PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 1
Liberty University
Question 1
Question
A 0.5 kg block of copper at 200
°
C is placed in 1 kg of water at 20
°
C. Assuming
no heat is lost to the surroundings, what will be the final temperature of the
system when thermal equilibrium is reached? (Specific heat capacity of copper
= 390 J/kg ·
°
C, specific heat capacity of water = 4186 J/kg ·
°
C)
Solution
Step 1: Calculate the heat gained by the water. The formula for calculating
heat transfer is Q=mc∆T, where Qis the heat transferred, mis the mass, c
is the specific heat capacity, and ∆Tis the change in temperature.
Given that the initial temperature of the water is 20
°
C and the final tem-
perature of the system is T
°
C (to be determined), the change in temperature
for the water is ∆T=T−20.
The heat gained by the water is:
Qwater =mwater ·cwater ·∆Twater
Qwater = 1 kg ·4186 J/kg ·
°
C·(T−20)
Step 2: Calculate the heat lost by the copper. Similar to the water, the heat
lost by the copper is:
Qcopper =mcopper ·ccopper ·∆Tcopper
Qcopper = 0.5 kg ·390 J/kg ·
°
C·(T−200)
Step 3: Set up the equation based on the conservation of energy. According
to the principle of conservation of energy, the heat lost by the copper must equal
the heat gained by the water:
Qcopper =Qwater
0.5 kg ·390 J/kg ·
°
C·(T−200) = 1 kg ·4186 J/kg ·
°
C·(T−20)
Step 4: Solve for the final temperature. Now, we can solve for the final
temperature T:
195(T−200) = 4186(T−20)
195T−39000 = 4186T−83720
39925 = 3991T
T= 10
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached will be 10
°
C.
Question 2
Question
A copper kettle of mass 0.5 kg contains 2 kg of water at 20
°
C. How much heat
(in Joules) must be supplied to raise the water to its boiling point (100
°
C)?
(Specific heat capacity of copper = 390 J/kg◦C; specific heat capacity of water
= 4200 J/kg◦C; latent heat of vaporization of water = 2.26 ×106J/kg)
Solution
Let’s first calculate the heat required to raise the temperature of water from
20
°
C to 100
°
C:
Step 1: Calculate the heat required to raise the temperature of water. The
formula to calculate heat is given by Q=mc∆T, where Qis the heat, mis the
mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Given: mass of water, mwater = 2 kg initial temperature, Tinitial = 20◦C
final temperature, Tfinal = 100◦C specific heat capacity of water, cwater =
4200 J/kg◦C
Substitute the values into the formula: Qwater =mwater ·cwater ·∆Twater
Calculate ∆Twater: ∆Twater =Tfinal −Tinitial = 100◦C−20◦C = 80◦C
Substitute the values to find Qwater:Qwater = 2 kg ·4200 J/kg◦C·80◦C
Qwater = 672000 J
Therefore, the heat required to raise the temperature of water from 20
°
C to
100
°
C is 672,000 J.
Next, let’s calculate the heat required to boil the water:
2
Step 2: Calculate the heat required to boil the water. The heat required to
change the state of a substance is given by Q=mL, where Qis the heat, mis
the mass, and Lis the latent heat of vaporization.
Given: mass of water, mwater = 2 kg latent heat of vaporization of water,
L= 2.26 ×106J/kg
Substitute the values into the formula: Qvap =mwater ·L
Calculate Qvap:Qvap = 2 kg ·2.26 ×106J/kg Qvap = 4.52 ×106J
Therefore, the heat required to boil the water is 4.52 ×106J.
Step 3: Calculate the total heat required. The total heat required is the
sum of the heat required to raise the temperature and the heat required to boil
the water.
Qtotal =Qwater +Qvap Qtotal = 672000 J + 4.52 ×106JQtotal = 5.192 ×106J
Therefore, the total heat that must be supplied to raise the water to its
boiling point is 5.192 ×106J.
Question 3
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. If the rod is exposed to an environment at 20◦C, calculate the time it
takes for the rod to cool down to 40◦C. Assume the thermal conductivity of
copper is 390 W/mK and the heat transfer coefficient between the rod and the
environment is 50 W/m2K.
Solution
Step 1: Calculate the initial temperature difference (∆Tinitial): The initial tem-
perature difference between the rod and the environment is:
∆Tinitial = 100◦C−20◦C = 80◦C
Step 2: Calculate the surface area of the rod (A): The surface area of the
rod can be calculated using the formula for the surface area of a cylinder:
A=π×diameter ×length = π×0.01 m ×2 m
Step 3: Calculate the heat transfer rate ( ˙
Q): The heat transfer rate can be
calculated using Newton’s Law of Cooling:
˙
Q=h×A×∆Tinitial
Step 4: Calculate the thermal resistance of the rod: The thermal resistance
(Rrod) of the rod can be calculated using the formula:
Rrod =length
k×A
3
Step 5: Calculate the time constant (τ) of the rod: The time constant of the
rod can be calculated using the formula:
τ=ρ×c×V/(h×A)
Step 6: Calculate the time it takes for the rod to cool down to 40◦C: The
time it takes for the rod to cool down to 40◦C can be calculated using the
formula for exponential decay:
T(t) = Tenvironment + (Tinitial −Tenvironment)×e−t/τ
Solving for twhen T(t) = 40◦C will give us the time it takes for the rod to
cool down to 40◦C.
Question 4
Question
A copper block of mass 500 g at a temperature of 100◦C is dropped into 1
kg of water at 20◦C in an insulated container. Assuming no heat is lost to
the surroundings, calculate the final temperature of the system once thermal
equilibrium is reached. (Specific heat capacity of copper = 0.385 J/g◦C, specific
heat capacity of water = 4.18 J/g◦C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature. Given: Mass of copper block, mcopper = 500 g Initial temperature
of copper block, Tcopper initial = 100◦C Specific heat capacity of copper, ccopper =
0.385 J/g◦C Final temperature of the system, Tfinal =x(To be determined)
The heat lost by the copper block can be calculated using the formula:
Qcopper =mcopper ·ccopper ·(Tcopper initial −Tfinal)
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. Given: Mass of water, mwater = 1000 g Initial temperature of
water, Twater initial = 20◦C Specific heat capacity of water, cwater = 4.18 J/g◦C
The heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·(Tfinal −Twater initial)
Step 3: Since no heat is lost to the surroundings:
Qcopper =Qwater
Step 4: Set up and solve the equation from Step 3 to find the final temper-
ature Tfinal:
mcopper ·ccopper ·(Tcopper initial −Tfinal) = mwater ·cwater ·(Tfinal −Twater initial)
4
500 ·0.385 ·(100 −x) = 1000 ·4.18 ·(x−20)
192.5·(100 −x) = 4180 ·(x−20)
19250 −192.5x= 4180x−83600
192.5x+ 4180x= 83600 + 19250
4372.5x= 102850
x=102850
4372.5≈23.52◦C
Therefore, the final temperature of the system once thermal equilibrium is
reached is approximately 23.52◦C.
Question 5
Question
A block of steel with a mass of 2 kg initially at a temperature of 100
°
C is
placed in a container with 1 kg of water at 20
°
C. Assuming no heat is lost to
the surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of steel = 460 J/kg
°
C, specific heat capacity of water =
4200 J/kg
°
C).
Solution
Step 1: Calculate the heat absorbed or released by the steel block: The heat
transfer equation is given by Q=mc∆T, where mis the mass of the substance,
cis the specific heat capacity, and ∆Tis the change in temperature.
Given: msteel = 2 kg csteel = 460 J/kg
°
CTinitial, steel = 100C Tfinal =
Tequilibrium (final equilibrium temperature)
The heat absorbed by the steel block is: Qsteel =msteel ·csteel ·(Tfinal −
Tinitial, steel)
Step 2: Calculate the heat absorbed or released by the water: Given: mwater =
1 kg cwater = 4200 J/kg
°
CTinitial, water = 20C Tfinal =Tequilibrium (final equilib-
rium temperature)
The heat absorbed by the water is: Qwater =mwater ·cwater ·(Tfinal −
Tinitial, water)
Step 3: At thermal equilibrium, the total heat gained by the steel block is
equal to the total heat lost by the water: Qsteel =Qwater
Substitute the expressions for Qsteel and Qwater and solve for Tfinal to find
the equilibrium temperature.
5
Question 6
Question
A solid aluminum cube with side length 10 cm is heated from an initial tem-
perature of 20
°
C to a final temperature of 80
°
C. The specific heat capacity of
aluminum is 0.897 J/g
°
C and the density of aluminum is 2.70 g/cm3. Determine
the amount of heat energy absorbed by the aluminum cube during this heating
process.
Solution
Step 1: Calculate the mass of the aluminum cube. The mass of the cube can
be found using the formula:
Mass = Density ×Volume
The volume Vof a cube with side length ais given by V=a3. Substituting
a= 10 cm,
V= (10 cm)3= 1000 cm3
Given that the density of aluminum is 2.70 g/cm3, the mass mof the cube
is:
m= 2.70 g/cm3×1000 cm3= 2700 g = 2.7 kg
Step 2: Calculate the heat energy absorbed. The heat energy absorbed by
an object can be calculated using the formula:
Heat energy = Mass ×Specific heat capacity ×∆T
where ∆Tis the change in temperature.
Given that the specific heat capacity of aluminum is 0.897 J/g
°
C, the mass
of the cube is 2.7 kg, and the change in temperature is 80C−20C= 60C, we
have:
Heat energy = 2.7 kg ×0.897 J/g
°
C×60C
Heat energy = 2.7 kg ×0.897 J/g
°
C×60 = 145.458 J
Therefore, the amount of heat energy absorbed by the aluminum cube during
the heating process is 145.458 Joules.
Question 7
Question
A cylindrical rod of length Land radius Ris made of a material with thermal
conductivity k. One end of the rod is maintained at a temperature T1while the
other end is maintained at a temperature T2(T1> T2). Find an expression for
the rate at which heat is conducted through the rod.
6
Solution
Let’s denote the rate of heat conduction through the rod as Q. To find an
expression for Q, we can use Fourier’s law of heat conduction, which states:
Q=−kAdT
dx
where Ais the cross-sectional area of the rod, dT /dx is the temperature
gradient along the length of the rod, and kis the thermal conductivity of the
material.
Step 1: Compute the temperature gradient
The temperature gradient can be expressed as the change in temperature
with respect to the change in distance. In this case, the temperature difference
across the rod is T1−T2while the distance over which this difference occurs is
L. Therefore, the temperature gradient is:
dT
dx =T1−T2
L
Step 2: Determine the cross-sectional area
The cross-sectional area of the rod can be calculated using the formula for
the area of a cylinder: A=πR2.
Step 3: Substitute into Fourier’s law
Substitute the values of A,dT
dx , and kinto Fourier’s law to find the rate of
heat conduction Q:
Q=−kAdT
dx =−k(πR2)T1−T2
L
Therefore, the rate at which heat is conducted through the rod is given by
Q=−kπR2T1−T2
L.
Question 8
Question
A copper rod of length 2 meters and diameter 4 cm is heated from 20
°
C to
100
°
C. If the coefficient of linear expansion of copper is 1.7×10−5◦C−1and
the thermal conductivity of copper is 400 W/mK, determine the amount of heat
conducted through the rod during this temperature change.
Solution
Step 1: Calculate the change in length of the copper rod. Given that the
coefficient of linear expansion of copper is 1.7×10−5◦C−1, we can use the
formula for linear expansion:
∆L=α·L·∆T
7
where ∆Lis the change in length, αis the coefficient of linear expansion, Lis
the original length, and ∆Tis the change in temperature. Substitute the values:
∆L= (1.7×10−5)·2·(100 −20)
∆L= 0.000272 m = 0.272 mm
Step 2: Calculate the cross-sectional area of the rod. The diameter of the
rod is 4 cm, so the radius ris half of that, i.e., 2 cm or 0.02 m. Therefore,
A=πr2=π(0.02)2= 1.2566 ×10−3m2
Step 3: Calculate the rate of flow of heat. The rate of flow of heat through
a material is given by Fourier’s Law:
Q=−k·A·∆T
∆x
where Qis the heat conducted, kis the thermal conductivity of the material,
Ais the cross-sectional area, ∆Tis the change in temperature, and ∆xis the
change in length. Substitute the values:
Q=−400 ×1.2566 ×10−3×100 −20
2
Q=−400 ×1.2566 ×10−3×40
Q=−20.11 W
The negative sign indicates that the heat is flowing in the opposite direction of
increasing temperature.
Therefore, the amount of heat conducted through the copper rod during this
temperature change is 20.11 J.
Question 9
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 100
°
C. It is placed in a water bath at 20
°
C. If the rod cools down to 80
°
C in
10 minutes, calculate the rate at which heat is transferred from the rod to the
water bath.
Given: - Thermal conductivity of copper: 400 W/mK - Heat capacity of
copper: 385 J/kgK - Density of copper: 8930 kg/m3- Specific heat capacity of
water: 4186 J/kgK - Density of water: 1000 kg/m3
8
Solution
Step 1: Calculate the cross-sectional area of the rod.
The cross-sectional area of the rod can be found using the formula for the
area of a circle: A=πr2, where ris the radius of the rod. Given that the
diameter of the rod is 2 cm, the radius ris 1 cm or 0.01 m. Therefore, the
cross-sectional area Ais:
A=π×(0.01)2=π×0.0001 = 0.000314m2
Step 2: Calculate the volume of the rod.
The volume of the rod can be calculated using the formula for the volume of
a cylinder: V=A×L, where Lis the length of the rod. Given that the length
of the rod is 1.5 m, the volume Vis:
V= 0.000314 ×1.5=0.000471m3
Step 3: Calculate the mass of the rod.
The mass of the rod can be calculated using the formula: m= density ×V.
Given that the density of copper is 8930 kg/m3, the mass mis:
m= 8930 ×0.000471 = 4.2093 kg
Step 4: 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 of the rod, cis the heat capacity of copper, and ∆Tis
the temperature change. Given that the initial temperature is 100
°
C, the final
temperature is 80
°
C, and the time interval is 10 minutes (or 600 seconds), the
temperature change ∆Tis 20
°
C.
∆Q= 4.2093 ×385 ×20 = 32232.03 J
Step 5: Calculate the rate of heat transfer.
The rate of heat transfer is given by: P=∆Q
∆t, where ∆tis the time interval.
Given that the time interval is 600 seconds, the rate of heat transfer Pis:
P=32232.03
600 = 53.72 W
Therefore, the rate at which heat is transferred from the rod to the water
bath is 53.72 W.
Question 10
Question
A 2 kg block of copper initially at a temperature of 100
°
C is placed in a container
of water initially at 20
°
C. If the final equilibrium temperature of the system is
30
°
C, calculate the mass of water in the container. Assume no heat is lost to the
surroundings and the specific heat capacity of copper is 390 J/kg
°
C and that of
water is 4186 J/kg
°
C.
9
Solution
Step 1: Calculate the heat lost by the copper block. Given: Mass of copper
block, mcopper = 2 kg Initial temperature of copper, Tcopper, initial = 100
°
C
Final temperature of the system, Tfinal = 30
°
C Specific heat capacity of copper,
ccopper = 390 J/kg
°
C
The heat lost by the copper block is given by the formula:
Qlost =mcopper ·ccopper ·(Tcopper, initial −Tfinal)
Substitute the given values:
Qlost = 2 kg ×390 J/kg
°
C×(100C−30C)
Qlost = 2 kg ×390 J/kg
°
C×70C
Qlost = 54600 J
Step 2: Calculate the heat gained by the water. Let the mass of water be
mwater kg. The heat gained by the water is given by the formula:
Qgained =mwater ×4186 ×(30 −20)
Substitute the given values:
54600 J = mwater ×4186 J/kg
°
C×10C
Solve for mwater:
mwater =54600 J
41860 J/kg
°
C
mwater ≈1.303 kg
Hence, the mass of water in the container is approximately 1.303 kg.
Question 11
Question
A copper rod of length 2 m and diameter 4 cm is heated from 20
°
C to 60
°
C.
If the coefficient of linear expansion of copper is 1.7×10−5/
°
C and its density
is 8.96 ×103kg/m3, determine the change in volume of the rod and the heat
absorbed by the rod during this process.
10
Solution
Step 1: Calculate the change in length of the rod due to heating.
Given: Initial temperature, T1= 20CFinal temperature, T2= 60CCoeffi-
cient of linear expansion, α= 1.7×10−5/
°
C Original length, L= 2 m
Using the formula for linear expansion, ∆L=α·L·∆T, where ∆T=T2−T1:
∆L= (1.7×10−5)·2·(60 −20)
∆L= 0.000034 m
Step 2: Calculate the change in volume of the rod.
The change in volume, ∆V, of a cylinder due to heating is given by ∆V=
π·(r2)·∆L, where ris the radius of the cylinder.
Given: Diameter, d= 4 cm Radius, r=d
2= 2 cm = 0.02 m
∆V=π·(0.022)·0.000034
∆V= 3.416 ×10−8m3
Step 3: Calculate the mass of the rod.
The mass, m, of the rod can be calculated using the formula m=V·ρ,
where ρis the density of copper. Given: Density, ρ= 8.96 ×103kg/m3
m= 2 ·π·(0.02)2·L·ρ
m= 2 ·π·(0.02)2·2·8.96 ×103
m= 0.002265 kg
Step 4: Calculate the heat absorbed by the rod.
The heat absorbed, Q, by the rod can be calculated using the formula Q=
mc∆T, where cis the specific heat capacity of copper.
The specific heat capacity of copper is approximately 387 J/kg
°
C. Therefore:
Q= 0.002265 ·387 ·(60 −20)
Q= 33.287 J
Therefore, the change in volume of the rod is 3.416 ×10−8m3and the heat
absorbed by the rod during this process is 33.287 J.
Question 12
Question
A solid copper sphere of radius 5 cm is heated until it reaches a temperature of
100
°
C. If the coefficient of volume expansion of copper is 5 ×10−5/◦C, what is
the change in the volume of the sphere when the temperature rises from 20
°
C
to 100
°
C?
11
Solution
Step 1: First, we need to calculate the original volume of the copper sphere at
20
°
C using the formula for the volume of a sphere: V=4
3πr3. Given that the
initial radius r= 5 cm, the initial volume is:
Vinitial =4
3×π×(5 cm)3=500
3πcm3
Step 2: Next, we need to calculate the change in temperature: ∆T=
100◦C−20◦C= 80◦C.
Step 3: The change in volume of the sphere can be calculated using the
formula:
∆V=β·Vinitial∆T
where βis the coefficient of volume expansion and ∆Tis the change in temper-
ature.
Step 4: Substitute β= 5 ×10−5/◦C,Vinitial =500
3πcm3, and ∆T= 80◦C
into the formula:
∆V= 5 ×10−5/◦C·500
3πcm3·80
°
C
Step 5: Calculate the change in volume:
∆V= 5 ×10−5·500
3π·80 cm3=200
3πcm3
Therefore, the change in the volume of the copper sphere when the temper-
ature rises from 20
°
C to 100
°
C is 200
3πcm3.
Question 13
Question
A 2 kg block of copper at an initial temperature of 100
°
C is dropped into a
container holding 1 kg of water at an initial temperature of 20
°
C. Assuming
there is no heat loss to the surroundings, what will be the final equilibrium
temperature of the system? (Specific heat capacity of copper is 0.385 J/g
°
C and
of water is 4.18 J/g
°
C, ignore the heat capacity of the container).
Solution
Step 1: Calculate the heat gained or lost by the copper block and water. For
the copper block: - Mass (m1) = 2 kg = 2000 g - Initial temperature (Tc1) =
100
°
C - Final temperature (Tcf ) = Teq (the equilibrium temperature) - Specific
heat capacity of copper (c1) = 0.385 J/g
°
C
The heat lost by the copper block is equal to the heat gained by the water,
since this is an isolated system. The heat lost by the copper block = heat gained
by the water
12
m1×c1×(Tcf −Tc1) = m2×c2×(Tcf −Tw1)
Substitute the values:
2000 ×0.385 ×(Teq −100) = 1000 ×4.18 ×(Teq −20)
Step 2: Solve for the equilibrium temperature.
770 ×(Teq −100) = 4180 ×(Teq −20)
770Teq −77000 = 4180Teq −83600
3410Teq = 6600
Teq =6600
3410 ≈1.9386
°
C
Therefore, the final equilibrium temperature of the system is approximately
1.94
°
C.
Question 14
Question
A copper block with a mass of 0.5 kg is heated from an initial temperature
of 20
°
C to a final temperature of 80
°
C. Calculate the amount of heat energy
required to raise the temperature of the block. (Specific heat capacity of copper
= 390 J/kg◦C)
Solution
Step 1: Calculate the change in temperature of the copper block.
∆T=Tf−Ti
∆T= 80◦C−20◦C
∆T= 60◦C
Step 2: Use the formula Q=mc∆Tto calculate the amount of heat energy
required.
Q= (0.5 kg)(390 J/kg◦C)(60◦C)
Q= 0.5×390 ×60 J
Q= 11700 J
Q= 11.7 kJ
Therefore, the amount of heat energy required to raise the temperature of
the copper block is 11.7 kJ.
13
Question 15
Question
A block of copper with a mass of 0.5 kg is heated from 20◦C to 100◦C. If the
specific heat capacity of copper is 387 J/kg◦C, how much heat is required to
accomplish this temperature change?
Solution
Step 1: Calculate the temperature change ∆T. Given: Initial temperature
T1= 20◦C Final temperature T2= 100◦C Temperature change ∆T=T2−T1
∆T= 100◦C−20◦C = 80◦C
Step 2: Use the formula Q=mc∆Tto calculate the heat required. Given:
Mass m= 0.5 kg Specific heat capacity c= 387 J/kg◦C Temperature change
∆T= 80◦C
Q=mc∆T
Q= (0.5 kg)(387 J/kg◦C)(80 C)
Q= 15540 J
Therefore, the amount of heat required to heat the copper block from 20◦C
to 100◦C is 15540 J.
Question 16
Question
A copper bar of length 50 cm and cross-sectional area 4 cm2is heated from 20◦C
to 80◦C. If the coefficient of linear expansion of copper is 1.7×10−5/K and the
Young’s modulus of copper is 1.39 ×1011 Pa, calculate the stress developed
in the bar due to the heating. Assume that the bar is free to expand in the
longitudinal direction but is constrained in the other two directions.
Solution
Step 1: First, calculate the change in length of the copper bar using the linear
expansion formula:
∆L=αL∆T
where ∆Lis the change in length, αis the coefficient of linear expansion, Lis
the initial length, and ∆Tis the change in temperature.
Given that α= 1.7×10−5/K, L= 50 cm, and ∆T= 80 −20 = 60 K, we
substitute these values into the formula to find:
∆L= (1.7×10−5)(50)(60) = 0.051
14
cm
Step 2: Next, calculate the tensile stress developed in the copper bar due to
the change in length using Hooke’s Law:
σ=Y∆L
L
where σis the stress, Yis the Young’s modulus, ∆Lis the change in length,
and Lis the initial length.
Substitute Y= 1.39 ×1011 Pa, ∆L= 0.051 cm, and L= 50 cm into the
formula to find:
σ= (1.39 ×1011)0.051
50 = 1.4308 ×108Pa = 143.08 MPa
Therefore, the stress developed in the copper bar due to the heating is 143.08
MPa.
Question 17
Question
A copper rod of length 2.0 m is initially at a temperature of 100
°
C. One end of
the rod is immersed in a cold-water bath at 0
°
C while the other end is kept in
contact with a hot plate at 150
°
C. If the thermal conductivity of copper is 400
Step 1: Calculate the rate of heat conduction through the rod using Fourier’s
law.
Rate of heat conduction = k·A·(Thot −Tcold)
L
where k= thermal conductivity of copper = 400 W/(mK)
A= cross-sectional area of the rod
Thot = temperature of the hot end = 150
°
C
Tcold = temperature of the cold end = 0
°
C
L= length of the rod = 2.0 m
Step 2: Calculate the time it takes for the temperature at the center of the
rod to reach 50
°
C using the heat equation.
Rate of temperature change = Rate of heat conduction
Heat capacity ·Mass
Time taken ∆t=Tcenter −Tcold
Rate of temperature change
15
Question 18
Question
A 1.50 kg block of iron is heated, causing its temperature to increase by 50.0
°
C.
Assuming the specific heat of iron is 450 J/kg ·
°
C, calculate the amount of heat
transferred to the block of iron.
Solution
Step 1: Identify the given values and the unknown. Let’s denote: m= mass
of the iron block = 1.50 kg ∆T= change in temperature = 50.0
°
Cc= specific
heat of iron = 450 J/kg ·
°
CQ= amount of heat transferred to the block of iron
(unknown)
Step 2: Use the formula to calculate the amount of heat transferred. The
formula for calculating the amount of heat transferred is:
Q=mc∆T
Step 3: Substitute the given values into the formula.
Q= (1.50 kg)(450 J/kg ·
°
C)(50.0
°
C)
Step 4: Perform the calculation to find the amount of heat transferred.
Q= 1.50 ×450 ×50.0
Q= 67,500 J
Therefore, the amount of heat transferred to the block of iron is 67,500 J.
Question 19
Question
A metal rod of length Land thermal conductivity kis kept between two reser-
voirs at different temperatures, T1and T2. The rod is initially at a uniform
temperature T0. If the ends of the rod are perfectly insulated, derive an expres-
sion for the temperature distribution T(x) along the length of the rod at time
t.
Solution
Step 1: The heat equation for the metal rod can be written as:
∂T
∂t =α∂2T
∂x2
16
where α=k
ρc is the thermal diffusivity, kis the thermal conductivity, ρis the
density, and cis the specific heat capacity.
Step 2: Given the initial condition T(x, 0) = T0and the boundary conditions
T(0, t) = T1and T(L, t) = T2, we can solve the heat equation using separation
of variables.
Step 3: Let T(x, t) = X(x)T(t) and substitute it into the heat equation to
get:
1
α
T′
T=X′′
X=−λ2
Step 4: Solve the ODEs 1
α
T′
T=−λ2and X′′ +λ2X= 0 to obtain the general
solution:
Tn(t) = Ane−αλ2
ntand Xn(x) = Bncos(λnx) + Cnsin(λnx)
where nrepresents the mode of the solution.
Step 5: Apply the boundary conditions to find λnand the coefficients Bn
and Cn. The temperature distribution T(x, t) is then given by the infinite series:
T(x, t) =
∞
X
n=1
Tn(t)Xn(x)
Step 6: The temperature distribution T(x) along the length of the rod at
time tcan be expressed as:
T(x, t) =
∞
X
n=1
(Ane−αλ2
nt)(Bncos(λnx) + Cnsin(λnx))
Question 20
Question
A metal block of mass 0.5 kg is initially at a temperature of 100
°
C. If 800 J
of heat is added to the block, what is the final temperature of the block? The
specific heat capacity of the metal is 400 J/kg
°
C.
Solution
Step 1: Identify the given values and the unknown. Let: - Mass of the metal
block, m= 0.5 kg - Initial temperature, Ti= 100
°
C - Heat added, Q= 800 J -
Specific heat capacity, c= 400 J/kg
°
C - Final temperature, Tf(unknown)
Step 2: Use the formula for heat energy. The heat energy transferred to an
object is given by:
Q=mc(Tf−Ti)
Step 3: Substitute the values into the formula. Plugging in the given values:
800 = 0.5×400 ×(Tf−100)
17
Step 4: Solve for the final temperature, Tf.
800 = 200 ×(Tf−100)
800 = 200Tf−20000
200Tf= 20800
Tf=20800
200
Tf= 104
°
C
The final temperature of the metal block after 800 J of heat is added is
104
°
C.
Question 21
Question
A copper rod of length 2 m and uniform cross-sectional area 0.01 m2has one
end kept in steam at 100◦C and the other end in ice at 0◦C. If the thermal
conductivity of copper is 390 W/mK, calculate the rate of heat transfer along
the rod.
Solution
Step 1: Calculate the temperature difference along the rod. The temperature
difference (∆T) along the rod can be calculated as:
∆T= 100◦C−0◦C= 100◦C= 100 K
Step 2: Calculate the rate of heat transfer using Fourier’s Law. The rate of
heat transfer (Q) through the rod can be calculated using Fourier’s Law:
Q=−kA∆T
L
where k= thermal conductivity of copper = 390 W/mK, A= cross-sectional
area of the rod = 0.01 m2, ∆T= temperature difference along the rod = 100 K,
and L= length of the rod = 2 m.
Substitute the given values into the formula:
Q=−(390 W/mK)(0.01 m2)100 K
2 m
Step 3: Calculate the rate of heat transfer.
Q=−3.9 kW/m ×50 K = −195 kW/m
Therefore, the rate of heat transfer along the rod is 195 kW/m , from steam
to ice.
18
Question 22
Question
A steel rod of length 2 m and cross-sectional area 4 ×10−4m2is initially at a
temperature of 100◦C. The rod is then heated until it reaches a temperature of
200◦C. If the thermal conductivity of steel is 50 W/(m·K), calculate the amount
of heat supplied to the rod during this process.
Solution
Step 1: Calculate the change in temperature of the rod. Given that the initial
temperature T1= 100◦C and the final temperature T2= 200◦C, the change in
temperature is:
∆T=T2−T1= 200◦C−100◦C = 100◦C
Step 2: Calculate the rate of heat transfer through the rod. The rate of heat
transfer (P) through the rod is given by Fourier’s law:
P=k·A·∆T
L
where k= thermal conductivity = 50 W/(m·K), A= cross-sectional area =
4×10−4m2, ∆T= change in temperature = 100◦C, and L= length = 2 m.
Plugging in the values, we get:
P= 50 ×4×10−4×100
2= 10 W
Step 3: Calculate the amount of heat supplied. The amount of heat supplied
(Q) can be calculated by multiplying the rate of heat transfer by the time:
Q=P×∆t
Since the time period over which the heat is supplied is not given, we cannot
calculate the exact amount of heat supplied without this information.
Question 23
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into 1 kg
of water at 20
°
C in a perfectly insulated container. Assuming no heat is lost
to the surroundings, calculate the final temperature of the system. The specific
heat capacity of copper is 0.385 J/g
°
C and of water is 4.18 J/g
°
C.
19
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 (500 g), - cis the specific heat
capacity of copper (0.385 J/g
°
C), - ∆Tis the change in temperature of the
copper block.
The change in temperature is calculated as the final temperature minus the
initial temperature:
∆Tcopper =Tfinal −Tinitial =Tfinal −100
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 (1000 g), - cis the specific heat capacity of
water (4.18 J/g
°
C), - ∆Tis the change in temperature of the water.
The change in temperature is calculated as the final temperature minus the
initial temperature:
∆Twater =Tfinal −Tinitial =Tfinal −20
Step 3: Setting up the energy balance equation: Since it is an isolated system
with no heat lost to the surroundings, the heat lost by the copper block is equal
to the heat gained by the water:
Qlost =Qgain
mc∆Tcopper =mc∆Twater
Step 4: Solve for the final temperature: Substitute the expressions for
∆Tcopper and ∆Twater into the energy balance equation:
500 ×0.385 ×(Tfinal −100) = 1000 ×4.18 ×(Tfinal −20)
Solve for Tfinal to find the final temperature of the system.
Question 24
Question
An aluminum block of mass 2 kg is heated to a temperature of 100◦C. If the
specific heat capacity of aluminum is 900 J/kg◦C, how much heat is required to
raise the temperature of the block to 200◦C?
20
Solution
Step 1: Calculate the change in temperature: Given: - Initial temperature
Tinitial = 100◦C - Final temperature Tfinal = 200◦C - Specific heat capacity
c= 900 J/kg◦C We can find the change in temperature (∆T) using the formula:
∆T=Tfinal −Tinitial = 200◦C−100◦C = 100◦C
Step 2: Calculate the heat required: The heat (Q) required to raise the
temperature of an object can be calculated using the formula:
Q=mc∆T
Where: - Mass m= 2 kg - Specific heat capacity c= 900 J/kg◦C - Change in
temperature ∆T= 100◦C
Substitute the given values into the formula:
Q= (2 kg)(900 J/kg◦C)(100◦C)
Step 3: Calculate the heat required:
Q= 2 ×900 ×100 J = 180,000 J
Therefore, the heat required to raise the temperature of the block to 200◦C
is 180,000 J.
Question 25
Question
A 2 kg block of copper at 100
°
C is dropped into 1 kg of water at 20
°
C in a
perfectly insulated container. Assuming no heat is lost to the surroundings,
calculate the final equilibrium temperature of the system. The specific heat
capacity of copper is 390 J/(kg ·K) and the specific heat capacity of water is
4186 J/(kg ·K).
Solution
Step 1: Calculate the heat lost by the block of copper as it cools down to reach
the final equilibrium temperature. The formula for heat transfer is: Q=mc∆T,
where Qis the heat transfer, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Given: - mass of copper, mcopper = 2 kg - specific heat capacity of copper,
ccopper = 390 J/(kg ·K) - initial temperature of copper, Tinitial,copper = 100C=
373 K - final equilibrium temperature of the system, Tfinal
The heat lost by the copper block is:
Qcopper =mcopper ·ccopper ·(Tfinal −Tinitial,copper)
21
Step 2: Calculate the heat gained by the water as it heats up to reach the
final equilibrium temperature. Similar to Step 1, we have: - mass of water,
mwater = 1 kg - specific heat capacity of water, cwater = 4186 J/(kg ·K) - initial
temperature of water, Tinitial,water = 20C= 293 K
The heat gained by the water is:
Qwater =mwater ·cwater ·(Tfinal −Tinitial,water)
Step 3: Since the system is perfectly insulated (no heat is lost to the sur-
roundings), the heat lost by the copper block is equal to the heat gained by the
water:
Qcopper =Qwater
mcopper ·ccopper ·(Tfinal −Tinitial,copper) = mwater ·cwater ·(Tfinal −Tinitial,water)
Step 4: Solve for Tfinal in the equation from Step 3 to find the final equilib-
rium temperature of the system.
2·390 ·(Tfinal −373) = 1 ·4186 ·(Tfinal −293)
780 ·Tfinal −292140 = 4186 ·Tfinal −1227398
3406 ·Tfinal = 935258
Tfinal =935258
3406 ≈274.74 K
Therefore, the final equilibrium temperature of the system is approximately
274.74 K.
Question 26
Question
A copper rod of length 2 m and diameter 2 cm is heated from 20
°
C to 120
°
C.
If the coefficient of linear expansion of copper is 1.7×10−5/C and the specific
heat capacity of copper is 0.385 J/g
°
C, determine the change in volume of the
rod and the heat supplied to it during this process.
Solution
Step 1: Calculate the change in length of the rod. Given that the coefficient
of linear expansion of copper is α= 1.7×10−5/C, we can use the formula for
linear expansion:
∆L=L·α·∆T
where Lis the initial length, αis the coefficient of linear expansion, and ∆Tis
the change in temperature. Substituting the values L= 2 m, α= 1.7×10−5/C,
∆T= 100C:
∆L= 2 m ·(1.7×10−5/C)·100C= 0.0034 m = 3.4 mm
22
Step 2: Calculate the change in volume of the rod. The change in volume
can be approximated as three times the change in length, since the expansion
is isotropic:
∆V= 3∆L
Substituting the value ∆L= 0.0034 m:
∆V= 3 ×0.0034 m = 0.0102 m3
Step 3: Calculate the mass of the rod. First, we need to calculate the radius
of the rod:
r=d
2=0.02 m
2= 0.01 m
Then, we can calculate the volume of the rod:
Vrod =πr2·L=π·(0.01)2·2=0.000628 m3
Given that the density of copper is 8.96 g/cm3, the mass of the rod can be
calculated as:
m=ρ·Vrod = 8.96 g/cm3·0.000628 m3= 0.005628 kg
Step 4: Calculate the heat supplied to the rod. The heat supplied is equal
to the change in internal energy of the rod, which can be calculated using the
formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change
in temperature. Substituting the values m= 0.005628 kg, c= 0.385 J/g
°
C,
∆T= 100C:
Q= 0.005628 kg ×0.385 J/g
°
C×100C= 2.17 J
Therefore, the change in volume of the rod is 0.0102 m3and the heat supplied
to it during this process is 2.17 J.
Question 27
Question
A 2.0 kg block of aluminum at 20
°
C is submerged in 3.0 kg of water at 40
°
C.
Assuming no heat is lost to the surroundings, what will be the final temperature
of the system when thermal equilibrium is reached? (Specific heat capacity of
aluminum = 900 J/kg ·K, specific heat capacity of water = 4186 J/kg ·K)
23
Solution
Step 1: Calculate the heat lost by the aluminum block when it cools down to
the final temperature. The heat lost by the aluminum block can be calculated
using the formula:
Qaluminum =maluminum ·caluminum ·(Tf−Ti)
where: Qaluminum is the heat lost, maluminum = 2.0 kg is the mass of aluminum,
caluminum = 900 J/kg ·K is the specific heat capacity of aluminum, Ti= 20C
is the initial temperature of aluminum, and Tfis the final temperature of the
system.
Substitute the values into the formula:
Qaluminum = 2.0 kg ×900 J/kg ·K×(Tf−20C)
Step 2: Calculate the heat gained by the water when it cools down to the
final temperature. The heat gained by the water can be calculated using the
formula:
Qwater =mwater ·cwater ·(Tf−Ti)
where: Qwater is the heat gained, mwater = 3.0 kg is the mass of water, cwater =
4186 J/kg ·K is the specific heat capacity of water, Ti= 40Cis the initial
temperature of water, and Tfis the final temperature of the system.
Substitute the values into the formula:
Qwater = 3.0 kg ×4186 J/kg ·K×(Tf−40C)
Step 3: At thermal equilibrium, the heat lost by the aluminum block is equal
to the heat gained by the water. Set the two equations equal to each other and
solve for Tf.
2.0×900 ×(Tf−20) = 3.0×4186 ×(Tf−40)
Step 4: Solve the equation for Tfto find the final temperature of the system.
1800 ×(Tf−20) = 12558 ×(Tf−40)
1800 ×Tf−36000 = 12558 ×Tf−502320
502320 −36000 = 12558 ×Tf−1800 ×Tf
466320 = 10758 ×Tf
Tf=466320
10758 ≈43.33C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 43.33C.
24
Question 28
Question
A 2 kg block of copper at 300 K is placed in contact with a 3 kg block of
aluminum at 400 K. The two blocks are well-insulated and the system reaches
thermal equilibrium. Assuming no heat is lost to the surroundings, determine
the final temperature of the system. The specific heat capacities of copper and
aluminum are cCu = 0.386 J/g·K and cAl = 0.897 J/g·K, respectively.
Solution
Step 1: Calculate the initial thermal energies of the copper and aluminum
blocks. The thermal energy Qof a substance is given by:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
For copper:
QCu, initial = 2 kg ·0.386 J/g ·K·(Tfinal −300 K)
For aluminum:
QAl, initial = 3 kg ·0.897 J/g ·K·(Tfinal −400 K)
Step 2: Since no heat is lost to the surroundings, the total thermal energy of
the system remains constant. At thermal equilibrium, the total initial thermal
energy equals the total final thermal energy.
QCu, initial +QAl, initial = 0
Step 3: Solve the equation for Tfinal. Substitute the expressions for QCu, initial
and QAl, initial into the equation:
2·0.386 ·(Tfinal −300) + 3 ·0.897 ·(Tfinal −400) = 0
Step 4: Solve for Tfinal.
0.772 ·Tfinal −0.772 ·300 + 2.691 ·Tfinal −3.582 = 0
3.463 ·Tfinal = 3.582 + 0.772 ·300
Tfinal =3.582 + 0.772 ·300
3.463
Tfinal =229.382
3.463
Tfinal ≈66.26 K
Therefore, the final temperature of the system is approximately 66.26 K.
25
Question 29
Question
A steel rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. The rod is heated until its temperature raises to 300◦C. Calculate the
amount of heat energy transferred to the rod. (Given: Specific heat capacity of
steel = 450 J/kg◦C and density of steel = 7850 kg/m3)
Solution
Step 1: Calculate 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
ris the radius and his the height. Given the diameter of the rod is 1 cm, the
radius ris 0.5 cm = 0.005 m. The height his 2 m.
Volume (V) = π(0.005)2×2 = π×0.000025 ×2≈1.57 ×10−4m3
The mass mof the steel rod can be calculated using the formula: m= density×
volume. Given the density of steel is 7850 kg/m3:
m= 7850 ×1.57 ×10−4≈1.23 kg
Step 2: Calculate the change in temperature and the amount of heat energy
transferred. The change in temperature ∆Tis given by:
∆T= 300 −100 = 200 ◦C
The amount of heat energy transferred Qis given by the formula:
Q=mc∆T
Given the specific heat capacity cof steel is 450 J/kg◦C:
Q= 1.23 ×450 ×200 = 110700 J
Therefore, the amount of heat energy transferred to the steel rod is 110700
J.
Question 30
Question
A piece of copper of mass 200 g at a temperature of 100
°
C is placed in a vessel
of water at 0
°
C. If the final temperature of the mixture is 10
°
C, determine the
mass of water in the vessel. The specific heat capacity of copper is 0.385 J/g
°
C
and the specific heat capacity of water is 4.18 J/g
°
C.
26
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
The heat lost by the copper is equal to the heat gained by the water. We can
use the formula:
m1c1(Tf−T1) = m2c2(T2−Tf)
where: m1= mass of copper (200 g) c1= specific heat capacity of copper
(0.385 J/g
°
C) T1= initial temperature of copper (100C)Tf= final temperature
of the mixture (10C)m2= mass of water c2= specific heat capacity of water
(4.18 J/g
°
C) T2= initial temperature of water (0C)
Substitute the values and solve for m2:
200 ×0.385 ×(10 −100) = m2×4.18 ×(0 −10)
−7740 = −41.8m2
m2=7740
41.8≈185.166 g
Therefore, the mass of water in the vessel is approximately 185.166 g.
Question 31
Question
A block of copper with a mass of 500 g is initially at a temperature of 100
°
C. It is
placed in a calorimeter containing 200 g of water at 20
°
C. If the final temperature
of the system is 30
°
C, determine the specific heat capacity of copper. Assume
no heat is lost to the surroundings.
Solution
Step 1: Identify the information given in the problem. The following values are
provided: - Mass of copper (mcopper) = 500 g - Initial temperature of copper
(Tinitial, copper) = 100
°
C - Mass of water in calorimeter (mwater) = 200 g - Initial
temperature of water (Tinitial, water) = 20
°
C - Final temperature of the system
(Tfinal) = 30
°
C - Specific heat capacity of water (cwater) = 4186 J/kg
°
C
Step 2: Calculate the heat absorbed by the water. The heat absorbed by
the water can be calculated using the formula:
Qwater =mwater ·cwater ·∆Twater
where ∆Twater is the change in temperature of the water and can be calculated
as:
∆Twater =Tfinal −Tinitial, water
27
Step 3: Calculate the heat lost by the copper. Assuming no heat is lost to
the surroundings, the heat lost by the copper is equal to the heat absorbed by
the water. This can be calculated using the formula:
Qcopper =−Qwater
Step 4: Use the specific heat capacity formula to find the specific heat ca-
pacity of copper. The specific heat capacity formula is:
Q=m·c·∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Step 5: Substitute the values into the specific heat capacity formula to find
ccopper. Substitute the known values into the formula:
mcopper ·ccopper ·∆Tcopper =−mwater ·cwater ·∆Twater
Solve for ccopper to find the specific heat capacity of copper.
Question 32
Question
A 2 kg block of copper at 100◦C is dropped into a container of 1 kg of water
at 20◦C. Assuming no heat is lost to the surroundings, what will be the final
temperature of the mixture? The specific heat capacity of copper is 390 J/kg◦C
and for water is 4186 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 by the copper block can be calculated using the
formula:
Q=mc∆T
where: Q= heat lost or gained, m= mass of the copper block, c= specific
heat capacity of copper, ∆T= change in temperature.
Given: m= 2 kg, c= 390 J/kg◦C, Tinitial = 100◦C, Tfinal =Tmixture (un-
known).
The change in temperature for the copper block is:
∆T=Tfinal −Tinitial
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. The heat gained by the water can also be calculated using the
formula:
Q=mc∆T
28
where: Q= heat lost or gained, m= mass of the water, c= specific heat
capacity of water, ∆T= change in temperature.
Given: m= 1 kg, c= 4186 J/kg◦C, Tinitial = 20◦C, Tfinal =Tmixture (un-
known).
The change in temperature for the water is:
∆T=Tfinal −Tinitial
Step 3: Apply the principle of conservation of energy to find the final tem-
perature of the mixture. Since there is no heat lost to the surroundings:
Heat lost by copper = Heat gained by water
mcCu(Tfinal −Tinitial,Cu) = mcH2O(Tfinal −Tinitial,H2O)
Substitute the given values and solve for Tfinal to find the final temperature
of the mixture.
Question 33
Question
A 0.5 kg copper block is heated from 20
°
C to 100
°
C. If the specific heat capacity
of copper is 390 J/kg◦C, how much heat is required?
Solution
Step 1: Calculate the change in temperature. Given that the initial temperature
Ti= 20Cand the final temperature Tf= 100C, the change in temperature is
given by:
∆T=Tf−Ti= 100C−20C= 80C
Step 2: Use the formula Q=mc∆Tto calculate the heat required. Substi-
tute the mass m= 0.5 kg, specific heat capacity c= 390 J/kg◦C, and change in
temperature ∆T= 80Cinto the formula:
Q= (0.5 kg) ×(390 J/kg◦C) ×80C
Q= 0.5×390 ×80 J
Q= 15600 J
Therefore, the amount of heat required to heat the copper block from 20
°
C
to 100
°
C is 15600 J.
29
Question 34
Question
A copper rod of length 1 m and cross-sectional area 4×10−4m2is heated at one
end by a small flame. The surface temperature of the heated end of the rod is
150◦C and the other end is kept at 30◦C. Given that the thermal conductivity
of copper is 390 W/(m
·
K) and the rod loses heat to the surrounding air at a
rate of 50 W, calculate the rate at which heat is conducted along the rod.
Solution
Step 1: Calculate the temperature difference along the rod.
∆T= 150◦C−30◦C = 120◦C
Step 2: Calculate the rate of heat conducted along the rod using Fourier’s
Law of Heat Conduction.
Rate of heat conducted = Thermal conductivity ×Area ×∆T
Length
Step 3: Substituting the given values into the formula:
Rate of heat conducted = 390 W/(m ·K) ×4×10−4m2×120 K
1 m
Step 4: Simplify the expression to find the rate of heat conducted along the
rod.
Rate of heat conducted = 187.2 W
Therefore, the rate at which heat is conducted along the rod is 187.2 W .
Question 35
Question
A steel rod of length 2.0 m and cross-sectional area 4.0 cm2is heated from 20◦C
to 100◦C. If the linear expansion coefficient of steel is 1.2×10−5K−1and its
specific heat capacity is 0.45 J/(g·K), calculate the increase in length of the rod
and the heat energy transferred to the rod during this process.
Solution
Step 1: Calculate the increase in length of the rod due to temperature change.
Given: Initial temperature (T1) = 20◦C = 20 K
Final temperature (T2) = 100◦C = 100 K
Change in temperature (∆T) = T2−T1= 100 K - 20 K = 80 K
30
Step 3: Set up the equation based on the conservation of energy. According
to the principle of conservation of energy, the heat lost by the copper must equal
the heat gained by the water:
Qcopper =Qwater
0.5 kg ·390 J/kg ·
°
C·(T−200) = 1 kg ·4186 J/kg ·
°
C·(T−20)
Step 4: Solve for the final temperature. Now, we can solve for the final
temperature T:
195(T−200) = 4186(T−20)
195T−39000 = 4186T−83720
39925 = 3991T
T= 10
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached will be 10
°
C.
Question 2
Question
A copper kettle of mass 0.5 kg contains 2 kg of water at 20
°
C. How much heat
(in Joules) must be supplied to raise the water to its boiling point (100
°
C)?
(Specific heat capacity of copper = 390 J/kg◦C; specific heat capacity of water
= 4200 J/kg◦C; latent heat of vaporization of water = 2.26 ×106J/kg)
Solution
Let’s first calculate the heat required to raise the temperature of water from
20
°
C to 100
°
C:
Step 1: Calculate the heat required to raise the temperature of water. The
formula to calculate heat is given by Q=mc∆T, where Qis the heat, mis the
mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Given: mass of water, mwater = 2 kg initial temperature, Tinitial = 20◦C
final temperature, Tfinal = 100◦C specific heat capacity of water, cwater =
4200 J/kg◦C
Substitute the values into the formula: Qwater =mwater ·cwater ·∆Twater
Calculate ∆Twater: ∆Twater =Tfinal −Tinitial = 100◦C−20◦C = 80◦C
Substitute the values to find Qwater:Qwater = 2 kg ·4200 J/kg◦C·80◦C
Qwater = 672000 J
Therefore, the heat required to raise the temperature of water from 20
°
C to
100
°
C is 672,000 J.
Next, let’s calculate the heat required to boil the water:
2
Step 2: Calculate the heat required to boil the water. The heat required to
change the state of a substance is given by Q=mL, where Qis the heat, mis
the mass, and Lis the latent heat of vaporization.
Given: mass of water, mwater = 2 kg latent heat of vaporization of water,
L= 2.26 ×106J/kg
Substitute the values into the formula: Qvap =mwater ·L
Calculate Qvap:Qvap = 2 kg ·2.26 ×106J/kg Qvap = 4.52 ×106J
Therefore, the heat required to boil the water is 4.52 ×106J.
Step 3: Calculate the total heat required. The total heat required is the
sum of the heat required to raise the temperature and the heat required to boil
the water.
Qtotal =Qwater +Qvap Qtotal = 672000 J + 4.52 ×106JQtotal = 5.192 ×106J
Therefore, the total heat that must be supplied to raise the water to its
boiling point is 5.192 ×106J.
Question 3
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. If the rod is exposed to an environment at 20◦C, calculate the time it
takes for the rod to cool down to 40◦C. Assume the thermal conductivity of
copper is 390 W/mK and the heat transfer coefficient between the rod and the
environment is 50 W/m2K.
Solution
Step 1: Calculate the initial temperature difference (∆Tinitial): The initial tem-
perature difference between the rod and the environment is:
∆Tinitial = 100◦C−20◦C = 80◦C
Step 2: Calculate the surface area of the rod (A): The surface area of the
rod can be calculated using the formula for the surface area of a cylinder:
A=π×diameter ×length = π×0.01 m ×2 m
Step 3: Calculate the heat transfer rate ( ˙
Q): The heat transfer rate can be
calculated using Newton’s Law of Cooling:
˙
Q=h×A×∆Tinitial
Step 4: Calculate the thermal resistance of the rod: The thermal resistance
(Rrod) of the rod can be calculated using the formula:
Rrod =length
k×A
3
Step 5: Calculate the time constant (τ) of the rod: The time constant of the
rod can be calculated using the formula:
τ=ρ×c×V/(h×A)
Step 6: Calculate the time it takes for the rod to cool down to 40◦C: The
time it takes for the rod to cool down to 40◦C can be calculated using the
formula for exponential decay:
T(t) = Tenvironment + (Tinitial −Tenvironment)×e−t/τ
Solving for twhen T(t) = 40◦C will give us the time it takes for the rod to
cool down to 40◦C.
Question 4
Question
A copper block of mass 500 g at a temperature of 100◦C is dropped into 1
kg of water at 20◦C in an insulated container. Assuming no heat is lost to
the surroundings, calculate the final temperature of the system once thermal
equilibrium is reached. (Specific heat capacity of copper = 0.385 J/g◦C, specific
heat capacity of water = 4.18 J/g◦C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature. Given: Mass of copper block, mcopper = 500 g Initial temperature
of copper block, Tcopper initial = 100◦C Specific heat capacity of copper, ccopper =
0.385 J/g◦C Final temperature of the system, Tfinal =x(To be determined)
The heat lost by the copper block can be calculated using the formula:
Qcopper =mcopper ·ccopper ·(Tcopper initial −Tfinal)
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. Given: Mass of water, mwater = 1000 g Initial temperature of
water, Twater initial = 20◦C Specific heat capacity of water, cwater = 4.18 J/g◦C
The heat gained by the water can be calculated using the formula:
Qwater =mwater ·cwater ·(Tfinal −Twater initial)
Step 3: Since no heat is lost to the surroundings:
Qcopper =Qwater
Step 4: Set up and solve the equation from Step 3 to find the final temper-
ature Tfinal:
mcopper ·ccopper ·(Tcopper initial −Tfinal) = mwater ·cwater ·(Tfinal −Twater initial)
4
500 ·0.385 ·(100 −x) = 1000 ·4.18 ·(x−20)
192.5·(100 −x) = 4180 ·(x−20)
19250 −192.5x= 4180x−83600
192.5x+ 4180x= 83600 + 19250
4372.5x= 102850
x=102850
4372.5≈23.52◦C
Therefore, the final temperature of the system once thermal equilibrium is
reached is approximately 23.52◦C.
Question 5
Question
A block of steel with a mass of 2 kg initially at a temperature of 100
°
C is
placed in a container with 1 kg of water at 20
°
C. Assuming no heat is lost to
the surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of steel = 460 J/kg
°
C, specific heat capacity of water =
4200 J/kg
°
C).
Solution
Step 1: Calculate the heat absorbed or released by the steel block: The heat
transfer equation is given by Q=mc∆T, where mis the mass of the substance,
cis the specific heat capacity, and ∆Tis the change in temperature.
Given: msteel = 2 kg csteel = 460 J/kg
°
CTinitial, steel = 100C Tfinal =
Tequilibrium (final equilibrium temperature)
The heat absorbed by the steel block is: Qsteel =msteel ·csteel ·(Tfinal −
Tinitial, steel)
Step 2: Calculate the heat absorbed or released by the water: Given: mwater =
1 kg cwater = 4200 J/kg
°
CTinitial, water = 20C Tfinal =Tequilibrium (final equilib-
rium temperature)
The heat absorbed by the water is: Qwater =mwater ·cwater ·(Tfinal −
Tinitial, water)
Step 3: At thermal equilibrium, the total heat gained by the steel block is
equal to the total heat lost by the water: Qsteel =Qwater
Substitute the expressions for Qsteel and Qwater and solve for Tfinal to find
the equilibrium temperature.
5
Question 6
Question
A solid aluminum cube with side length 10 cm is heated from an initial tem-
perature of 20
°
C to a final temperature of 80
°
C. The specific heat capacity of
aluminum is 0.897 J/g
°
C and the density of aluminum is 2.70 g/cm3. Determine
the amount of heat energy absorbed by the aluminum cube during this heating
process.
Solution
Step 1: Calculate the mass of the aluminum cube. The mass of the cube can
be found using the formula:
Mass = Density ×Volume
The volume Vof a cube with side length ais given by V=a3. Substituting
a= 10 cm,
V= (10 cm)3= 1000 cm3
Given that the density of aluminum is 2.70 g/cm3, the mass mof the cube
is:
m= 2.70 g/cm3×1000 cm3= 2700 g = 2.7 kg
Step 2: Calculate the heat energy absorbed. The heat energy absorbed by
an object can be calculated using the formula:
Heat energy = Mass ×Specific heat capacity ×∆T
where ∆Tis the change in temperature.
Given that the specific heat capacity of aluminum is 0.897 J/g
°
C, the mass
of the cube is 2.7 kg, and the change in temperature is 80C−20C= 60C, we
have:
Heat energy = 2.7 kg ×0.897 J/g
°
C×60C
Heat energy = 2.7 kg ×0.897 J/g
°
C×60 = 145.458 J
Therefore, the amount of heat energy absorbed by the aluminum cube during
the heating process is 145.458 Joules.
Question 7
Question
A cylindrical rod of length Land radius Ris made of a material with thermal
conductivity k. One end of the rod is maintained at a temperature T1while the
other end is maintained at a temperature T2(T1> T2). Find an expression for
the rate at which heat is conducted through the rod.
6
Solution
Let’s denote the rate of heat conduction through the rod as Q. To find an
expression for Q, we can use Fourier’s law of heat conduction, which states:
Q=−kAdT
dx
where Ais the cross-sectional area of the rod, dT /dx is the temperature
gradient along the length of the rod, and kis the thermal conductivity of the
material.
Step 1: Compute the temperature gradient
The temperature gradient can be expressed as the change in temperature
with respect to the change in distance. In this case, the temperature difference
across the rod is T1−T2while the distance over which this difference occurs is
L. Therefore, the temperature gradient is:
dT
dx =T1−T2
L
Step 2: Determine the cross-sectional area
The cross-sectional area of the rod can be calculated using the formula for
the area of a cylinder: A=πR2.
Step 3: Substitute into Fourier’s law
Substitute the values of A,dT
dx , and kinto Fourier’s law to find the rate of
heat conduction Q:
Q=−kAdT
dx =−k(πR2)T1−T2
L
Therefore, the rate at which heat is conducted through the rod is given by
Q=−kπR2T1−T2
L.
Question 8
Question
A copper rod of length 2 meters and diameter 4 cm is heated from 20
°
C to
100
°
C. If the coefficient of linear expansion of copper is 1.7×10−5◦C−1and
the thermal conductivity of copper is 400 W/mK, determine the amount of heat
conducted through the rod during this temperature change.
Solution
Step 1: Calculate the change in length of the copper rod. Given that the
coefficient of linear expansion of copper is 1.7×10−5◦C−1, we can use the
formula for linear expansion:
∆L=α·L·∆T
7
where ∆Lis the change in length, αis the coefficient of linear expansion, Lis
the original length, and ∆Tis the change in temperature. Substitute the values:
∆L= (1.7×10−5)·2·(100 −20)
∆L= 0.000272 m = 0.272 mm
Step 2: Calculate the cross-sectional area of the rod. The diameter of the
rod is 4 cm, so the radius ris half of that, i.e., 2 cm or 0.02 m. Therefore,
A=πr2=π(0.02)2= 1.2566 ×10−3m2
Step 3: Calculate the rate of flow of heat. The rate of flow of heat through
a material is given by Fourier’s Law:
Q=−k·A·∆T
∆x
where Qis the heat conducted, kis the thermal conductivity of the material,
Ais the cross-sectional area, ∆Tis the change in temperature, and ∆xis the
change in length. Substitute the values:
Q=−400 ×1.2566 ×10−3×100 −20
2
Q=−400 ×1.2566 ×10−3×40
Q=−20.11 W
The negative sign indicates that the heat is flowing in the opposite direction of
increasing temperature.
Therefore, the amount of heat conducted through the copper rod during this
temperature change is 20.11 J.
Question 9
Question
A copper rod of length 1.5 m and diameter 2 cm is initially at a temperature
of 100
°
C. It is placed in a water bath at 20
°
C. If the rod cools down to 80
°
C in
10 minutes, calculate the rate at which heat is transferred from the rod to the
water bath.
Given: - Thermal conductivity of copper: 400 W/mK - Heat capacity of
copper: 385 J/kgK - Density of copper: 8930 kg/m3- Specific heat capacity of
water: 4186 J/kgK - Density of water: 1000 kg/m3
8
Solution
Step 1: Calculate the cross-sectional area of the rod.
The cross-sectional area of the rod can be found using the formula for the
area of a circle: A=πr2, where ris the radius of the rod. Given that the
diameter of the rod is 2 cm, the radius ris 1 cm or 0.01 m. Therefore, the
cross-sectional area Ais:
A=π×(0.01)2=π×0.0001 = 0.000314m2
Step 2: Calculate the volume of the rod.
The volume of the rod can be calculated using the formula for the volume of
a cylinder: V=A×L, where Lis the length of the rod. Given that the length
of the rod is 1.5 m, the volume Vis:
V= 0.000314 ×1.5=0.000471m3
Step 3: Calculate the mass of the rod.
The mass of the rod can be calculated using the formula: m= density ×V.
Given that the density of copper is 8930 kg/m3, the mass mis:
m= 8930 ×0.000471 = 4.2093 kg
Step 4: 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 of the rod, cis the heat capacity of copper, and ∆Tis
the temperature change. Given that the initial temperature is 100
°
C, the final
temperature is 80
°
C, and the time interval is 10 minutes (or 600 seconds), the
temperature change ∆Tis 20
°
C.
∆Q= 4.2093 ×385 ×20 = 32232.03 J
Step 5: Calculate the rate of heat transfer.
The rate of heat transfer is given by: P=∆Q
∆t, where ∆tis the time interval.
Given that the time interval is 600 seconds, the rate of heat transfer Pis:
P=32232.03
600 = 53.72 W
Therefore, the rate at which heat is transferred from the rod to the water
bath is 53.72 W.
Question 10
Question
A 2 kg block of copper initially at a temperature of 100
°
C is placed in a container
of water initially at 20
°
C. If the final equilibrium temperature of the system is
30
°
C, calculate the mass of water in the container. Assume no heat is lost to the
surroundings and the specific heat capacity of copper is 390 J/kg
°
C and that of
water is 4186 J/kg
°
C.
9
Solution
Step 1: Calculate the heat lost by the copper block. Given: Mass of copper
block, mcopper = 2 kg Initial temperature of copper, Tcopper, initial = 100
°
C
Final temperature of the system, Tfinal = 30
°
C Specific heat capacity of copper,
ccopper = 390 J/kg
°
C
The heat lost by the copper block is given by the formula:
Qlost =mcopper ·ccopper ·(Tcopper, initial −Tfinal)
Substitute the given values:
Qlost = 2 kg ×390 J/kg
°
C×(100C−30C)
Qlost = 2 kg ×390 J/kg
°
C×70C
Qlost = 54600 J
Step 2: Calculate the heat gained by the water. Let the mass of water be
mwater kg. The heat gained by the water is given by the formula:
Qgained =mwater ×4186 ×(30 −20)
Substitute the given values:
54600 J = mwater ×4186 J/kg
°
C×10C
Solve for mwater:
mwater =54600 J
41860 J/kg
°
C
mwater ≈1.303 kg
Hence, the mass of water in the container is approximately 1.303 kg.
Question 11
Question
A copper rod of length 2 m and diameter 4 cm is heated from 20
°
C to 60
°
C.
If the coefficient of linear expansion of copper is 1.7×10−5/
°
C and its density
is 8.96 ×103kg/m3, determine the change in volume of the rod and the heat
absorbed by the rod during this process.
10
Solution
Step 1: Calculate the change in length of the rod due to heating.
Given: Initial temperature, T1= 20CFinal temperature, T2= 60CCoeffi-
cient of linear expansion, α= 1.7×10−5/
°
C Original length, L= 2 m
Using the formula for linear expansion, ∆L=α·L·∆T, where ∆T=T2−T1:
∆L= (1.7×10−5)·2·(60 −20)
∆L= 0.000034 m
Step 2: Calculate the change in volume of the rod.
The change in volume, ∆V, of a cylinder due to heating is given by ∆V=
π·(r2)·∆L, where ris the radius of the cylinder.
Given: Diameter, d= 4 cm Radius, r=d
2= 2 cm = 0.02 m
∆V=π·(0.022)·0.000034
∆V= 3.416 ×10−8m3
Step 3: Calculate the mass of the rod.
The mass, m, of the rod can be calculated using the formula m=V·ρ,
where ρis the density of copper. Given: Density, ρ= 8.96 ×103kg/m3
m= 2 ·π·(0.02)2·L·ρ
m= 2 ·π·(0.02)2·2·8.96 ×103
m= 0.002265 kg
Step 4: Calculate the heat absorbed by the rod.
The heat absorbed, Q, by the rod can be calculated using the formula Q=
mc∆T, where cis the specific heat capacity of copper.
The specific heat capacity of copper is approximately 387 J/kg
°
C. Therefore:
Q= 0.002265 ·387 ·(60 −20)
Q= 33.287 J
Therefore, the change in volume of the rod is 3.416 ×10−8m3and the heat
absorbed by the rod during this process is 33.287 J.
Question 12
Question
A solid copper sphere of radius 5 cm is heated until it reaches a temperature of
100
°
C. If the coefficient of volume expansion of copper is 5 ×10−5/◦C, what is
the change in the volume of the sphere when the temperature rises from 20
°
C
to 100
°
C?
11
Solution
Step 1: First, we need to calculate the original volume of the copper sphere at
20
°
C using the formula for the volume of a sphere: V=4
3πr3. Given that the
initial radius r= 5 cm, the initial volume is:
Vinitial =4
3×π×(5 cm)3=500
3πcm3
Step 2: Next, we need to calculate the change in temperature: ∆T=
100◦C−20◦C= 80◦C.
Step 3: The change in volume of the sphere can be calculated using the
formula:
∆V=β·Vinitial∆T
where βis the coefficient of volume expansion and ∆Tis the change in temper-
ature.
Step 4: Substitute β= 5 ×10−5/◦C,Vinitial =500
3πcm3, and ∆T= 80◦C
into the formula:
∆V= 5 ×10−5/◦C·500
3πcm3·80
°
C
Step 5: Calculate the change in volume:
∆V= 5 ×10−5·500
3π·80 cm3=200
3πcm3
Therefore, the change in the volume of the copper sphere when the temper-
ature rises from 20
°
C to 100
°
C is 200
3πcm3.
Question 13
Question
A 2 kg block of copper at an initial temperature of 100
°
C is dropped into a
container holding 1 kg of water at an initial temperature of 20
°
C. Assuming
there is no heat loss to the surroundings, what will be the final equilibrium
temperature of the system? (Specific heat capacity of copper is 0.385 J/g
°
C and
of water is 4.18 J/g
°
C, ignore the heat capacity of the container).
Solution
Step 1: Calculate the heat gained or lost by the copper block and water. For
the copper block: - Mass (m1) = 2 kg = 2000 g - Initial temperature (Tc1) =
100
°
C - Final temperature (Tcf ) = Teq (the equilibrium temperature) - Specific
heat capacity of copper (c1) = 0.385 J/g
°
C
The heat lost by the copper block is equal to the heat gained by the water,
since this is an isolated system. The heat lost by the copper block = heat gained
by the water
12
m1×c1×(Tcf −Tc1) = m2×c2×(Tcf −Tw1)
Substitute the values:
2000 ×0.385 ×(Teq −100) = 1000 ×4.18 ×(Teq −20)
Step 2: Solve for the equilibrium temperature.
770 ×(Teq −100) = 4180 ×(Teq −20)
770Teq −77000 = 4180Teq −83600
3410Teq = 6600
Teq =6600
3410 ≈1.9386
°
C
Therefore, the final equilibrium temperature of the system is approximately
1.94
°
C.
Question 14
Question
A copper block with a mass of 0.5 kg is heated from an initial temperature
of 20
°
C to a final temperature of 80
°
C. Calculate the amount of heat energy
required to raise the temperature of the block. (Specific heat capacity of copper
= 390 J/kg◦C)
Solution
Step 1: Calculate the change in temperature of the copper block.
∆T=Tf−Ti
∆T= 80◦C−20◦C
∆T= 60◦C
Step 2: Use the formula Q=mc∆Tto calculate the amount of heat energy
required.
Q= (0.5 kg)(390 J/kg◦C)(60◦C)
Q= 0.5×390 ×60 J
Q= 11700 J
Q= 11.7 kJ
Therefore, the amount of heat energy required to raise the temperature of
the copper block is 11.7 kJ.
13
Question 15
Question
A block of copper with a mass of 0.5 kg is heated from 20◦C to 100◦C. If the
specific heat capacity of copper is 387 J/kg◦C, how much heat is required to
accomplish this temperature change?
Solution
Step 1: Calculate the temperature change ∆T. Given: Initial temperature
T1= 20◦C Final temperature T2= 100◦C Temperature change ∆T=T2−T1
∆T= 100◦C−20◦C = 80◦C
Step 2: Use the formula Q=mc∆Tto calculate the heat required. Given:
Mass m= 0.5 kg Specific heat capacity c= 387 J/kg◦C Temperature change
∆T= 80◦C
Q=mc∆T
Q= (0.5 kg)(387 J/kg◦C)(80 C)
Q= 15540 J
Therefore, the amount of heat required to heat the copper block from 20◦C
to 100◦C is 15540 J.
Question 16
Question
A copper bar of length 50 cm and cross-sectional area 4 cm2is heated from 20◦C
to 80◦C. If the coefficient of linear expansion of copper is 1.7×10−5/K and the
Young’s modulus of copper is 1.39 ×1011 Pa, calculate the stress developed
in the bar due to the heating. Assume that the bar is free to expand in the
longitudinal direction but is constrained in the other two directions.
Solution
Step 1: First, calculate the change in length of the copper bar using the linear
expansion formula:
∆L=αL∆T
where ∆Lis the change in length, αis the coefficient of linear expansion, Lis
the initial length, and ∆Tis the change in temperature.
Given that α= 1.7×10−5/K, L= 50 cm, and ∆T= 80 −20 = 60 K, we
substitute these values into the formula to find:
∆L= (1.7×10−5)(50)(60) = 0.051
14
cm
Step 2: Next, calculate the tensile stress developed in the copper bar due to
the change in length using Hooke’s Law:
σ=Y∆L
L
where σis the stress, Yis the Young’s modulus, ∆Lis the change in length,
and Lis the initial length.
Substitute Y= 1.39 ×1011 Pa, ∆L= 0.051 cm, and L= 50 cm into the
formula to find:
σ= (1.39 ×1011)0.051
50 = 1.4308 ×108Pa = 143.08 MPa
Therefore, the stress developed in the copper bar due to the heating is 143.08
MPa.
Question 17
Question
A copper rod of length 2.0 m is initially at a temperature of 100
°
C. One end of
the rod is immersed in a cold-water bath at 0
°
C while the other end is kept in
contact with a hot plate at 150
°
C. If the thermal conductivity of copper is 400
Step 1: Calculate the rate of heat conduction through the rod using Fourier’s
law.
Rate of heat conduction = k·A·(Thot −Tcold)
L
where k= thermal conductivity of copper = 400 W/(mK)
A= cross-sectional area of the rod
Thot = temperature of the hot end = 150
°
C
Tcold = temperature of the cold end = 0
°
C
L= length of the rod = 2.0 m
Step 2: Calculate the time it takes for the temperature at the center of the
rod to reach 50
°
C using the heat equation.
Rate of temperature change = Rate of heat conduction
Heat capacity ·Mass
Time taken ∆t=Tcenter −Tcold
Rate of temperature change
15
Question 18
Question
A 1.50 kg block of iron is heated, causing its temperature to increase by 50.0
°
C.
Assuming the specific heat of iron is 450 J/kg ·
°
C, calculate the amount of heat
transferred to the block of iron.
Solution
Step 1: Identify the given values and the unknown. Let’s denote: m= mass
of the iron block = 1.50 kg ∆T= change in temperature = 50.0
°
Cc= specific
heat of iron = 450 J/kg ·
°
CQ= amount of heat transferred to the block of iron
(unknown)
Step 2: Use the formula to calculate the amount of heat transferred. The
formula for calculating the amount of heat transferred is:
Q=mc∆T
Step 3: Substitute the given values into the formula.
Q= (1.50 kg)(450 J/kg ·
°
C)(50.0
°
C)
Step 4: Perform the calculation to find the amount of heat transferred.
Q= 1.50 ×450 ×50.0
Q= 67,500 J
Therefore, the amount of heat transferred to the block of iron is 67,500 J.
Question 19
Question
A metal rod of length Land thermal conductivity kis kept between two reser-
voirs at different temperatures, T1and T2. The rod is initially at a uniform
temperature T0. If the ends of the rod are perfectly insulated, derive an expres-
sion for the temperature distribution T(x) along the length of the rod at time
t.
Solution
Step 1: The heat equation for the metal rod can be written as:
∂T
∂t =α∂2T
∂x2
16
where α=k
ρc is the thermal diffusivity, kis the thermal conductivity, ρis the
density, and cis the specific heat capacity.
Step 2: Given the initial condition T(x, 0) = T0and the boundary conditions
T(0, t) = T1and T(L, t) = T2, we can solve the heat equation using separation
of variables.
Step 3: Let T(x, t) = X(x)T(t) and substitute it into the heat equation to
get:
1
α
T′
T=X′′
X=−λ2
Step 4: Solve the ODEs 1
α
T′
T=−λ2and X′′ +λ2X= 0 to obtain the general
solution:
Tn(t) = Ane−αλ2
ntand Xn(x) = Bncos(λnx) + Cnsin(λnx)
where nrepresents the mode of the solution.
Step 5: Apply the boundary conditions to find λnand the coefficients Bn
and Cn. The temperature distribution T(x, t) is then given by the infinite series:
T(x, t) =
∞
X
n=1
Tn(t)Xn(x)
Step 6: The temperature distribution T(x) along the length of the rod at
time tcan be expressed as:
T(x, t) =
∞
X
n=1
(Ane−αλ2
nt)(Bncos(λnx) + Cnsin(λnx))
Question 20
Question
A metal block of mass 0.5 kg is initially at a temperature of 100
°
C. If 800 J
of heat is added to the block, what is the final temperature of the block? The
specific heat capacity of the metal is 400 J/kg
°
C.
Solution
Step 1: Identify the given values and the unknown. Let: - Mass of the metal
block, m= 0.5 kg - Initial temperature, Ti= 100
°
C - Heat added, Q= 800 J -
Specific heat capacity, c= 400 J/kg
°
C - Final temperature, Tf(unknown)
Step 2: Use the formula for heat energy. The heat energy transferred to an
object is given by:
Q=mc(Tf−Ti)
Step 3: Substitute the values into the formula. Plugging in the given values:
800 = 0.5×400 ×(Tf−100)
17
Step 4: Solve for the final temperature, Tf.
800 = 200 ×(Tf−100)
800 = 200Tf−20000
200Tf= 20800
Tf=20800
200
Tf= 104
°
C
The final temperature of the metal block after 800 J of heat is added is
104
°
C.
Question 21
Question
A copper rod of length 2 m and uniform cross-sectional area 0.01 m2has one
end kept in steam at 100◦C and the other end in ice at 0◦C. If the thermal
conductivity of copper is 390 W/mK, calculate the rate of heat transfer along
the rod.
Solution
Step 1: Calculate the temperature difference along the rod. The temperature
difference (∆T) along the rod can be calculated as:
∆T= 100◦C−0◦C= 100◦C= 100 K
Step 2: Calculate the rate of heat transfer using Fourier’s Law. The rate of
heat transfer (Q) through the rod can be calculated using Fourier’s Law:
Q=−kA∆T
L
where k= thermal conductivity of copper = 390 W/mK, A= cross-sectional
area of the rod = 0.01 m2, ∆T= temperature difference along the rod = 100 K,
and L= length of the rod = 2 m.
Substitute the given values into the formula:
Q=−(390 W/mK)(0.01 m2)100 K
2 m
Step 3: Calculate the rate of heat transfer.
Q=−3.9 kW/m ×50 K = −195 kW/m
Therefore, the rate of heat transfer along the rod is 195 kW/m , from steam
to ice.
18
Question 22
Question
A steel rod of length 2 m and cross-sectional area 4 ×10−4m2is initially at a
temperature of 100◦C. The rod is then heated until it reaches a temperature of
200◦C. If the thermal conductivity of steel is 50 W/(m·K), calculate the amount
of heat supplied to the rod during this process.
Solution
Step 1: Calculate the change in temperature of the rod. Given that the initial
temperature T1= 100◦C and the final temperature T2= 200◦C, the change in
temperature is:
∆T=T2−T1= 200◦C−100◦C = 100◦C
Step 2: Calculate the rate of heat transfer through the rod. The rate of heat
transfer (P) through the rod is given by Fourier’s law:
P=k·A·∆T
L
where k= thermal conductivity = 50 W/(m·K), A= cross-sectional area =
4×10−4m2, ∆T= change in temperature = 100◦C, and L= length = 2 m.
Plugging in the values, we get:
P= 50 ×4×10−4×100
2= 10 W
Step 3: Calculate the amount of heat supplied. The amount of heat supplied
(Q) can be calculated by multiplying the rate of heat transfer by the time:
Q=P×∆t
Since the time period over which the heat is supplied is not given, we cannot
calculate the exact amount of heat supplied without this information.
Question 23
Question
A copper block of mass 500 g at a temperature of 100
°
C is dropped into 1 kg
of water at 20
°
C in a perfectly insulated container. Assuming no heat is lost
to the surroundings, calculate the final temperature of the system. The specific
heat capacity of copper is 0.385 J/g
°
C and of water is 4.18 J/g
°
C.
19
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 (500 g), - cis the specific heat
capacity of copper (0.385 J/g
°
C), - ∆Tis the change in temperature of the
copper block.
The change in temperature is calculated as the final temperature minus the
initial temperature:
∆Tcopper =Tfinal −Tinitial =Tfinal −100
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 (1000 g), - cis the specific heat capacity of
water (4.18 J/g
°
C), - ∆Tis the change in temperature of the water.
The change in temperature is calculated as the final temperature minus the
initial temperature:
∆Twater =Tfinal −Tinitial =Tfinal −20
Step 3: Setting up the energy balance equation: Since it is an isolated system
with no heat lost to the surroundings, the heat lost by the copper block is equal
to the heat gained by the water:
Qlost =Qgain
mc∆Tcopper =mc∆Twater
Step 4: Solve for the final temperature: Substitute the expressions for
∆Tcopper and ∆Twater into the energy balance equation:
500 ×0.385 ×(Tfinal −100) = 1000 ×4.18 ×(Tfinal −20)
Solve for Tfinal to find the final temperature of the system.
Question 24
Question
An aluminum block of mass 2 kg is heated to a temperature of 100◦C. If the
specific heat capacity of aluminum is 900 J/kg◦C, how much heat is required to
raise the temperature of the block to 200◦C?
20
Solution
Step 1: Calculate the change in temperature: Given: - Initial temperature
Tinitial = 100◦C - Final temperature Tfinal = 200◦C - Specific heat capacity
c= 900 J/kg◦C We can find the change in temperature (∆T) using the formula:
∆T=Tfinal −Tinitial = 200◦C−100◦C = 100◦C
Step 2: Calculate the heat required: The heat (Q) required to raise the
temperature of an object can be calculated using the formula:
Q=mc∆T
Where: - Mass m= 2 kg - Specific heat capacity c= 900 J/kg◦C - Change in
temperature ∆T= 100◦C
Substitute the given values into the formula:
Q= (2 kg)(900 J/kg◦C)(100◦C)
Step 3: Calculate the heat required:
Q= 2 ×900 ×100 J = 180,000 J
Therefore, the heat required to raise the temperature of the block to 200◦C
is 180,000 J.
Question 25
Question
A 2 kg block of copper at 100
°
C is dropped into 1 kg of water at 20
°
C in a
perfectly insulated container. Assuming no heat is lost to the surroundings,
calculate the final equilibrium temperature of the system. The specific heat
capacity of copper is 390 J/(kg ·K) and the specific heat capacity of water is
4186 J/(kg ·K).
Solution
Step 1: Calculate the heat lost by the block of copper as it cools down to reach
the final equilibrium temperature. The formula for heat transfer is: Q=mc∆T,
where Qis the heat transfer, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Given: - mass of copper, mcopper = 2 kg - specific heat capacity of copper,
ccopper = 390 J/(kg ·K) - initial temperature of copper, Tinitial,copper = 100C=
373 K - final equilibrium temperature of the system, Tfinal
The heat lost by the copper block is:
Qcopper =mcopper ·ccopper ·(Tfinal −Tinitial,copper)
21
Step 2: Calculate the heat gained by the water as it heats up to reach the
final equilibrium temperature. Similar to Step 1, we have: - mass of water,
mwater = 1 kg - specific heat capacity of water, cwater = 4186 J/(kg ·K) - initial
temperature of water, Tinitial,water = 20C= 293 K
The heat gained by the water is:
Qwater =mwater ·cwater ·(Tfinal −Tinitial,water)
Step 3: Since the system is perfectly insulated (no heat is lost to the sur-
roundings), the heat lost by the copper block is equal to the heat gained by the
water:
Qcopper =Qwater
mcopper ·ccopper ·(Tfinal −Tinitial,copper) = mwater ·cwater ·(Tfinal −Tinitial,water)
Step 4: Solve for Tfinal in the equation from Step 3 to find the final equilib-
rium temperature of the system.
2·390 ·(Tfinal −373) = 1 ·4186 ·(Tfinal −293)
780 ·Tfinal −292140 = 4186 ·Tfinal −1227398
3406 ·Tfinal = 935258
Tfinal =935258
3406 ≈274.74 K
Therefore, the final equilibrium temperature of the system is approximately
274.74 K.
Question 26
Question
A copper rod of length 2 m and diameter 2 cm is heated from 20
°
C to 120
°
C.
If the coefficient of linear expansion of copper is 1.7×10−5/C and the specific
heat capacity of copper is 0.385 J/g
°
C, determine the change in volume of the
rod and the heat supplied to it during this process.
Solution
Step 1: Calculate the change in length of the rod. Given that the coefficient
of linear expansion of copper is α= 1.7×10−5/C, we can use the formula for
linear expansion:
∆L=L·α·∆T
where Lis the initial length, αis the coefficient of linear expansion, and ∆Tis
the change in temperature. Substituting the values L= 2 m, α= 1.7×10−5/C,
∆T= 100C:
∆L= 2 m ·(1.7×10−5/C)·100C= 0.0034 m = 3.4 mm
22
Step 2: Calculate the change in volume of the rod. The change in volume
can be approximated as three times the change in length, since the expansion
is isotropic:
∆V= 3∆L
Substituting the value ∆L= 0.0034 m:
∆V= 3 ×0.0034 m = 0.0102 m3
Step 3: Calculate the mass of the rod. First, we need to calculate the radius
of the rod:
r=d
2=0.02 m
2= 0.01 m
Then, we can calculate the volume of the rod:
Vrod =πr2·L=π·(0.01)2·2=0.000628 m3
Given that the density of copper is 8.96 g/cm3, the mass of the rod can be
calculated as:
m=ρ·Vrod = 8.96 g/cm3·0.000628 m3= 0.005628 kg
Step 4: Calculate the heat supplied to the rod. The heat supplied is equal
to the change in internal energy of the rod, which can be calculated using the
formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change
in temperature. Substituting the values m= 0.005628 kg, c= 0.385 J/g
°
C,
∆T= 100C:
Q= 0.005628 kg ×0.385 J/g
°
C×100C= 2.17 J
Therefore, the change in volume of the rod is 0.0102 m3and the heat supplied
to it during this process is 2.17 J.
Question 27
Question
A 2.0 kg block of aluminum at 20
°
C is submerged in 3.0 kg of water at 40
°
C.
Assuming no heat is lost to the surroundings, what will be the final temperature
of the system when thermal equilibrium is reached? (Specific heat capacity of
aluminum = 900 J/kg ·K, specific heat capacity of water = 4186 J/kg ·K)
23
Solution
Step 1: Calculate the heat lost by the aluminum block when it cools down to
the final temperature. The heat lost by the aluminum block can be calculated
using the formula:
Qaluminum =maluminum ·caluminum ·(Tf−Ti)
where: Qaluminum is the heat lost, maluminum = 2.0 kg is the mass of aluminum,
caluminum = 900 J/kg ·K is the specific heat capacity of aluminum, Ti= 20C
is the initial temperature of aluminum, and Tfis the final temperature of the
system.
Substitute the values into the formula:
Qaluminum = 2.0 kg ×900 J/kg ·K×(Tf−20C)
Step 2: Calculate the heat gained by the water when it cools down to the
final temperature. The heat gained by the water can be calculated using the
formula:
Qwater =mwater ·cwater ·(Tf−Ti)
where: Qwater is the heat gained, mwater = 3.0 kg is the mass of water, cwater =
4186 J/kg ·K is the specific heat capacity of water, Ti= 40Cis the initial
temperature of water, and Tfis the final temperature of the system.
Substitute the values into the formula:
Qwater = 3.0 kg ×4186 J/kg ·K×(Tf−40C)
Step 3: At thermal equilibrium, the heat lost by the aluminum block is equal
to the heat gained by the water. Set the two equations equal to each other and
solve for Tf.
2.0×900 ×(Tf−20) = 3.0×4186 ×(Tf−40)
Step 4: Solve the equation for Tfto find the final temperature of the system.
1800 ×(Tf−20) = 12558 ×(Tf−40)
1800 ×Tf−36000 = 12558 ×Tf−502320
502320 −36000 = 12558 ×Tf−1800 ×Tf
466320 = 10758 ×Tf
Tf=466320
10758 ≈43.33C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 43.33C.
24
Question 28
Question
A 2 kg block of copper at 300 K is placed in contact with a 3 kg block of
aluminum at 400 K. The two blocks are well-insulated and the system reaches
thermal equilibrium. Assuming no heat is lost to the surroundings, determine
the final temperature of the system. The specific heat capacities of copper and
aluminum are cCu = 0.386 J/g·K and cAl = 0.897 J/g·K, respectively.
Solution
Step 1: Calculate the initial thermal energies of the copper and aluminum
blocks. The thermal energy Qof a substance is given by:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
For copper:
QCu, initial = 2 kg ·0.386 J/g ·K·(Tfinal −300 K)
For aluminum:
QAl, initial = 3 kg ·0.897 J/g ·K·(Tfinal −400 K)
Step 2: Since no heat is lost to the surroundings, the total thermal energy of
the system remains constant. At thermal equilibrium, the total initial thermal
energy equals the total final thermal energy.
QCu, initial +QAl, initial = 0
Step 3: Solve the equation for Tfinal. Substitute the expressions for QCu, initial
and QAl, initial into the equation:
2·0.386 ·(Tfinal −300) + 3 ·0.897 ·(Tfinal −400) = 0
Step 4: Solve for Tfinal.
0.772 ·Tfinal −0.772 ·300 + 2.691 ·Tfinal −3.582 = 0
3.463 ·Tfinal = 3.582 + 0.772 ·300
Tfinal =3.582 + 0.772 ·300
3.463
Tfinal =229.382
3.463
Tfinal ≈66.26 K
Therefore, the final temperature of the system is approximately 66.26 K.
25
Question 29
Question
A steel rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. The rod is heated until its temperature raises to 300◦C. Calculate the
amount of heat energy transferred to the rod. (Given: Specific heat capacity of
steel = 450 J/kg◦C and density of steel = 7850 kg/m3)
Solution
Step 1: Calculate 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
ris the radius and his the height. Given the diameter of the rod is 1 cm, the
radius ris 0.5 cm = 0.005 m. The height his 2 m.
Volume (V) = π(0.005)2×2 = π×0.000025 ×2≈1.57 ×10−4m3
The mass mof the steel rod can be calculated using the formula: m= density×
volume. Given the density of steel is 7850 kg/m3:
m= 7850 ×1.57 ×10−4≈1.23 kg
Step 2: Calculate the change in temperature and the amount of heat energy
transferred. The change in temperature ∆Tis given by:
∆T= 300 −100 = 200 ◦C
The amount of heat energy transferred Qis given by the formula:
Q=mc∆T
Given the specific heat capacity cof steel is 450 J/kg◦C:
Q= 1.23 ×450 ×200 = 110700 J
Therefore, the amount of heat energy transferred to the steel rod is 110700
J.
Question 30
Question
A piece of copper of mass 200 g at a temperature of 100
°
C is placed in a vessel
of water at 0
°
C. If the final temperature of the mixture is 10
°
C, determine the
mass of water in the vessel. The specific heat capacity of copper is 0.385 J/g
°
C
and the specific heat capacity of water is 4.18 J/g
°
C.
26
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
The heat lost by the copper is equal to the heat gained by the water. We can
use the formula:
m1c1(Tf−T1) = m2c2(T2−Tf)
where: m1= mass of copper (200 g) c1= specific heat capacity of copper
(0.385 J/g
°
C) T1= initial temperature of copper (100C)Tf= final temperature
of the mixture (10C)m2= mass of water c2= specific heat capacity of water
(4.18 J/g
°
C) T2= initial temperature of water (0C)
Substitute the values and solve for m2:
200 ×0.385 ×(10 −100) = m2×4.18 ×(0 −10)
−7740 = −41.8m2
m2=7740
41.8≈185.166 g
Therefore, the mass of water in the vessel is approximately 185.166 g.
Question 31
Question
A block of copper with a mass of 500 g is initially at a temperature of 100
°
C. It is
placed in a calorimeter containing 200 g of water at 20
°
C. If the final temperature
of the system is 30
°
C, determine the specific heat capacity of copper. Assume
no heat is lost to the surroundings.
Solution
Step 1: Identify the information given in the problem. The following values are
provided: - Mass of copper (mcopper) = 500 g - Initial temperature of copper
(Tinitial, copper) = 100
°
C - Mass of water in calorimeter (mwater) = 200 g - Initial
temperature of water (Tinitial, water) = 20
°
C - Final temperature of the system
(Tfinal) = 30
°
C - Specific heat capacity of water (cwater) = 4186 J/kg
°
C
Step 2: Calculate the heat absorbed by the water. The heat absorbed by
the water can be calculated using the formula:
Qwater =mwater ·cwater ·∆Twater
where ∆Twater is the change in temperature of the water and can be calculated
as:
∆Twater =Tfinal −Tinitial, water
27
Step 3: Calculate the heat lost by the copper. Assuming no heat is lost to
the surroundings, the heat lost by the copper is equal to the heat absorbed by
the water. This can be calculated using the formula:
Qcopper =−Qwater
Step 4: Use the specific heat capacity formula to find the specific heat ca-
pacity of copper. The specific heat capacity formula is:
Q=m·c·∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Step 5: Substitute the values into the specific heat capacity formula to find
ccopper. Substitute the known values into the formula:
mcopper ·ccopper ·∆Tcopper =−mwater ·cwater ·∆Twater
Solve for ccopper to find the specific heat capacity of copper.
Question 32
Question
A 2 kg block of copper at 100◦C is dropped into a container of 1 kg of water
at 20◦C. Assuming no heat is lost to the surroundings, what will be the final
temperature of the mixture? The specific heat capacity of copper is 390 J/kg◦C
and for water is 4186 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 by the copper block can be calculated using the
formula:
Q=mc∆T
where: Q= heat lost or gained, m= mass of the copper block, c= specific
heat capacity of copper, ∆T= change in temperature.
Given: m= 2 kg, c= 390 J/kg◦C, Tinitial = 100◦C, Tfinal =Tmixture (un-
known).
The change in temperature for the copper block is:
∆T=Tfinal −Tinitial
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. The heat gained by the water can also be calculated using the
formula:
Q=mc∆T
28
where: Q= heat lost or gained, m= mass of the water, c= specific heat
capacity of water, ∆T= change in temperature.
Given: m= 1 kg, c= 4186 J/kg◦C, Tinitial = 20◦C, Tfinal =Tmixture (un-
known).
The change in temperature for the water is:
∆T=Tfinal −Tinitial
Step 3: Apply the principle of conservation of energy to find the final tem-
perature of the mixture. Since there is no heat lost to the surroundings:
Heat lost by copper = Heat gained by water
mcCu(Tfinal −Tinitial,Cu) = mcH2O(Tfinal −Tinitial,H2O)
Substitute the given values and solve for Tfinal to find the final temperature
of the mixture.
Question 33
Question
A 0.5 kg copper block is heated from 20
°
C to 100
°
C. If the specific heat capacity
of copper is 390 J/kg◦C, how much heat is required?
Solution
Step 1: Calculate the change in temperature. Given that the initial temperature
Ti= 20Cand the final temperature Tf= 100C, the change in temperature is
given by:
∆T=Tf−Ti= 100C−20C= 80C
Step 2: Use the formula Q=mc∆Tto calculate the heat required. Substi-
tute the mass m= 0.5 kg, specific heat capacity c= 390 J/kg◦C, and change in
temperature ∆T= 80Cinto the formula:
Q= (0.5 kg) ×(390 J/kg◦C) ×80C
Q= 0.5×390 ×80 J
Q= 15600 J
Therefore, the amount of heat required to heat the copper block from 20
°
C
to 100
°
C is 15600 J.
29
Question 34
Question
A copper rod of length 1 m and cross-sectional area 4×10−4m2is heated at one
end by a small flame. The surface temperature of the heated end of the rod is
150◦C and the other end is kept at 30◦C. Given that the thermal conductivity
of copper is 390 W/(m
·
K) and the rod loses heat to the surrounding air at a
rate of 50 W, calculate the rate at which heat is conducted along the rod.
Solution
Step 1: Calculate the temperature difference along the rod.
∆T= 150◦C−30◦C = 120◦C
Step 2: Calculate the rate of heat conducted along the rod using Fourier’s
Law of Heat Conduction.
Rate of heat conducted = Thermal conductivity ×Area ×∆T
Length
Step 3: Substituting the given values into the formula:
Rate of heat conducted = 390 W/(m ·K) ×4×10−4m2×120 K
1 m
Step 4: Simplify the expression to find the rate of heat conducted along the
rod.
Rate of heat conducted = 187.2 W
Therefore, the rate at which heat is conducted along the rod is 187.2 W .
Question 35
Question
A steel rod of length 2.0 m and cross-sectional area 4.0 cm2is heated from 20◦C
to 100◦C. If the linear expansion coefficient of steel is 1.2×10−5K−1and its
specific heat capacity is 0.45 J/(g·K), calculate the increase in length of the rod
and the heat energy transferred to the rod during this process.
Solution
Step 1: Calculate the increase in length of the rod due to temperature change.
Given: Initial temperature (T1) = 20◦C = 20 K
Final temperature (T2) = 100◦C = 100 K
Change in temperature (∆T) = T2−T1= 100 K - 20 K = 80 K
30
Linear expansion coefficient (α) = 1.2×10−5K−1
Initial length of the rod (L0) = 2.0 m
The change in length of the rod is given by the formula:
∆L=α·L0·∆T
∆L= (1.2×10−5K−1)·(2.0 m) ·(80 K)
∆L= 0.00192 m = 1.92 mm
Therefore, the increase in length of the rod is 1.92 mm.
Step 2: Calculate the heat energy transferred to the rod. Specific heat
capacity of steel = 0.45 J/(g·K) = 450 J/(kg·K)
Cross-sectional area of the rod = 4.0 cm2= 4.0×10−4m2
Density of steel = 7900 kg/m3
The mass of the steel rod can be calculated using:
Volume = Area ×Length = (4.0×10−4m2)×(2.0 m) = 8.0×10−4m3
Mass = Volume ×Density = (8.0×10−4m3)×(7900 kg/m3) = 6.32 kg
The heat energy transferred is given by the formula:
Q= Mass ×Specific heat capacity ×∆T
Q= (6.32 kg) ×(450 J/kg ·K) ×(80 K)
Q= 227520 J
Therefore, the heat energy transferred to the rod during this process is
227520 J.
31
