat capacity of aluminum, - Tinitial,aluminum = 0 ◦C is the initial
temperature of aluminum.
Step 2: Since the heat lost by the copper block should be equal to the heat
gained by the aluminum block, we have
mcopper ·ccopper ·(Tf−Tinitial,copper) = maluminum ·caluminum ·(Tf−Tinitial,aluminum)
Step 3: Solve for Tf:
2×390 ×(Tf−200) = 1 ×900 ×(Tf−0)
780 ×Tf−156000 = 900 ×Tf
120 ×Tf= 156000
Tf= 1300 ◦C
Therefore, the final temperature of the two-block system when thermal equi-
librium is reached is 1300◦C.
22
Question 25
Question
A copper rod of length 2.5 m and diameter 2.5 cm is initially at a temperature of
100
°
C. One end of the rod is immersed in a large ice bath at 0
°
C, while the other
end is in a steam chamber at 100
°
C. Assuming steady-state conditions, calculate
the rate at which heat is conducted along the rod. (Thermal conductivity of
copper = 400 W/mK)
Solution
Step 1: Calculate the cross-sectional area of the rod. The diameter of the rod
is 2.5 cm, so the radius ris given by:
r=2.5 cm
2= 0.0125 m
The cross-sectional area Ais given by:
A=πr2=π(0.0125)2= 4.91 ×10−4m2
Step 2: Calculate the temperature difference across the rod. The tempera-
ture difference ∆Tacross the rod is given by:
∆T= (100 −0)
°
C = 100
°
C
Step 3: Calculate the rate of heat conduction. The rate of heat conduction
Qalong the rod can be calculated using the formula:
Q=k·A·∆T
L
where kis the thermal conductivity of copper, Ais the cross-sectional area, ∆T
is the temperature difference, and Lis the length of the rod.
Substitute the given values into the formula:
Q=400 W/mK ·4.91 ×10−4m2·100 K
2.5 m
Q=196.4 W ·m−1·K·K
2.5 m = 78.56 W
Therefore, the rate at which heat is conducted along the rod is 78.56 W.
Question 26
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into a con-
tainer of water at 20
°
C. If the final temperature of the system is 30
°
C, calculate
the mass of water in the container. Assume no heat is lost to the surround-
ings. The specific heat capacity of copper is 0.385 J/g
°
C and that of water is
4.18 J/g
°
C.
23
Solution
Step 1: Calculate the heat lost by the copper block. Let’s denote: - mc= 500 g
as the mass of the copper block, - Tinitial = 100
°
C as the initial temperature
of the copper block, - Tfinal = 30
°
C as the final temperature of the system, -
ccopper = 0.385 J/g
°
C as the specific heat capacity of copper.
The heat lost by the copper block can be calculated using the formula:
Qcopper =mc·ccopper ·(Tinitial −Tfinal)
Substitute the given values into the formula:
Qcopper = 500 g ×0.385 J/g
°
C×(100
°
C−30
°
C)
Qcopper = 500 ×0.385 ×70
Qcopper = 13,475 J
Step 2: Calculate the heat gained by the water. Let’s denote: - mwas the
mass of water in the container, - cwater = 4.18 J/g
°
C as the specific heat capacity
of water.
The heat gained by the water can be calculated using the formula:
Qwater =mw·cwater ·(Tfinal −20
°
C)
Substitute the given values into the formula and the heat lost by the copper
block:
Qwater =mw×4.18 ×(30 −20)
Qwater = 10.54 ×mw
Since there is no heat lost to the surroundings,
Qwater =Qcopper
10.54 ×mw= 13,475
mw=13,475
10.54
mw≈1,279.77 g
Therefore, the mass of water in the container is approximately 1,279.77 g.
Question 27
Question
A metal bar of length 0.5 m and uniform cross-sectional area is subjected to a
temperature change from 20
°
C to 220
°
C. The bar is clamped at both ends and
prevented from expanding. If the Young’s modulus of the material is 2 ×1011
N/m
²
and the linear coefficient of thermal expansion is 1.2×10−5per
°
C,
calculate the stress developed in the bar due to the temperature change.
24
Solution
Step 1: Calculate the change in temperature. Given that the initial temperature
is 20
°
C and the final temperature is 220
°
C, the change in temperature is:
∆T=Tfinal −Tinitial = 220C−20C= 200C
Step 2: Calculate the thermal strain in the metal bar. The thermal strain is
given by:
ε=α·∆T
where αis the linear coefficient of thermal expansion and ∆Tis the change in
temperature. Substitute the values:
ε= 1.2×10−5×200 = 0.0024
Step 3: Calculate the stress developed in the bar. The stress developed in
the bar can be calculated using Hooke’s Law, which states:
σ=E·ε
where Eis the Young’s modulus and εis the strain. Substitute the values:
σ= 2 ×1011 ×0.0024 = 4.8×108N/m
²
Therefore, the stress developed in the metal bar due to the temperature
change is 4.8×108N/m
²
.
Question 28
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into a
container of water at 20
°
C. The mass of water is 2 kg. If the specific heat
capacity of copper is 0.385 J/g
°
C and that of water is 4.18 J/g
°
C, calculate the
final temperature of the system when thermal equilibrium is reached. Assume
no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper block. The formula to calculate
heat lost or gained is Q=mc∆T, where: - Qis the heat lost or gained - mis
the mass of the material - cis the specific heat capacity of the material - ∆Tis
the change in temperature
Given that the initial temperature of the copper block is 100
°
C and the final
temperature is T
°
C (let’s assume it decreases), we have ∆T= 100 −T.
Substitute the values into the formula: Qcopper = 0.5 kg ×0.385 J/g
°
C×
(100 −T)
25
Step 2: Calculate the heat gained by the water. Using the same formula as
above, we have: Qwater = 2 kg ×4.18 J/g
°
C×(T−20)
Step 3: Set up the energy conservation equation. Since there is no heat lost
to the surroundings, the heat lost by the copper block must be equal to the heat
gained by the water. Therefore: Qcopper =Qwater
Step 4: Solve for the final temperature. Set Qcopper =Qwater and solve for
T: 0.5×0.385 ×(100 −T)=2×4.18 ×(T−20)
Solve for Tto find the final temperature of the system.
Question 29
Question
A physics lab consists of a 2 kg block of copper with a specific heat capacity of
385 J
kg·◦Cand a 3 kg block of aluminum with a specific heat capacity of 900 J
kg·◦C.
Initially, the copper block is at a temperature of 150◦Cand the aluminum block
is at a temperature of 50◦C. The two blocks are brought into thermal contact
and allowed to reach thermal equilibrium. Assuming no heat is lost to the
surroundings, what will be the final temperature of the system? (Specific heat
capacity of both blocks remains constant over the temperature range.)
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
aluminum block when they reach thermal equilibrium.
The heat lost by the copper block Qcopper =mc∆T, where m= 2 kg,
c= 385 J
kg·◦C, and ∆T=Tfinal −150.
The heat gained by the aluminum block Qaluminum =mc∆T, where m= 3
kg, c= 900 J
kg·◦C, and ∆T=Tfinal −50.
Since no heat is lost to the surroundings, Qcopper =Qaluminum.
So, 2 ×385 ×(Tfinal −150) = 3 ×900 ×(Tfinal −50).
Step 2: Solve the equation to find the final temperature Tfinal.
Solving the equation above, we get:
770(Tfinal −150) = 2700(Tfinal −50)
Expanding and simplifying,
770Tfinal −115500 = 2700Tfinal −135000
1930Tfinal = 19500
Tfinal =19500
1930 ≈10.1◦C
Therefore, the final temperature of the system when the two blocks reach
thermal equilibrium will be approximately 10.1◦C.
26
Question 30
Question
A copper block with a mass of 0.5 kg is heated until its temperature increases
by 50
°
C. If the specific heat capacity of copper is 390 J/kg ·K, how much heat
is required?
Solution
Step 1: Identify the given values and the unknown.
Mass of the copper block, m= 0.5 kg
Temperature increase, ∆T= 50◦C
Specific heat capacity of copper, c= 390 J/kg ·K
Unknown: Heat required, Q
Step 2: Use the formula for heat energy Q=mc∆T.
Heat energy, Q= 0.5 kg ×390 J/kg ·K×50◦C
Step 3: Calculate the heat required.
Q= 0.5×390 ×50 = 9750 J
Therefore, the amount of heat required to increase the temperature of the
copper block by 50◦C is 9750 J.
Question 31
Question
At a temperature of 200K, a certain material has a specific heat capacity of
0.5 J/g·K. If 500 J of heat is applied to 50g of this material, what is the final
temperature of the material?
Solution
Step 1: Calculate the heat energy absorbed by the material. Step 2: Use the
formula Q=mc∆Tto find the final temperature.
Step 1: Calculate the heat energy absorbed by the material. The heat
energy Qabsorbed by the material is given as 500 J.
Step 2: Use the formula Q=mc∆Tto find the final temperature. We can
rearrange the formula to solve for the final temperature ∆T:
Q=mc∆T=⇒∆T=Q
mc
27
Substitute the given values Q= 500 J, m= 50 g (convert to kg by dividing by
1000), and c= 0.5 J/g·K into the formula:
∆T=500
50/1000 ×0.5=500
0.025 = 20000 K
Therefore, the final temperature of the material is 20000 K.
Question 32
Question
A copper rod of length 2.0 m is heated from 20
°
C to 80
°
C. If the coefficient of
linear expansion for copper is 3.9×10−5
°
C−1, find the change in length of the
rod.
Solution
Step 1: Calculate the initial length of the copper rod using the formula:
L0=L(1 + α∆T)
where: - L0is the initial length of the rod, - Lis the final length of the rod, - αis
the coefficient of linear expansion for copper, - ∆Tis the change in temperature.
Given that L= 2.0 m, α= 3.9×10−5
°
C−1, ∆T= 80
°
C, we can substitute
these values into the formula:
L0= 2.0(1 + 3.9×10−5×80)
L0= 2.0(1 + 0.00312)
L0= 2.0×1.00312
L0= 2.00624 m
Step 2: Calculate the change in length of the rod using the formula:
∆L=L−L0
where: - ∆Lis the change in length of the rod, - L0is the initial length of the
rod, - Lis the final length of the rod.
Substitute the values into the formula:
∆L= 2.0−2.00624
∆L=−0.00624 m
Therefore, the change in length of the copper rod is -0.00624 m.
28
Question 33
Question
A copper rod of length 2.0 m and cross-sectional area 2.0 cm2is used to transfer
heat from a furnace to a room at 20
°
C. The ends of the rod are maintained at
100
°
C and 50
°
C respectively. If the thermal conductivity of copper is 390 W/m·
K, calculate the rate at which heat is transferred along the rod.
Solution
Step 1: Calculate the temperature difference across the rod. Given: Initial
temperature of the room, T1= 20◦C Temperature at one end of the rod, T2=
100◦C Temperature at the other end of the rod, T3= 50◦C
The temperature difference across the rod is:
∆T=T2−T3= 100◦C−50◦C = 50◦C
Step 2: Calculate the rate of heat transfer. The rate of heat transfer along
the rod can be calculated using Fourier’s Law of heat conduction:
q=−kAdT
dx
where: q= rate of heat transfer per unit time, k= thermal conductivity of
copper, A= cross-sectional area of the rod, dT
dx = temperature gradient along
the rod.
Given: k= 390 W/m ·K, A= 2.0×10−4m2, ∆T= 50◦C = 50 K, Length
of the rod, L= 2.0 m
We need to find dT
dx in order to calculate q. To do this, we divide the
temperature difference by the length of the rod:
dT
dx =∆T
L=50
2.0= 25 K/m
Substitute the values into Fourier’s Law to find the rate of heat transfer:
q=−kAdT
dx =−390 ×2.0×10−4×25 = −1.95 W
Therefore, the rate at which heat is transferred along the rod is 1.95 W.
Question 34
Question
A copper rod of length 2 meters and cross-sectional area 0.01 m2has one end
at 100◦C and the other end at 200◦C. The coefficient of thermal conductivity
of copper is 370 W/mK. Determine the rate of heat conduction along the rod.
29
Solution
Step 1: Calculate the temperature gradient. Since the rod is made of copper,
we can use Fourier’s law of heat conduction:
Q=−kA∆T
L
where: Q= rate of heat conduction, k= coefficient of thermal conductivity of
copper, A= cross-sectional area of the rod, ∆T= change in temperature, and
L= length of the rod.
Given: k= 370 W/mK, A= 0.01 m2, ∆T= 200◦C−100◦C= 100◦C=
100K,L= 2 m.
Plug in the values and solve for Q:
Q=−370 ×0.01 ×100
2=−1.85 W
Therefore, the rate of heat conduction along the rod is 1.85 W.
Question 35
Question
A copper sphere initially at a temperature of 100
°
C is dropped into a calorimeter
containing 500 g of water at 20
°
C. The final temperature of the system is found
to be 22
°
C. If the specific heat capacity of water is 4.18 J/g◦C, find the mass of
the copper sphere. Assume that the heat exchange between the copper sphere
and water is the only process taking place, and all heat lost by the copper sphere
is gained by the water.
Solution
Step 1: Calculate the heat lost by the copper sphere and heat gained by the
water. The heat lost by the copper sphere is equal to the heat gained by the
water:
msphere ·csphere ·(Tf−Ti) = mwater ·cwater ·(Tf−Ti)
where msphere = mass of copper sphere (g), csphere = specific heat capacity
of copper (J/g
°
C), Ti= initial temperature of copper sphere (
°
C), Tf= final
temperature of both water and sphere (
°
C), mwater = mass of water (g), and
cwater = specific heat capacity of water (J/g
°
C).
Step 2: Substitute the given values into the equation. Substitute msphere =
m,csphere = 0.386 J/g◦C, Ti= 100◦C, Tf= 22◦C, mwater = 500 g, and cwater =
4.18 J/g◦C into the equation:
m·0.386 ·(22 −100) = 500 ·4.18 ·(22 −20)
30
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum block is equal to the heat gained by the water. Set both heat equations
equal to each other:
500 g ×cJ/g
°
C×78C= 500 g ×4.18 J/g
°
C×2C
Solve for the specific heat capacity cof aluminum.
Question 2
Question
A copper container holds 0.5 kg of water at 20◦C. How much heat is required
to raise the temperature of the water to 90◦C? Given: Specific heat capacity
of water = 4200 J/kg◦C, Specific heat capacity of copper = 380 J/kg◦C, and
neglect any heat loss to the surroundings.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
20◦C to 90◦C. Step 2: Calculate the heat required to raise the temperature of
the copper container from 20◦C to 90◦C. Step 3: Add the heat required for
the water and the heat required for the copper container to find the total heat
required.
Step 1: The heat required to raise the temperature of the water is given by
the formula:
Qwater =mc∆T
where: m= 0.5 kg (mass of water), c= 4200 J/kg◦C (specific heat capacity of
water), ∆T= 90◦C−20◦C = 70◦C (change in temperature).
Substitute the values into the formula:
Qwater = 0.5×4200 ×70
Qwater = 147000 J
So, the heat required to raise the temperature of the water is 147000 J.
Step 2: The heat required to raise the temperature of the copper container
is given by the formula:
Qcopper =mc∆T
where: m= mass of the copper container, c= 380 J/kg◦C (specific heat capacity
of copper), ∆T= 70◦C (change in temperature).
As the mass of the copper container is not given, we will assume it is very
large so that it effectively absorbs all the heat energy from the water. Therefore,
the heat required to raise the temperature of the copper container will be equal
to the heat lost by the water, which is 147000 J.
2
Step 3: The total heat required is the sum of the heat required for the
water and the copper container:
Qtotal =Qwater +Qcopper
Qtotal = 147000 + 147000
Qtotal = 294000 J
Therefore, the total heat required to raise the temperature of the water to
90◦C is 294000 J.
Question 3
Question
A metal bar of length Land thermal conductivity kis initially at a uniform
temperature of T0throughout. One end of the bar is then placed in an ice bath
at 0◦C and the other end is placed in a boiling water at 100◦C. The bar reaches
a steady state where the temperature at xmeters from the ice bath, T(x), is
given by the equation T(x) = (T0−0◦C) 1−x
L+ 0◦C. Calculate the rate of
heat transfer through the bar in terms of the given variables.
Solution
Step 1: To find the rate of heat transfer, we need to calculate the heat flux
density qat any point xalong the metal bar. The heat flux density is given by
Fourier’s law:
q=−kAdT
dx ,
where Ais the cross-sectional area of the metal bar.
Step 2: We can differentiate the equation for temperature T(x) to find dT
dx :
dT
dx =−T0−0◦C
L.
Step 3: Substituting the expression for dT
dx into Fourier’s law, we get
q=kAT0−0◦C
L.
Step 4: The heat transfer rate Qthrough the bar is given by
Q=qA =kA2T0−0◦C
L.
Therefore, the rate of heat transfer through the bar in terms of the given
variables is
Q=kA2T0−0◦C
L.
3
Question 4
Question
A copper rod of length 1 m has a temperature of 100◦C at one end and 0◦C
at the other end. If the rod is insulated along its length, what is the average
temperature of the rod?
Solution
To find the average temperature of the rod, we need to consider the temperature
distribution along the length of the rod.
Step 1: Calculate the change in temperature along the length of the rod.
The change in temperature along the length of the rod is given by the temper-
ature difference between the two ends: ∆T= 100◦C - 0◦C = 100◦C.
Step 2: Find the average temperature. The average temperature of the rod
can be calculated using the formula: Average temperature = Initial temperature
+∆T
2. Substitute the values: Average temperature = 0◦C + 100
2= 50◦C.
Therefore, the average temperature of the rod is 50◦C.
Question 5
Question
A copper cylinder of mass 500 g at a temperature of 100
°
C is dropped into 400
g of water at 20
°
C. If the final temperature of the mixture is 30
°
C, determine
the specific heat capacity of copper. Assume that all the heat lost by the copper
is gained by the water and that there are no heat losses to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper cylinder. Let the specific heat
capacity of copper be denoted by ccand the initial temperature of the copper
be denoted by Tc1. The heat lost by the copper cylinder can be calculated using
the formula:
Qlost =mc·cc·(Tfinal −Tc1)
where mcis the mass of the copper cylinder and Tfinal is the final temperature of
the mixture. Substitute the given values: mc= 500 g = 0.5 kg, Tc1= 100◦C =
373 K, and Tfinal = 30◦C = 303 K.
Qlost = 0.5·cc·(303 −373)
Step 2: Calculate the heat gained by the water. Let the specific heat capacity
of water be denoted by cwand the initial temperature of the water be denoted
by Tw1= 20◦C = 293 K. The heat gained by the water can be calculated using
the formula:
Qgain =mw·cw·(Tfinal −Tw1)
4
where mwis the mass of the water. Given that the mass of water, mw= 400 g =
0.4 kg, and the final temperature, Tfinal = 30◦C = 303 K.
Qgain = 0.4·cw·(303 −293)
Step 3: Since there is no heat loss to the surroundings, the heat lost by
the copper must be equal to the heat gained by the water. Equating the two
equations from Step 1 and Step 2:
0.5·cc·(303 −373) = 0.4·cw·(303 −293)
Step 4: Solve for cc(specific heat capacity of copper). Simplify and solve
the equation from Step 3 to find the specific heat capacity of copper, cc.
Question 6
Question
A piece of aluminum initially at 100◦C is dropped into a container of water at
20◦C. The mass of the aluminum is 500 g and the mass of the water is 2 kg.
Assuming no heat is lost to the surroundings, calculate the final equilibrium
temperature of the system. The specific heat capacity of aluminum is 0.9 J/g◦C
and that of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat gained or lost by the aluminum. The heat lost by
the aluminum is equal to the heat gained by the water, assuming no heat is lost
to the surroundings. Let Tfbe the final equilibrium temperature of the system.
The heat lost by the aluminum is given by:
Qalu =malu ·calu ·(Tf−Tinitial, alu)
where: malu = mass of aluminum = 0.5 kg = 500 g, calu = specific heat capacity
of aluminum = 0.9 J/g◦C, Tinitial, alu = initial temperature of aluminum =
100◦C, Tf= final equilibrium temperature of the system, Qalu = heat lost by
aluminum.
Substitute the given values into the formula:
Qalu = 500 g ·0.9 J/g◦C·(Tf−100◦C)
Qalu = 450 ·(Tf−100)
Step 2: Calculate the heat gained by the water. The heat gained by the
water is given by:
Qwater =mwater ·cwater ·(Tf−Tinitial, water)
5
where: mwater = mass of water = 2 kg = 2000 g, cwater = specific heat capacity
of water = 4.18 J/g◦C, Tinitial, water = initial temperature of water = 20◦C, Tf
= final equilibrium temperature of the system, Qwater = heat gained by water.
Substitute the given values into the formula:
Qwater = 2000 g ·4.18 J/g◦C·(Tf−20◦C)
Qwater = 8360 ·(Tf−20)
Step 3: Set up the equation. Since no heat is lost to the surroundings,
Qalu =Qwater.
450 ·(Tf−100) = 8360 ·(Tf−20)
Step 4: Solve for Tf.
450Tf−45000 = 8360Tf−167200
450Tf−8360Tf=−167200 + 45000
7910Tf=−122200
Tf=−122200
7910
Tf≈15.44◦C
Therefore, the final equilibrium temperature of the system is approximately
15.44◦C.
Question 7
Question
A block of copper with a mass of 500 g undergoes a temperature change of 50
°
C.
How much heat energy is required to cause this temperature change? (Specific
heat capacity of copper is 0.385 J/g
°
C)
Solution
Step 1: First, calculate the amount of heat energy using the formula:
Q=mc∆T
where: Q= heat energy (in Joules), m= mass (in grams) = 500 g, c=
specific heat capacity of copper = 0.385 J/g
°
C, ∆T= temperature change =
50
°
C.
Step 2: Substitute the values into the formula:
Q= (500 g)(0.385 J/g
°
C)(50C)
Step 3: Calculate the heat energy:
6
Q= 500 ×0.385 ×50
Q= 9625 J
Therefore, the amount of heat energy required to cause a temperature change
of 50
°
C in a 500 g block of copper is 9625 J.
Question 8
Question
A gas at pressure P1and volume V1is heated from an initial temperature T1to
a final temperature T2. The gas follows the equation of state P V =nRT where
nis the number of moles and Ris the ideal gas constant. If the final volume of
the gas is V2, find the work done on the gas during the process.
Solution
Step 1: First, let’s find the final pressure P2of the gas using the ideal gas law:
P1V1=nRT1and P2V2=nRT2
Since the number of moles nand the ideal gas constant Rare constant, we have:
P1V1
T1
=P2V2
T2
=⇒P2=P1V1T2
V2T1
Step 2: The work done on the gas during the process can be calculated using
the formula:
W=ZV2
V1
P dV
Substitute the expression for Pin terms of Vinto the integral:
W=ZV2
V1
P1V1T2
V2T1
dV =P1V1T2
V2T1ZV2
V1
dV
W=P1V1T2
T1
[ln(V2)−ln(V1)] = P1V1T2ln V2
V1
Therefore, the work done on the gas during the process is P1V1T2ln V2
V1.
Question 9
Question
A copper cylinder of length 50 cm and diameter 10 cm is heated from 20
°
C to
80
°
C. If the coefficient of linear expansion for copper is 1.7×10−5per degree
Celsius, determine the change in volume of the cylinder.
7
Solution
Step 1: Find the change in length of the cylinder. Step 2: Use the change in
length to find the change in volume.
Step 1: The formula for linear expansion is given by:
∆L=α·L·∆T
where: ∆L= change in length, α= coefficient of linear expansion, L= original
length, and ∆T= change in temperature.
Substitute the given values:
∆L= (1.7×10−5)·(50 cm) ·(80 −20)
°
C
∆L= 0.000085 cm ×60
°
C
∆L= 0.0051 cm
Step 2: The formula for volume expansion is given by:
∆V= 3α·V·∆T
where: ∆V= change in volume, α= coefficient of linear expansion, V= original
volume, and ∆T= change in temperature.
First, calculate the original volume of the cylinder:
V=πD2L
4=π(10 cm)2·50 cm
4
V= 1250πcm3
Now, substitute the values to find the change in volume:
∆V= 3(1.7×10−5)·(1250π)·60
∆V= 0.000051 ·7500π
∆V= 0.38πcm3
Therefore, the change in volume of the copper cylinder is 0.38πcm3.
Question 10
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 100
°
C. It is then placed in an ice bath at 0
°
C. If the specific heat capacity
of copper is 390 J/kg ·K and its density is 8900 kg/m3, calculate the amount of
heat transferred to or from the rod to bring it to thermal equilibrium with the
ice bath.
8
Solution
Step 1: Calculate the mass of the copper rod.
Given that the rod is cylindrical, we can find its volume using the formula for
the volume of a cylinder: V=πr2h, where ris the radius and his the height.
The mass of the rod can be calculated using the density formula: m=ρV ,
where ρis the density of copper.
Given: Diameter = 2 cm = 0.02 m (radius, r= 0.01 m), Height, h= 1.5 m,
Density, ρ= 8900 kg/m3.
Plugging in these values, we get:
V=π(0.01 m)2×1.5 m
V=π×10−4×1.5
V= 1.5π×10−4m3
m= 8900 kg/m3×1.5π×10−4m3
m= 1.3345 kg
Therefore, the mass of the copper rod is 1.3345 kg.
Step 2: Calculate the heat lost by the rod.
The heat lost by 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: Initial temperature, Tinitial = 100
°
C, Final temperature, Tfinal = 0
°
C,
Specific heat capacity, c= 390 J/kg ·K.
Plugging in these values, we get:
∆T=Tfinal −Tinitial = 0 −100 = −100 K
Q= 1.3345 ×390 ×(−100)
Q=−519465 J
Therefore, the heat lost by the rod is 519465 J.
Question 11
Question
A copper sphere with a radius of 10 cm is initially at a temperature of 100◦C. It
is then placed in a large vat of water at 20◦C. Assuming the only heat transfer is
between the sphere and the water, calculate the time it takes for the temperature
of the sphere to reach 25◦C. The specific heat capacity of copper is 0.385 J/g◦C,
and the density of copper is 8.96 g/cm3. Assume the sphere loses heat primarily
through radiation with a constant loss rate of 5W.
9
Solution
Step 1: Determine the mass of the copper sphere. Given that density of copper
is 8.96 g/cm3and the sphere has a radius of 10 cm, the volume of the sphere
can be calculated using the formula for the volume of a sphere:
V=4
3πr3
Substitute the given radius:
V=4
3π(10 cm)3=4000
3πcm3
Then, the mass of the sphere can be calculated using the density:
Mass = Density ×Volume = 8.96 ×4000
3πg≈37642 g
Step 2: Calculate the energy needed to change the temperature of the sphere
from 100◦C to 25◦C. The heat energy required to change the temperature can
be calculated as:
Q=mc∆T
Where: - m= 37642 g is the mass of the copper sphere, - c= 0.385 J/g◦C is
the specific heat capacity of copper, and - ∆T= 100 −25 = 75 ◦C is the change
in temperature. Substitute the values:
Q= 37642 ×0.385 ×75 = 1099957 J
Step 3: Calculate the time it takes for the temperature to decrease by 5◦C.
The rate at which the sphere loses heat energy through radiation is given as 5
W. This means:
P=∆Q
∆t
Where ∆Q= 1099957 J is the heat energy to be lost, and P= 5 W is the
power. Solve for ∆t:
5 = 1099957
∆t
∆t=1099957
5≈219991.4 seconds
Therefore, it will take approximately 219991.4 seconds for the temperature
of the copper sphere to decrease from 100◦C to 25◦C.
Question 12
Question
A 5 kg iron bar at 100◦C is dropped into 10 liters of water at 20◦C. Assuming
no heat is lost to the surroundings, calculate the final temperature of the iron
and water system. (Specific heat capacity of iron is 450 J/kg◦C, specific heat
capacity of water is 4186 J/kg◦C, and the density of water is 1000 kg/m3.)
10
Solution
Step 1: Find the heat gained by the iron bar as it cools down from 100◦C to
the final temperature. The formula for heat transfer is given by:
Q=mc∆T
where: Q= heat transferred, m= mass, c= specific heat capacity, and ∆T=
change in temperature.
For the iron bar: m= 5 kg, ciron = 450 J/kg◦C, Initial temperature Ti= 100
◦C, Final temperature Tf=T◦C, and ∆T=T−Ti.
The heat lost by the iron bar is equal to the heat gained by the water,
therefore:
mciron(T−Ti) = mcwater(T−Twater)
Substitute in the given values:
5×450(T−100) = 10 ×4186(T−20)
Step 2: Solve for Tto find the final temperature of the iron and water
system. Expand and solve the equation:
2250T−225000 = 41860T−837200
39610T= 612200
T≈15.47◦C
Therefore, the final temperature of the iron and water system is approxi-
mately 15.47◦C.
Question 13
Question
A metal rod of length Land thermal conductivity kis heated at one end to a
temperature T1and kept at a constant temperature of T1while the other end
is kept at a constant temperature of T2such that (T1> T2). The rod loses heat
to the surroundings at a rate of Qper unit time. If the rod is in a steady state,
determine the temperature distribution along the rod and express the heat flux
qin terms of T1,T2,L, and Q.
Solution
Step 1: Set up the heat conduction equation
d2T
dx2= 0
11
where T(x) is the temperature distribution along the rod and xis the distance
from the end at temperature T1.
Step 2: Integrate the heat conduction equation twice Integrating once, we
get dT
dx =c1
where c1is an arbitrary constant.
Integrating again, we get
T(x) = c1x+c2
where c2is another arbitrary constant.
Step 3: Apply boundary conditions At x= 0, T=T1. So,
T(0) = c2=T1
At x=L,T=T2. So,
T(L) = c1L+T1=T2
c1=T2−T1
L
Step 4: Express the heat flux The heat flux qis given by Fourier’s law as
q=−kdT
dx
Plugging in our expression for dT
dx ,
q=−kc1=kT1−T2
L
Therefore, the temperature distribution along the rod is given by
T(x) = T2−T1
Lx+T1
and the heat flux qis given by
q=kT1−T2
L
Question 14
Question
A metal cylinder of mass 2 kg and specific heat capacity 420 J/(kg·K) is heated
to a temperature of 100◦C. It is then placed in a calorimeter containing 1 kg of
water initially at 20◦C. If the final temperature of the system is 30◦C and no
heat is lost to the surroundings, find the specific heat capacity of the calorimeter.
12
Solution
Step 1: Calculate the heat absorbed by the metal cylinder.
Qmetal =mc∆T
= 2 kg ×420 J/(kg·K) ×(100 −30) K
= 2 ×420 ×70
= 58800 J
Step 2: Calculate the heat lost by the metal cylinder in order to raise the
temperature of the water and calorimeter to the final temperature of 30◦C.
Qmetal =Qwater +Qcalorimeter
58800 = (1 kg ×cwater ×(30 −20) K) + (mcalorimeter ×ccalorimeter ×(30 −20) K)
= 10cwater + 10ccalorimeter
Step 3: From the information given, we know that the specific heat capacity
of water is 4186 J/(kg·K). Substituting this into the equation above, we can
solve for the specific heat capacity of the calorimeter.
58800 = 10(4186) + 10ccalorimeter
10ccalorimeter = 58800 −10(4186)
10ccalorimeter = 50540
ccalorimeter =50540
10
ccalorimeter = 5054 J/(kg·K)
Therefore, the specific heat capacity of the calorimeter is 5054 J/(kg·K).
Question 15
Question
A metal rod of length 2 m and diameter 5 cm is heated from 20 ◦C to 80 ◦C.
If the coefficient of linear expansion for the metal is 1.8×10−5◦C−1and the
specific heat capacity is 450 J/kg ·◦C, calculate the amount of heat transferred
to the rod.
Solution
Step 1: Determine the change in length of the metal rod due to heating. Given:
Initial temperature, Ti= 20 ◦C Final temperature, Tf= 80 ◦C Coefficient of
linear expansion, α= 1.8×10−5◦C−1Original length, L0= 2 m
The change in length, ∆L, of the rod can be calculated using the formula:
∆L=L0·α·(Tf−Ti)
13
Substitute the given values:
∆L= 2 ·1.8×10−5·(80 −20) = 0.000 12 m
Step 2: Calculate the volume of the metal rod. The radius, r, of the rod is
half of the diameter:
r=diameter
2=5
2cm = 0.025 m
Volume of the rod, V=πr2L
V=π(0.025)2·2 = π·0.00125 ·2 = 0.0025 m3
Step 3: Calculate the mass of the metal rod. Given: Density of the metal
= 8000 kg/m3Mass, m= density ×volume
m= 8000 ×0.0025 = 20 kg
Step 4: Determine the amount of heat transferred to the rod. The heat
transferred, Q, is given by the formula:
Q=mc∆T
where cis the specific heat capacity and ∆Tis the temperature change.
Q= 20 ×450 ×(80 −20) = 720 000 J
Therefore, the amount of heat transferred to the rod is 720 000 J.
Question 16
Question
A copper rod of length 2 m and diameter 1 cm is heated from 20◦C to 100◦C.
Given that the coefficient of linear expansion of copper is 1.7×10−5C−1, deter-
mine the change in the volume of the rod.
Solution
Step 1: Find the initial volume of the rod. The initial volume of the copper rod
can be calculated as the volume of a cylinder:
Vinitial =πr2h
where ris the initial radius, his the length of the rod, and Vinitial is the initial
volume.
Given that the initial diameter is 1 cm, the initial radius ris 0.5 cm =
0.005 m, and the length his 2 m, we can calculate:
Vinitial =π(0.005 m)2×2 m
14
Vinitial =π×0.000025 ×2
Vinitial = 0.00005πm3
Step 2: Find the final volume of the rod. The final volume of the rod is
obtained by considering the increase in length of the rod due to heating. The
change in length can be calculated using the formula:
∆L=αL0∆T
where αis the coefficient of linear expansion, L0is the initial length, and ∆T
is the change in temperature.
Given that α= 1.7×10−5C−1,L0= 2 m, and ∆T= 80◦C (as the rod is
heated from 20◦C to 100◦C), we can find:
∆L= 1.7×10−5×2×80
∆L= 0.00272 m
The final volume can now be calculated using the change in length:
Vfinal =π(0.005 m)2×(2 + 0.00272) m
Vfinal =π×0.000025 ×2.00272
Vfinal ≈0.0000501πm3
Step 3: Determine the change in volume. The change in volume (∆V) is
calculated by finding the difference between the final volume and the initial
volume:
∆V=Vfinal −Vinitial
∆V= 0.0000501π−0.00005π
∆V≈0.0000001πm3
Therefore, the change in volume of the copper rod when heated from 20◦C
to 100◦C is approximately 0.0000001πm3.
Question 17
Question
A copper block with a mass of 500 g is heated from 25
°
C to 75
°
C. The specific
heat capacity of copper is 0.386 J/g
°
C. How much heat energy was supplied to
the copper block?
15
Solution
Step 1: Calculate the change in temperature of the copper block. Given: Initial
temperature, T1= 25CFinal temperature, T2= 75C
The change in temperature, ∆T=T2−T1= 75C−25C= 50C.
Step 2: Calculate the heat energy supplied to the copper block. The formula
for calculating heat energy is:
Q=mc∆T
where: Q= heat energy supplied (in joules) m= mass of the object (in grams)
c= specific heat capacity (in J/g
°
C) ∆T= change in temperature (in
°
C)
Given: m= 500 g c= 0.386 J/g
°
C ∆T= 50
°
C
Substitute the values into the formula:
Q= 500 g ×0.386 J/g
°
C×50
°
C
Q= 500 ×0.386 ×50
Q= 9650 joules
Therefore, the amount of heat energy supplied to the copper block is 9650
joules.
Question 18
Question
A metal rod of length 1 m has one end kept at a temperature of 100
°
C and the
other end at 0
°
C. The thermal conductivity of the rod is 200 W/mK and its
area of cross-section is 10−4m
²
. Calculate the rate of transfer of heat along the
rod.
Solution
Step 1: Calculate the temperature gradient along the rod. Given: - Length of
the rod, L= 1 m - Area of cross-section, A= 10−4m
²
- Temperature at one
end, T1= 100C= 373 K - Temperature at the other end, T2= 0C= 273 K -
Thermal conductivity, k= 200 W/mK
The temperature gradient dT
dx along the rod is given by:
dT
dx =T1−T2
L
dT
dx =373 −273
1
16
dT
dx = 100K/m
Step 2: Calculate the rate of transfer of heat along the rod. The rate of
transfer of heat, dQ
dt , is given by Fourier’s Law:
dQ
dt =−kAdT
dx
dQ
dt =−200 ×10−4×100
dQ
dt =−2 W
Therefore, the rate of transfer of heat along the rod is 2 W.
Question 19
Question
A copper cylinder of mass 0.5 kg is initially at a temperature of 200
°
C. It is
placed in a container of water at 20
°
C, and after some time, the temperature
of the copper cylinder and water reach a common temperature of 30
°
C. If the
specific heat capacity of copper is 390 J/kg·
°
C and that of water is 4200 J/kg·
°
C,
determine the mass of water in the container.
Solution
Step 1: Calculate the heat lost by the copper cylinder and the heat gained by
the water. The heat lost by the copper cylinder is equal to the heat gained by
the water:
mcCu∆TCu =mcwater∆Twater
where: mCu = 0.5 kg (mass of copper), cCu = 390 J/kg·
°
C (specific heat capacity
of copper), ∆TCu = 200C−30C= 170C(temperature change of copper),
cwater = 4200 J/kg·
°
C (specific heat capacity of water), ∆Twater = 30C−20C=
10C(temperature change of water).
Step 2: Substitute the values into the equation and solve for the mass of
water.
0.5×390 ×170 = m×4200 ×10
85,500 = 42,000m
m=85,500
42,000 ≈2.02 kg
Therefore, the mass of water in the container is approximately 2.02 kg.
17
Question 20
Question
A copper bar of length 2 m and cross-sectional area 0.004 m2is heated from
20
°
C to 75
°
C. If the thermal conductivity of copper is 400 W/(m·K) and the
specific heat capacity of copper is 390 J/(kg ·K), calculate the heat transfer in
the bar.
Solution
Step 1: Calculate the volume of the copper bar. The volume of the bar is given
by:
V= length ×cross-sectional area = 2 m ×0.004 m2= 0.008 m3
Step 2: Calculate the mass of the copper bar. The density of copper is
approximately 8900 kg/m3. Therefore, the mass of the copper bar can be found
as:
m= density ×volume = 8900 kg/m3×0.008 m3= 71.2kg
Step 3: Calculate the change in temperature. The change in temperature is
given by:
∆T=Tf−Ti= 75◦C−20◦C= 55◦C
Step 4: Calculate the heat transfer using the formula:
Q=mc∆T
where: - mis the mass of the copper bar, - cis the specific heat capacity of
copper, and - ∆Tis the change in temperature.
Plugging in the values:
Q= 71.2kg ×390 J/(kg ·K)×55 K= 1,394,200 J
Step 5: Convert the heat transfer to kilowatt-hours (kWh). Since 1 J=
2.78 ×10−7kW h, we have:
Q= 1,394,200 J×2.78 ×10−7kW h/J ≈0.387 kW h
Therefore, the heat transfer in the bar is approximately 0.387 kWh.
Question 21
Question
A copper vessel of mass 0.5 kg contains 2 kg of water at 20
°
C. If 0.5 kg of ice at
-10
°
C is added to the vessel, calculate the final temperature of the system when
thermal equilibrium is reached. Assume all heat losses to the surroundings are
negligible.
Given: Specific heat capacity of copper, cCu = 390 J/kg◦C Specific heat
capacity of water, cw= 4200 J/kg◦C Heat of fusion of ice, L= 3.36 ×105J/kg
18
Solution
Let the final temperature of the system be T
°
C.
Step 1: Calculate the heat lost by the copper vessel to reach the
final temperature T
°
CThe heat lost by the copper vessel is given by the
formula:
QCu =mCu ·cCu ·(T−20)
Substitute the given values:
QCu = 0.5 kg ×390 J/kg◦C×(T−20)
Step 2: Calculate the heat gained by the water The heat gained by
the water is given by the formula:
Qw=mw·cw·(T−20)
Where mw= 2 kg (mass of water).
Substitute the given values:
Qw= 2 kg ×4200 J/kg◦C×(T−20)
Step 3: Calculate the heat gained by the ice to reach 0
°
CThe heat
gained by the ice to reach 0
°
C is given by the formula:
Qice =mice ·L
Where mice = 0.5 kg (mass of ice).
Substitute the given values:
Qice = 0.5 kg ×3.36 ×105J/kg = 1.68 ×105J
Step 4: Write the heat balance equation At thermal equilibrium, the
total heat lost and gained must be equal.
QCu =Qw+Qice
Substitute the calculated values:
0.5·390 ·(T−20) = 2 ·4200 ·(T−20) + 1.68 ×105
Step 5: Solve for the final temperature TSolve the equation obtained
in Step 4 to find the final temperature T.
Question 22
Question
A steel rod of length 2.5 m and diameter 2 cm is initially at a temperature of
100◦C. It is then placed in a large body of ice at 0◦C. Assuming the specific heat
capacity of steel is 450 J/kg·K and its density is 7850 kg/m3, calculate the heat
transferred from the rod to the ice as the rod cools down to the temperature of
the ice.
19
Solution
Step 1: First, we need to find the mass of the steel rod. The volume of the steel
rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where r=d
2is the radius and his the length of the rod. Thus,
V=π2 cm
22
×2.5 m
V=π×1 cm2×2.5 m
V= 2.5πcm3≈7.85 cm3
Converting the volume to m3:
V= 7.85 cm3×1 m
100 cm3
V= 7.85 ×10−6m3
The mass of the steel rod can be calculated using the formula:
mass = density ×volume
mass = 7850 kg/m3×7.85 ×10−6m3
mass ≈0.0617 kg
Step 2: Next, we calculate the heat transferred from the rod to the ice using
the formula:
Q=mc∆T
where: - Qis the heat transferred, - mis the mass of the steel rod, - cis the
specific heat capacity of steel, and - ∆Tis the change in temperature.
Given that the final temperature of the rod is 0◦C and the initial temperature
is 100◦C, ∆T= 100 K.
Substitute the values into the formula:
Q= 0.0617 kg ×450 J/kg ·K×100 K
Q= 2781.75 J
Therefore, the heat transferred from the rod to the ice as it cools down to
the temperature of the ice is 2781.75 Joules.
20
Question 23
Question
A metal rod of length 50 cm and diameter 2 cm is initially at a temperature of
100
°
C. If the rod is placed in a room at 20
°
C, what is the rate at which heat is
transferred from the rod to the room? Assume the thermal conductivity of the
metal is 200 W/(m*K) and its specific heat capacity is 0.5 J/(g*K). (Hint: The
surface area of the rod can be calculated using the formula for the surface area
of a cylinder.)
Solution
Step 1: Calculate the surface area of the rod. The surface area Aof a cylinder
can be calculated using the formula:
A= 2πrh + 2πr2
where ris the radius and his the height of the cylinder (in this case, the length
of the rod). Given that the diameter is 2 cm, the radius ris 1 cm or 0.01 m and
the height his 50 cm or 0.5 m. Plugging in these values, we get:
A= 2π×0.01 ×0.5+2π×0.012
A= 0.02π+ 0.0002π
A= 0.0202π≈0.0635 m2
Step 2: Calculate the rate of heat transfer. The rate of heat transfer can be
calculated using the formula:
Q=kA∆T
L
where Qis the rate of heat transfer, kis the thermal conductivity, Ais the
surface area, ∆Tis the temperature difference, and Lis the thickness of the
material. Given that the metal rod is uniform in temperature, we can consider
Lto be negligible. The temperature difference ∆Tis 100−20 = 80
°
C. Plugging
in the values, we get:
Q= 200 ×0.0635 ×80
0.5
Q= 200 ×0.0635 ×160
Q= 2048 W
Therefore, the rate at which heat is transferred from the rod to the room is
2048 W.
21
Question 24
Question
A 2 kg block of copper at 200◦C is dropped into a 1 kg block of aluminum at
0◦C. If the specific heat capacities of copper and aluminum are 390 J/(kg·K)
and 900 J/(kg·K) respectively, determine the final temperature of the two-block
system when thermal equilibrium is reached. Assume no heat is lost to the
surroundings.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
aluminum block.
The heat lost by the copper block is given by
Qcopper =mcopper ·ccopper ·(Tf−Tinitial,copper)
where: - mcopper = 2 kg is the mass of copper, - ccopper = 390 J/(kg·K) is the
specific heat capacity of copper, - Tfis the final temperature of the two-block
system, - Tinitial,copper = 200 ◦C is the initial temperature of copper.
The heat gained by the aluminum block is given by
Qaluminum =maluminum ·caluminum ·(Tf−Tinitial,aluminum)
where: - maluminum = 1 kg is the mass of aluminum, - caluminum = 900 J/(kg·K)
is the specific heat capacity of aluminum, - Tinitial,aluminum = 0 ◦C is the initial
temperature of aluminum.
Step 2: Since the heat lost by the copper block should be equal to the heat
gained by the aluminum block, we have
mcopper ·ccopper ·(Tf−Tinitial,copper) = maluminum ·caluminum ·(Tf−Tinitial,aluminum)
Step 3: Solve for Tf:
2×390 ×(Tf−200) = 1 ×900 ×(Tf−0)
780 ×Tf−156000 = 900 ×Tf
120 ×Tf= 156000
Tf= 1300 ◦C
Therefore, the final temperature of the two-block system when thermal equi-
librium is reached is 1300◦C.
22
Question 25
Question
A copper rod of length 2.5 m and diameter 2.5 cm is initially at a temperature of
100
°
C. One end of the rod is immersed in a large ice bath at 0
°
C, while the other
end is in a steam chamber at 100
°
C. Assuming steady-state conditions, calculate
the rate at which heat is conducted along the rod. (Thermal conductivity of
copper = 400 W/mK)
Solution
Step 1: Calculate the cross-sectional area of the rod. The diameter of the rod
is 2.5 cm, so the radius ris given by:
r=2.5 cm
2= 0.0125 m
The cross-sectional area Ais given by:
A=πr2=π(0.0125)2= 4.91 ×10−4m2
Step 2: Calculate the temperature difference across the rod. The tempera-
ture difference ∆Tacross the rod is given by:
∆T= (100 −0)
°
C = 100
°
C
Step 3: Calculate the rate of heat conduction. The rate of heat conduction
Qalong the rod can be calculated using the formula:
Q=k·A·∆T
L
where kis the thermal conductivity of copper, Ais the cross-sectional area, ∆T
is the temperature difference, and Lis the length of the rod.
Substitute the given values into the formula:
Q=400 W/mK ·4.91 ×10−4m2·100 K
2.5 m
Q=196.4 W ·m−1·K·K
2.5 m = 78.56 W
Therefore, the rate at which heat is conducted along the rod is 78.56 W.
Question 26
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into a con-
tainer of water at 20
°
C. If the final temperature of the system is 30
°
C, calculate
the mass of water in the container. Assume no heat is lost to the surround-
ings. The specific heat capacity of copper is 0.385 J/g
°
C and that of water is
4.18 J/g
°
C.
23
Solution
Step 1: Calculate the heat lost by the copper block. Let’s denote: - mc= 500 g
as the mass of the copper block, - Tinitial = 100
°
C as the initial temperature
of the copper block, - Tfinal = 30
°
C as the final temperature of the system, -
ccopper = 0.385 J/g
°
C as the specific heat capacity of copper.
The heat lost by the copper block can be calculated using the formula:
Qcopper =mc·ccopper ·(Tinitial −Tfinal)
Substitute the given values into the formula:
Qcopper = 500 g ×0.385 J/g
°
C×(100
°
C−30
°
C)
Qcopper = 500 ×0.385 ×70
Qcopper = 13,475 J
Step 2: Calculate the heat gained by the water. Let’s denote: - mwas the
mass of water in the container, - cwater = 4.18 J/g
°
C as the specific heat capacity
of water.
The heat gained by the water can be calculated using the formula:
Qwater =mw·cwater ·(Tfinal −20
°
C)
Substitute the given values into the formula and the heat lost by the copper
block:
Qwater =mw×4.18 ×(30 −20)
Qwater = 10.54 ×mw
Since there is no heat lost to the surroundings,
Qwater =Qcopper
10.54 ×mw= 13,475
mw=13,475
10.54
mw≈1,279.77 g
Therefore, the mass of water in the container is approximately 1,279.77 g.
Question 27
Question
A metal bar of length 0.5 m and uniform cross-sectional area is subjected to a
temperature change from 20
°
C to 220
°
C. The bar is clamped at both ends and
prevented from expanding. If the Young’s modulus of the material is 2 ×1011
N/m
²
and the linear coefficient of thermal expansion is 1.2×10−5per
°
C,
calculate the stress developed in the bar due to the temperature change.
24
Solution
Step 1: Calculate the change in temperature. Given that the initial temperature
is 20
°
C and the final temperature is 220
°
C, the change in temperature is:
∆T=Tfinal −Tinitial = 220C−20C= 200C
Step 2: Calculate the thermal strain in the metal bar. The thermal strain is
given by:
ε=α·∆T
where αis the linear coefficient of thermal expansion and ∆Tis the change in
temperature. Substitute the values:
ε= 1.2×10−5×200 = 0.0024
Step 3: Calculate the stress developed in the bar. The stress developed in
the bar can be calculated using Hooke’s Law, which states:
σ=E·ε
where Eis the Young’s modulus and εis the strain. Substitute the values:
σ= 2 ×1011 ×0.0024 = 4.8×108N/m
²
Therefore, the stress developed in the metal bar due to the temperature
change is 4.8×108N/m
²
.
Question 28
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into a
container of water at 20
°
C. The mass of water is 2 kg. If the specific heat
capacity of copper is 0.385 J/g
°
C and that of water is 4.18 J/g
°
C, calculate the
final temperature of the system when thermal equilibrium is reached. Assume
no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper block. The formula to calculate
heat lost or gained is Q=mc∆T, where: - Qis the heat lost or gained - mis
the mass of the material - cis the specific heat capacity of the material - ∆Tis
the change in temperature
Given that the initial temperature of the copper block is 100
°
C and the final
temperature is T
°
C (let’s assume it decreases), we have ∆T= 100 −T.
Substitute the values into the formula: Qcopper = 0.5 kg ×0.385 J/g
°
C×
(100 −T)
25
Step 2: Calculate the heat gained by the water. Using the same formula as
above, we have: Qwater = 2 kg ×4.18 J/g
°
C×(T−20)
Step 3: Set up the energy conservation equation. Since there is no heat lost
to the surroundings, the heat lost by the copper block must be equal to the heat
gained by the water. Therefore: Qcopper =Qwater
Step 4: Solve for the final temperature. Set Qcopper =Qwater and solve for
T: 0.5×0.385 ×(100 −T)=2×4.18 ×(T−20)
Solve for Tto find the final temperature of the system.
Question 29
Question
A physics lab consists of a 2 kg block of copper with a specific heat capacity of
385 J
kg·◦Cand a 3 kg block of aluminum with a specific heat capacity of 900 J
kg·◦C.
Initially, the copper block is at a temperature of 150◦Cand the aluminum block
is at a temperature of 50◦C. The two blocks are brought into thermal contact
and allowed to reach thermal equilibrium. Assuming no heat is lost to the
surroundings, what will be the final temperature of the system? (Specific heat
capacity of both blocks remains constant over the temperature range.)
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
aluminum block when they reach thermal equilibrium.
The heat lost by the copper block Qcopper =mc∆T, where m= 2 kg,
c= 385 J
kg·◦C, and ∆T=Tfinal −150.
The heat gained by the aluminum block Qaluminum =mc∆T, where m= 3
kg, c= 900 J
kg·◦C, and ∆T=Tfinal −50.
Since no heat is lost to the surroundings, Qcopper =Qaluminum.
So, 2 ×385 ×(Tfinal −150) = 3 ×900 ×(Tfinal −50).
Step 2: Solve the equation to find the final temperature Tfinal.
Solving the equation above, we get:
770(Tfinal −150) = 2700(Tfinal −50)
Expanding and simplifying,
770Tfinal −115500 = 2700Tfinal −135000
1930Tfinal = 19500
Tfinal =19500
1930 ≈10.1◦C
Therefore, the final temperature of the system when the two blocks reach
thermal equilibrium will be approximately 10.1◦C.
26
Question 30
Question
A copper block with a mass of 0.5 kg is heated until its temperature increases
by 50
°
C. If the specific heat capacity of copper is 390 J/kg ·K, how much heat
is required?
Solution
Step 1: Identify the given values and the unknown.
Mass of the copper block, m= 0.5 kg
Temperature increase, ∆T= 50◦C
Specific heat capacity of copper, c= 390 J/kg ·K
Unknown: Heat required, Q
Step 2: Use the formula for heat energy Q=mc∆T.
Heat energy, Q= 0.5 kg ×390 J/kg ·K×50◦C
Step 3: Calculate the heat required.
Q= 0.5×390 ×50 = 9750 J
Therefore, the amount of heat required to increase the temperature of the
copper block by 50◦C is 9750 J.
Question 31
Question
At a temperature of 200K, a certain material has a specific heat capacity of
0.5 J/g·K. If 500 J of heat is applied to 50g of this material, what is the final
temperature of the material?
Solution
Step 1: Calculate the heat energy absorbed by the material. Step 2: Use the
formula Q=mc∆Tto find the final temperature.
Step 1: Calculate the heat energy absorbed by the material. The heat
energy Qabsorbed by the material is given as 500 J.
Step 2: Use the formula Q=mc∆Tto find the final temperature. We can
rearrange the formula to solve for the final temperature ∆T:
Q=mc∆T=⇒∆T=Q
mc
27
Substitute the given values Q= 500 J, m= 50 g (convert to kg by dividing by
1000), and c= 0.5 J/g·K into the formula:
∆T=500
50/1000 ×0.5=500
0.025 = 20000 K
Therefore, the final temperature of the material is 20000 K.
Question 32
Question
A copper rod of length 2.0 m is heated from 20
°
C to 80
°
C. If the coefficient of
linear expansion for copper is 3.9×10−5
°
C−1, find the change in length of the
rod.
Solution
Step 1: Calculate the initial length of the copper rod using the formula:
L0=L(1 + α∆T)
where: - L0is the initial length of the rod, - Lis the final length of the rod, - αis
the coefficient of linear expansion for copper, - ∆Tis the change in temperature.
Given that L= 2.0 m, α= 3.9×10−5
°
C−1, ∆T= 80
°
C, we can substitute
these values into the formula:
L0= 2.0(1 + 3.9×10−5×80)
L0= 2.0(1 + 0.00312)
L0= 2.0×1.00312
L0= 2.00624 m
Step 2: Calculate the change in length of the rod using the formula:
∆L=L−L0
where: - ∆Lis the change in length of the rod, - L0is the initial length of the
rod, - Lis the final length of the rod.
Substitute the values into the formula:
∆L= 2.0−2.00624
∆L=−0.00624 m
Therefore, the change in length of the copper rod is -0.00624 m.
28
Question 33
Question
A copper rod of length 2.0 m and cross-sectional area 2.0 cm2is used to transfer
heat from a furnace to a room at 20
°
C. The ends of the rod are maintained at
100
°
C and 50
°
C respectively. If the thermal conductivity of copper is 390 W/m·
K, calculate the rate at which heat is transferred along the rod.
Solution
Step 1: Calculate the temperature difference across the rod. Given: Initial
temperature of the room, T1= 20◦C Temperature at one end of the rod, T2=
100◦C Temperature at the other end of the rod, T3= 50◦C
The temperature difference across the rod is:
∆T=T2−T3= 100◦C−50◦C = 50◦C
Step 2: Calculate the rate of heat transfer. The rate of heat transfer along
the rod can be calculated using Fourier’s Law of heat conduction:
q=−kAdT
dx
where: q= rate of heat transfer per unit time, k= thermal conductivity of
copper, A= cross-sectional area of the rod, dT
dx = temperature gradient along
the rod.
Given: k= 390 W/m ·K, A= 2.0×10−4m2, ∆T= 50◦C = 50 K, Length
of the rod, L= 2.0 m
We need to find dT
dx in order to calculate q. To do this, we divide the
temperature difference by the length of the rod:
dT
dx =∆T
L=50
2.0= 25 K/m
Substitute the values into Fourier’s Law to find the rate of heat transfer:
q=−kAdT
dx =−390 ×2.0×10−4×25 = −1.95 W
Therefore, the rate at which heat is transferred along the rod is 1.95 W.
Question 34
Question
A copper rod of length 2 meters and cross-sectional area 0.01 m2has one end
at 100◦C and the other end at 200◦C. The coefficient of thermal conductivity
of copper is 370 W/mK. Determine the rate of heat conduction along the rod.
29
Solution
Step 1: Calculate the temperature gradient. Since the rod is made of copper,
we can use Fourier’s law of heat conduction:
Q=−kA∆T
L
where: Q= rate of heat conduction, k= coefficient of thermal conductivity of
copper, A= cross-sectional area of the rod, ∆T= change in temperature, and
L= length of the rod.
Given: k= 370 W/mK, A= 0.01 m2, ∆T= 200◦C−100◦C= 100◦C=
100K,L= 2 m.
Plug in the values and solve for Q:
Q=−370 ×0.01 ×100
2=−1.85 W
Therefore, the rate of heat conduction along the rod is 1.85 W.
Question 35
Question
A copper sphere initially at a temperature of 100
°
C is dropped into a calorimeter
containing 500 g of water at 20
°
C. The final temperature of the system is found
to be 22
°
C. If the specific heat capacity of water is 4.18 J/g◦C, find the mass of
the copper sphere. Assume that the heat exchange between the copper sphere
and water is the only process taking place, and all heat lost by the copper sphere
is gained by the water.
Solution
Step 1: Calculate the heat lost by the copper sphere and heat gained by the
water. The heat lost by the copper sphere is equal to the heat gained by the
water:
msphere ·csphere ·(Tf−Ti) = mwater ·cwater ·(Tf−Ti)
where msphere = mass of copper sphere (g), csphere = specific heat capacity
of copper (J/g
°
C), Ti= initial temperature of copper sphere (
°
C), Tf= final
temperature of both water and sphere (
°
C), mwater = mass of water (g), and
cwater = specific heat capacity of water (J/g
°
C).
Step 2: Substitute the given values into the equation. Substitute msphere =
m,csphere = 0.386 J/g◦C, Ti= 100◦C, Tf= 22◦C, mwater = 500 g, and cwater =
4.18 J/g◦C into the equation:
m·0.386 ·(22 −100) = 500 ·4.18 ·(22 −20)
30
Step 3: Solve the equation for the mass of the copper sphere, m.
m·0.386 ·(−78) = 500 ·4.18 ·2
−30.108m= 4180
m=4180
−30.108 ≈ −138.79
Since mass cannot be negative, there is an error in the calculation. Let’s
check the calculations and correct any mistakes.
31
