PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 6
Liberty University
Question 1
Question
A metal rod of length Land thermal conductivity kis initially at a uniform
temperature of T1along its length. One end of the rod is then held in a bath
of ice at a temperature of 0◦C, while the other end is held in a bath of boiling
water at 100◦C. The rod reaches a steady state where its temperature varies
linearly from 0◦C to 100◦C. Determine the temperature as a function of distance
xalong the rod in this steady state.
Solution
Let T(x) denote the temperature at a distance xalong the rod, where 0 ≤x≤L.
Step 1: Apply Fourier’s law of heat conduction to the rod. According to
Fourier’s law of heat conduction, the rate of heat transfer Qthrough a material
is proportional to the temperature gradient dT
dx and the cross-sectional area A,
and inversely proportional to the thickness dx of the material. Mathematically,
this can be expressed as:
Q=−kAdT
dx or dQ
dx =−kA d
dx(T)
Step 2: Apply conservation of energy to the rod. Since the rod is in steady
state with no heat accumulation, the rate of heat transfer into an infinitesimally
small section of the rod at xmust be equal to the rate of heat transfer out of
that section. Mathematically:
dQin
dx =dQout
dx
Step 3: Integrate the differential equation. Integrating the equation ob-
tained in Step 1 gives:
ZQx
0
dQ =−kA Zx
0
d
dx(T)dx
Qx=−kA(Tx−T1)
Step 4: Solve the differential equation. Solving the differential equation
and applying the boundary conditions T(0) = 100◦C and T(L) = 0◦C:
T(x) = T1+ (T2−T1)1−x
L
Hence, the temperature as a function of distance xalong the rod in the
steady state is given by T(x) = T1+ (100◦C−T1)1−x
L.
Question 2
Question
A copper rod of length 2.0 m and diameter 1.0 cm is initially at a temperature
of 100◦C. The rod is placed in an ice-water mixture at 0◦C. If the rod loses
800 J of heat to the surroundings, calculate the final temperature of the rod.
Assume that the only significant heat transfer is along the length of the rod.
Solution
Step 1: Calculate the initial temperature difference between the rod and the
ice-water mixture. Given that the initial temperature of the rod is 100◦C and
the ice-water mixture is at 0◦C, the initial temperature difference (∆Ti) is:
∆Ti= 100 −0 = 100◦C
Step 2: Calculate the cross-sectional area of the rod. The radius of the rod
is r=1.0 cm
2= 0.005 m. The cross-sectional area (A) of the rod is:
A=πr2=π(0.005)2= 7.85 ×10−5m2
Step 3: Calculate the volume of the rod. The volume of the rod (V) is:
V=A×L= 7.85 ×10−5×2.0=1.57 ×10−4m3
Step 4: Calculate the mass of the rod. The density of copper is ρ=
8930 kg/m3. The mass of the rod (m) is:
m=ρ×V= 8930 ×1.57 ×10−4= 1.40 kg
Step 5: Calculate the specific heat capacity of copper. The specific heat
capacity of copper (c) is 390 J/(kg·K).
2
Step 6: Calculate the initial heat content of the rod. The initial heat content
of the rod (Qi) is given by:
Qi=mc∆Ti= 1.40 ×390 ×100 = 54,600 J
Step 7: Calculate the final temperature of the rod. Since the rod loses 800
J of heat to the surroundings and no work is done, the final heat content of the
rod (Qf) is Qi−800.
Qf= 54,600 −800 = 53,800 J
Step 8: Calculate the final temperature of the rod. The final temperature
difference (∆Tf) can be calculated using:
Qf=mc∆Tf
∆Tf=Qf
mc =53,800
1.40 ×390100.9◦C
Therefore, the final temperature of the rod is approximately 100.9◦C.
Question 3
Question
A piece of copper of mass 150 g at 150
°
C is placed in 500 g of water at 20
°
C.
Assuming no heat is lost to the surroundings, what is the final temperature of
the system when thermal equilibrium is reached? The specific heat capacity of
copper is 0.387 J/g
°
C and that of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat gained by the copper as it cools down to the final
temperature. The heat gained is equal to the heat lost by the water.
Heat lost by copper = Heat gained by water
mcopper ×ccopper ×(Tfinal −150) = mwater ×cwater ×(Tfinal −20)
Step 2: Substitute the known values into the equation.
150 ×0.387 ×(Tfinal −150) = 500 ×4.18 ×(Tfinal −20)
Step 3: Simplify the equation and solve for Tfinal.
58.05Tfinal −58.05 ×150 = 2090Tfinal −2090 ×20
58.05Tfinal −8707.5 = 2090Tfinal −41800
2090Tfinal −58.05Tfinal = 41800 −8707.5
2031.95Tfinal = 33092.5
Tfinal =33092.5
2031.95 ≈16.28
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 16.28
°
C.
3
Question 4
Question
A piece of copper weighing 300 g at a temperature of 100
°
C is placed in a vessel
containing 200 g of water at 20
°
C. If the final temperature of the mixture is
25
°
C, calculate the specific heat capacity of copper. (Specific heat capacity of
water = 4200 J/kg
°
C)
Solution
Step 1: Calculate the heat lost by the copper piece when it cools from 100
°
C to
25
°
C. The formula for heat lost or gained is given by:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the substance (in kg), -
cis the specific heat capacity of the substance (in J/kg
°
C), - ∆Tis the change
in temperature (in
°
C).
Given that the mass of the copper, m= 0.3 kg, initial temperature, Ti=
100C, final temperature, Tf= 25C, and specific heat capacity of water, cwater =
4200 J/kg
°
C.
Substitute these values into the formula to find the heat lost by the copper:
Qcopper = (0.3)(ccopper)(100 −25)
Step 2: Calculate the heat gained by the water when it heats up from 20
°
C
to 25
°
C. The heat gained by the water is equal to the heat lost by the copper
(assuming no heat is lost to the surroundings). Therefore, we can write:
Qwater = (0.2)(4200)(25 −20)
Step 3: Set the heat lost by the copper equal to the heat gained by the water
and solve for ccopper:
Qcopper =Qwater
Step 4: Substitute the expressions for Qcopper and Qwater into the equation
from Step 3 and solve for ccopper. This will give the specific heat capacity of
copper.
Question 5
Question
A copper rod of length 2 m and cross-sectional area 4 cm2is initially at a uniform
temperature of 200◦C. The rod is then placed in a water bath at 20◦C. If the
thermal conductivity of copper is 400 W/(m·K) and the heat transfer coefficient
between copper and water is 1000 W/(m2·K), calculate the time required for the
rod to reach a temperature of 50◦C. Assume one-dimensional heat conduction
along the rod.
4
Solution
Step 1: First, we calculate the thermal resistance of the rod using the formula:
R=L
k·A
where: - Ris the thermal resistance, - Lis the length of the rod, - kis the thermal
conductivity of the material (copper in this case), - Ais the cross-sectional area
of the rod.
Plugging in the values:
R=2
400 ×4×10−4
R=2
0.16 = 12.5 K/W
Step 2: Next, we calculate the overall heat transfer coefficient (U) using the
formula:
U=1
hi
+A
k+1
ho−1
where: - hiis the heat transfer coefficient inside the rod, - hois the heat transfer
coefficient at the outer surface of the rod, - Ais the cross-sectional area of the
rod.
Given that hiand hoare the same due to one-dimensional heat conduction,
we get:
U=A
k+A
k−1
=2
400 +2
400−1
=1
0.01 = 100 W/(m2·K)
Step 3: Using the thermal resistance-capacitance method, the temperature
at any time tcan be calculated using the following equation:
T(t) = T∞+ (T0−T∞)·e−t
τ
where: - T(t) is the temperature at time t, - T∞is the ambient temperature, -
T0is the initial temperature, - τ=C·ρ·V
A·Uis the time constant derived from the
thermal properties of the object.
Step 4: Substituting the given values into the equation, we have:
50 = 20 + (200 −20) ·e
−t
C·ρ·V
A·U
30 = 180 ·e
−t
C·ρ·V
A·U
Step 5: Solving for the time constant τ:
t
τ= ln 180
30 = ln(6)
t= ln(6) ·τ
Thus, the time required for the rod to reach a temperature of 50◦C is ln(6) ·
C·ρ·V
A·U.
5
Question 6
Question
A sample of gas is heated at constant volume until the pressure doubles. If the
initial pressure was 1 atm, what is the final pressure of the gas in atm?
Solution
Step 1: Use the ideal gas law to relate the initial and final pressures of the gas.
- The ideal gas law is given by: P V =nRT , where Pis the pressure, Vis the
volume, nis the number of moles of gas, Ris the ideal gas constant, and Tis
the temperature in Kelvin.
Step 2: Since the volume is constant, the ideal gas law simplifies to P1=
nRT1and P2=nRT2, where P1and P2are the initial and final pressures, T1is
the initial temperature, and T2is the final temperature.
Step 3: Since the gas is heated at constant volume, the number of moles and
the gas constant are constant, so we have P2
P1=T2
T1.
Step 4: If the pressure doubles, then the final pressure P2= 2P1, and using
the relation from step 3, we have 2P1
1=T2
T1.
Step 5: Solving for the final temperature T2, we have T2= 2T1.
Step 6: Recall that temperature must be in Kelvin, so if the initial temper-
ature T1= 300 K, then the final temperature T2= 2 ×300 = 600 K.
Step 7: Recall the ideal gas law P V =nRT and use it to find the final pres-
sure P2. - Since P2=nRT2and T2= 600 K, and assuming n= 1 mole and R=
0.0821 atm L/(mol K), we have P2= (1 mol)(0.0821 atm L/(mol K))(600 K). -
Therefore, P2= 49.26 atm.
Final Answer
The final pressure of the gas is 49.26 atm.
Question 7
Question
A copper rod of length 0.5 m and diameter 0.02 m is used to conduct heat from
a furnace to a heat exchanger. The temperature of the hot end of the rod is 350
K and the temperature of the cold end is 300 K. The thermal conductivity of
copper is 400 W/(m K). Calculate the rate of heat conduction through the rod.
6
Solution
Step 1: Calculate the cross-sectional area of the rod. The cross-sectional area
Aof the rod can be calculated using the formula for the area of a circle:
A=πd2
4
where dis the diameter of the rod. Substituting d= 0.02 m into the formula
gives:
A=π×(0.02)2
4= 3.14 ×10−4m2
Step 2: Calculate the temperature difference across the rod. The tempera-
ture difference ∆Tacross the rod is given by:
∆T= 350 K −300 K = 50 K
Step 3: Calculate the rate of heat conduction. The rate of heat conduction
Qthrough 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
of the rod, ∆Tis the temperature difference, and Lis the length of the rod.
Substituting the given values gives:
Q=400 ×3.14 ×10−4×50
0.5= 40 W
Therefore, the rate of heat conduction through the rod is 40 W.
Question 8
Question
A 2 kg block of aluminum is initially at a temperature of 20
°
C. How much heat
is needed to raise the temperature of the block to 150
°
C? The specific heat
capacity of aluminum is 900 J/kg◦C.
Solution
Step 1: Determine the change in temperature. Given: Initial temperature,
Ti= 20◦C Final temperature, Tf= 150◦C
The change in temperature, ∆T=Tf−Ti= 150◦C−20◦C = 130◦C
Step 2: Calculate the heat required. The formula for heat (Q) is given by:
Q=mc∆T
7
where: m= mass of the aluminum block = 2 kg c= specific heat capacity of
aluminum = 900 J/kg◦C ∆T= change in temperature = 130
°
C
Substitute the values into the formula:
Q= 2 kg ×900 J/kg◦C×130◦C
Q= 2 ×900 ×130 J
Q= 234000 J
Therefore, the amount of heat needed to raise the temperature of the block
to 150
°
C is 234,000 J.
Question 9
Question
A copper sphere of radius 5 cm is heated from 20
°
C to 200
°
C. Calculate the heat
energy required to achieve this temperature change. The specific heat capacity
of copper is 0.385 J/g
°
C.
Solution
Step 1: Calculate the mass of the copper sphere using its volume and density.
The volume of a sphere is given by the formula V=4
3πr3, where ris the radius.
Given r= 5 cm, we have V=4
3π(5)3=500π
3cm3. The density of copper is
approximately 8.96 g/cm
³
. Using the formula Density = Mass
Volume , we can find
the mass of the sphere: 8.96 = m
500π
3
⇒m= 8.96 ×500π
3=4478π
75 g.
Step 2: Calculate the increase in temperature. The change in temperature
is given by ∆T=Tfinal −Tinitial = 200 −20 = 180
°
C.
Step 3: Calculate the heat energy using the formula Q=mc∆T, where mis
the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Substitute the values m=4478π
75 g, c= 0.385 J/g
°
C, and ∆T= 180
°
C into
the formula: Q=4478π
75 (0.385)(180) Q=4478π×0.385×180
75 Q=4478π×69
75 Q=
308922π
75 Q≈4104.3 J.
Therefore, the heat energy required to achieve the temperature change is
approximately 4104.3 J.
Question 10
Question
A 1 kg block of copper is initially at a temperature of 100◦C. The block is
then placed in contact with a large ice bath at 0◦C. Assuming no heat is lost
to the surroundings, calculate the final temperature of the block when thermal
equilibrium is reached. The specific heat capacity of copper is c= 385 J/kg◦C.
8
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100◦C
to the final temperature: The heat lost by the copper block is given by the
formula:
Q=mc∆T
where: - mis the mass of the block (1 kg), - cis the specific heat capacity of
copper (385 J/kg◦C), - ∆Tis the change in temperature.
The change in temperature is 100◦C - Tf(Tfis the final temperature).
Therefore, the heat lost by the copper block is:
Q= 1 kg ×385 J/kg◦C×(100◦C−Tf)
Step 2: Calculate the heat gained by the copper block as it warms up in the
ice bath: The heat gained by the copper block is given by the formula:
Q=mc∆T
The change in temperature is Tf−0◦C. Therefore, the heat gained by the
copper block is:
Q= 1 kg ×385 J/kg◦C×(Tf−0◦C)
Step 3: Since no heat is lost to the surroundings, the heat lost by the block
is equal to the heat gained by the block. Equating the two expressions for heat
and solving for Tf:
1×385 ×(100 −Tf)=1×385 ×(Tf−0)
38500 −385Tf= 385Tf
385Tf+ 385Tf= 38500
770Tf= 38500
Tf=38500
770 = 50◦C
Therefore, the final temperature of the block when thermal equilibrium is
reached is 50◦C.
Question 11
Question
A copper block of mass 0.5 kg at a temperature of 100
°
C is placed in a calorime-
ter containing 1 kg of water at 20
°
C. The final equilibrium temperature of the
system is 30
°
C. Assuming no heat is lost to the surroundings, calculate the
specific heat capacity of the copper block.
9
Solution
Step 1: Calculate the heat lost by the copper block when it cools down from
100
°
C to 30
°
C. The heat lost by the copper block is given by the equation:
Qlost =mc∆T
where: m= mass of the copper block = 0.5 kg c= specific heat capacity of
copper ∆T= change in temperature = (100
°
C - 30
°
C) = 70
°
C
Substitute the values into the equation:
Qlost = 0.5×c×70
Step 2: Calculate the heat gained by the water when it warms up from 20
°
C
to 30
°
C. The heat gained by the water is given by the equation:
Qgained =mc∆T
where: m= mass of the water = 1 kg c= specific heat capacity of water (known,
c= 4186 J/kg
°
C) ∆T= change in temperature = (30
°
C - 20
°
C) = 10
°
C
Substitute the values into the equation:
Qgained = 1 ×4186 ×10
Step 3: Since heat lost by the copper block = heat gained by the water, we
have:
0.5×c×70 = 1 ×4186 ×10
Solve for cto find the specific heat capacity of the copper block.
Question 12
Question
A piece of metal of mass 0.5 kg is heated to 100◦C and then placed in 2 kg
of water at 20◦C. If the final temperature of the system is 25◦C, calculate the
specific heat capacity of the metal. Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat gained by the water when the metal is placed in it.
The heat gained by the water can be calculated using the formula:
Qwater =mwatercwater∆T
where: - mwater = 2 kg is the mass of water - cwater = 4186 J/kg◦C is the
specific heat capacity of water - ∆T=Tfinal −Tinitial = 25 −20 = 5 ◦C is the
change in temperature
Substitute these values into the formula to find Qwater.
10
Step 2: Calculate the heat lost by the metal. The heat lost by the metal can
be calculated using the formula:
Qmetal =mmetalcmetal∆T
where: - mmetal = 0.5 kg is the mass of the metal (given) - cmetal is the specific
heat capacity of the metal (to be found) - ∆T=Tfinal −Tinitial = 100 −25 = 75
◦C is the change in temperature
Substitute these values into the formula to find Qmetal.
Step 3: Equate the heat gained by the water to the heat lost by the metal.
Since we are assuming no heat is lost to the surroundings, we can say:
Qwater =Qmetal
Step 4: Solve for the specific heat capacity of the metal, cmetal. Set up an
equation using the values calculated in Steps 1 and 2, then solve for cmetal.
Question 13
Question
A copper container contains 200 g of water at 20
°
C. If 100 g of ice at -10
°
C
is added to the water, what will be the final temperature of the mixture, as-
suming no heat is lost to the surroundings? (Specific heat capacity of water =
4.18 J/g
°
C, specific heat capacity of copper = 0.385 J/g
°
C, heat of fusion for ice
= 334 J/g)
Solution
Step 1: Calculate the heat absorbed by the water to reach its final temperature.
Let the final temperature of the mixture be T
°
C. The heat lost by the ice is
equal to the heat gained by the water:
mice ·cice ·(0 −T) = mwater ·cwater ·(T−20)
Substitute the given values:
100 g ·334 J/g ·(0 −T) = 200 g ·4.18 J/g
°
C·(T−20)
Step 2: Solve the equation to find the final temperature, T.
−33400 ·T= 836 ·(T−20)
−33400 ·T= 836T−16720
−34236T=−16720
T=−16720
−34236 ≈0.49
°
C
Thus, the final temperature of the mixture will be approximately 0.49
°
C.
11
Question 14
Question
A 200 g block of copper at 150
°
C is placed in 500 g of water at 25
°
C in a perfectly
insulated container. Assuming no heat is lost to the surroundings, what is the
final temperature of the system? (Specific heat capacity of copper = 0.39 J/g
°
C,
specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the copper block and the water.
The heat gained or lost (Q) by an object can be calculated using the formula:
Q=mc∆T
where mis the mass of the object, cis the specific heat capacity, and ∆Tis
the change in temperature.
For the copper block:
Qcopper = (200 g)(0.39 J/g
°
C)(Tfinal −150)
For the water:
Qwater = (500 g)(4.18 J/g
°
C)(Tfinal −25)
Since no heat is lost to the surroundings, the heat gained by the copper
block must be equal to the heat lost by the water:
Qcopper =−Qwater
Step 2: Set up the equation based on the principle of conservation of energy.
(200)(0.39)(Tfinal −150) = −(500)(4.18)(Tfinal −25)
Step 3: Solve for the final temperature (Tfinal).
78(Tfinal −150) = −2090(Tfinal −25)
78Tfinal −11700 = −2090Tfinal + 52250
2168Tfinal = 63950
Tfinal = 29.5C
Therefore, the final temperature of the system is 29.5
°
C.
12
Question 15
Question
An aluminum cube with a side length of 10 cm is heated from an initial tem-
perature of 20
°
C to a final temperature of 80
°
C. Calculate the amount of heat
transferred to the cube, given that the specific heat capacity of aluminum is
0.903 J/g
°
C.
Solution
Step 1: Calculate the mass of the aluminum cube. Given that the side length
of the cube is 10 cm, the volume of the cube is V= (10 cm)3= 1000 cm3=
1000 mL = 1000 g (since 1 mL of water = 1 g). The density of aluminum is
approximately 2.70 g/cm3, therefore the mass of the cube is 1000 g.
Step 2: Calculate the change in temperature. The change in temperature is
∆T=Tf−Ti= 80C−20C= 60C.
Step 3: Calculate the amount of heat transferred. The amount of heat
transferred can be calculated using the formula:
Q=mc∆T
where: - Qis the amount of heat transferred, - mis the mass of the aluminum
cube, - cis the specific heat capacity of aluminum, - ∆Tis the change in
temperature.
Plugging in the values:
Q= (1000 g)(0.903 J/g
°
C)(60C)
Q= 54,180 J
Therefore, the amount of heat transferred to the aluminum cube is 54,180
J.
Question 16
Question
A copper rod of length 2 m and cross-sectional area 0.1 m2is heated from 20◦C
to 80◦C. If the thermal conductivity of copper is 400 W/(m*K), determine the
rate at which heat flows through the rod.
Solution
Step 1: Calculate the temperature difference, ∆T. Given: Initial temperature,
T1= 20◦C Final temperature, T2= 80◦C Temperature difference:
∆T=T2−T1= 80 −20 = 60◦C
13
Step 2: Calculate the rate of heat flow, ˙
Q. The rate of heat flow is given by
Fourier’s law of heat conduction:
˙
Q=k·A·∆T
L
where: k= thermal conductivity of copper = 400 W/(m*K) A= cross-sectional
area of the rod = 0.1 m2L= length of the rod = 2 m Plugging in the values:
˙
Q=400 ×0.1×60
2
˙
Q= 1200 W
Therefore, the rate at which heat flows through the rod is 1200 W.
Question 17
Question
A copper cube with sides of length 10 cm is initially at a temperature of 200
°
C.
It is then heated until its temperature rises to 250
°
C. If the specific heat of
copper is 0.385 J/g
°
C, calculate the amount of heat energy required to heat the
cube.
Solution
Step 1: Calculate the mass of the copper cube. Given that the density of copper
is 8.96 g/cm3, we can find the mass of the cube using the formula:
mass = density ×volume
mass = 8.96 g/cm3×(10 cm)3
mass = 8.96 g/cm3×1000 cm3
mass = 8960 g = 8.96 kg
Step 2: Calculate the change in temperature. The change in temperature of
the cube is:
∆T=Tfinal −Tinitial
∆T= 250◦C−200◦C = 50◦C
Step 3: Calculate the amount of heat energy required. The formula to
calculate the heat energy is:
Q=mc∆T
where: Q= heat energy, m= mass of the copper cube, c= specific heat of
copper, ∆T= change in temperature.
14
Plugging in the values:
Q= 8.96 kg ×0.385 J/g
°
C×50
°
C
Q= 8.96 ×385 ×50 J = 17304 J
Therefore, the amount of heat energy required to heat the copper cube is
17304 J.
Question 18
Question
An aluminum block with a mass of 2 kg and a specific heat capacity of 900 J/kg◦C
is initially at a temperature of 100◦C. It is submerged in 3 kg of water at 20◦C.
Assuming no heat is exchanged with the surroundings, calculate the final equi-
librium temperature of the system.
Solution
Step 1: Calculate the heat transfer from the aluminum block to the water using
the formula:
Q=mc∆T
where: - Qis the heat transfer - mis the mass - cis the specific heat capacity
- ∆Tis the change in temperature
For the aluminum block:
∆Taluminum =Tfinal −Tinitial =Tfinal −100
Qaluminum = 2 ×900 ×(Tfinal −100)
Step 2: Calculate the heat transfer to the water using the same formula:
∆Twater =Tfinal −Tinitial =Tfinal −20
Qwater = 3 ×4186 ×(Tfinal −20)
Step 3: Since heat is conserved (no heat exchange with surroundings), the
heat lost by the aluminum block is equal to the heat gained by the water:
2×900 ×(Tfinal −100) = 3 ×4186 ×(Tfinal −20)
Step 4: Solve for Tfinal by simplifying the equation from Step 3. This involves
expanding and rearranging terms to isolate Tfinal. Once done, solve for Tfinal to
find the equilibrium temperature of the system.
15
Question 19
Question
A metal rod of length 1 m and uniform cross-sectional area of 0.01 m
²
is initially
at a temperature of 300 K. The rod is heated until it reaches a final temperature
of 400 K. If the specific heat capacity of the metal is 500 J/kg
·
K and the density
is 8000 kg/m
³
, calculate the amount of heat energy required to increase the
temperature of the rod.
Solution
Step 1: Firstly, we need to calculate the mass of the metal rod. Given that the
density of the metal is 8000 kg/m
³
and the volume of the rod is 0.01 m
²
×1 m =
0.01 m
³
, we have:
Mass = Density ×Volume = 8000 kg/m
³
×0.01 m
³
= 80 kg
Step 2: Next, we can calculate the change in temperature of the rod.
∆T=Tf−Ti= 400 K −300 K = 100 K
Step 3: Now, we can calculate the amount of heat energy required to increase
the temperature of the rod using the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity, and ∆Tis the
change in temperature. Plugging in the values, we get:
Q= 80 kg ×500 J/kg
·
K×100 K = 400,000 J
Therefore, the amount of heat energy required to increase the temperature
of the rod is 400,000 J.
Question 20
Question
A copper ball of mass 500 g at a temperature of 100
°
C is dropped into a container
of water at 20
°
C. If the final temperature of the system is 25
°
C and the specific
heat capacity of copper is 0.385 J/g
°
C, determine the mass of water in the
container. Assume no heat is lost to the surroundings.
16
Solution
Step 1: Use the formula for thermal energy to find the heat lost by the copper
ball and heat gained by the water: The thermal energy formula for a substance
is q=mc∆T, where - qis the heat energy in Joules (J), - mis the mass in grams
(g), - cis the specific heat capacity in J/g
°
C, - ∆Tis the change in temperature
in
°
C.
For the copper ball: qcopper =mcopperccopper∆Tcopper
For the water: qwater =mwatercwater∆Twater
Given that the final temperature is 25
°
C (which means ∆Tcopper = 75C
and ∆Twater = 5C), we can rewrite the thermal energy equation as follows:
qcopper = 500 ×0.385 ×75 qwater =mwater ×4.18 ×5
Step 2: Since the system is isolated and no heat is lost to the surroundings,
the heat lost by the copper ball must be equal to the heat gained by the water:
mcopperccopper∆Tcopper =mwatercwater∆Twater
Substitute the values and solve for mwater: 500×0.385×75 = mwater×4.18×5
1443.75 = 20.9mwater mwater =1443.75
20.9
Therefore, the mass of water in the container is approximately 69.11 g .
Question 21
Question
A 2 kg aluminum block initially at 100
°
C is dropped into a tank containing 10 kg
of water at 20
°
C. Assuming no heat is lost to the surroundings, calculate the final
temperature of the system when thermal equilibrium is reached. The specific
heat capacity of aluminum is 900 J/kg◦C and that of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the aluminum block as it cools down to the
final temperature. The heat lost by the aluminum block can be calculated using
the formula:
Q=mc∆T
where: - m= 2 kg is the mass of the aluminum block, - c= 900 J/kg◦C is the
specific heat capacity of aluminum, and - ∆Tis the temperature change of the
aluminum block.
Given that the initial temperature of the aluminum block is 100
°
C and the
final temperature is T
°
C, we have ∆T= 100 −T= 100 −T
°
C.
Therefore, the heat lost by the aluminum block is:
Qaluminum = 2 ×900 ×(100 −T)
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. The heat gained by the water can be calculated using the same
17
formula as above:
Q=mc∆T
where: - m= 10 kg is the mass of the water, - c= 4200 J/kg◦C is the specific
heat capacity of water, and - ∆Tis the temperature change of the water.
Given that the initial temperature of the water is 20
°
C and the final tem-
perature is T
°
C, we have ∆T=T−20
°
C.
Therefore, the heat gained by the water is:
Qwater = 10 ×4200 ×(T−20)
Step 3: Set up the equation for heat transfer at thermal equilibrium. Since
the total heat lost by the aluminum block must be equal to the total heat gained
by the water at thermal equilibrium, we can set up the equation:
Qaluminum =Qwater
2×900 ×(100 −T) = 10 ×4200 ×(T−20)
Step 4: Solve for the final temperature T.
1800 ×(100 −T) = 42000 ×(T−20)
180000 −1800T= 42000T−840000
222000 = 6000T
T= 37C
Therefore, the final temperature of the system when thermal equilibrium is
reached is 37C.
Question 22
Question
A copper rod of length 1.5 m and cross-sectional area 4 cm2has one end kept
at 100◦C and the other end at 50◦C. If the thermal conductivity of copper is
400 W/(m·K), determine the rate at which heat is conducted along the rod.
Solution
Step 1: Calculate the temperature difference along the rod. Given that the
temperature at one end of the rod is 100◦C and at the other end is 50◦C, the
temperature difference (∆T) is:
∆T= (100 −50)◦C = 50◦C
Step 2: Calculate the rate of heat conduction using Fourier’s law of heat
conduction which states: q=−kA ∆T
L. Given that the thermal conductivity k
18
of copper is 400 W/(m·K), the cross-sectional area Ais 4 cm2= 4 ×10−4m2,
and the length Lof the rod is 1.5 m, we can substitute these values into the
formula:
q=−400 ×4×10−4×50
1.5
Step 3: Calculate the rate of heat conduction.
q=−0.4×4×10−4×33.3 = −0.05328 W
Therefore, the rate at which heat is conducted along the rod is 0.05328 W.
Question 23
Question
A copper ball of mass 400 g is heated to 150◦C and then dropped into a vessel
containing 800 g of water at 20◦C. If the final temperature of the mixture is
30◦C, determine the specific heat capacity of the copper ball. Assume no heat
is lost to the surroundings.
Solution
Step 1: Find the heat absorbed by the copper ball from the initial temperature
to the final temperature.
Q1=mc∆T
Q1= (0.4 kg)(386 J/kg◦C)(30◦C−150◦C)
Q1=−46480 J
Step 2: Find the heat released by the copper ball to raise the temperature
of the water from 20◦C to 30◦C.
Q2=mc∆T
Q2= (0.8 kg)(4186 J/kg◦C)(30◦C−20◦C)
Q2= 33488 J
Step 3: As no heat is lost to the surroundings, the heat absorbed by the
copper ball equals the heat released by it, so we have:
Q1=−Q2
Step 4: Equating Q1and Q2and solving for the specific heat capacity of the
copper ball, c:
−46480 = 33488
c=46480
33488
c≈1.387 J/kg◦C
19
Therefore, the specific heat capacity of the copper ball is approximately
1.387 J/kg◦C.
Question 24
Question
A copper ball at 100◦C is dropped into a large vat of water at 20◦C. If the mass
of the copper ball is 0.5 kg and the heat capacity of copper is 390 J/kg·K, and
the specific heat capacity of water is 4186 J/kg·K, what is the final temperature
of the copper ball and water when they reach thermal equilibrium?
Solution
Step 1: Calculate the heat lost by the copper ball as it cools down to the final
temperature. The heat lost is given by the equation:
Qlost =−mc∆T,
where: - mis the mass of the copper ball (0.5 kg), - cis the heat capacity of
copper (390 J/kg·K), - ∆Tis the change in temperature of the copper ball.
The change in temperature is given by:
∆T=Tfinal −Tinitial.
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. The heat gained is given by the equation:
Qgained =mc∆T,
where: - mis the mass of the water (equal to the mass of the copper ball, 0.5
kg), - cis the specific heat capacity of water (4186 J/kg·K), - ∆Tis the change
in temperature of the water.
Step 3: Set the heat lost equal to the heat gained to find the final tempera-
ture.
Qlost =Qgained.
Do you want to proceed with calculating the final temperature using the
steps above?
Question 25
Question
A 0.5 kg aluminum block initially at 100◦C is placed in a 1 kg water bath at
20◦C. Assume no heat is lost to the surroundings. Given that the specific heat
capacity of aluminum is 900 J/kg◦C and the specific heat capacity of water is
4186 J/kg◦C, determine the final equilibrium temperature of the system.
20
Solution
Step 1: Identify the information given in the question.
In this problem, we are given: - Mass of the aluminum block, mAl = 0.5 kg
- Initial temperature of the aluminum block, Ti,Al = 100◦C - Specific heat
capacity of aluminum, CAl = 900 J/kg◦C - Mass of the water bath, mw= 1 kg
- Initial temperature of the water bath, Ti,w= 20◦C - Specific heat capacity of
water, Cw= 4186 J/kg◦C
Step 2: Determine the heat lost by the aluminum block and gained by the
water bath.
The heat lost by the aluminum block is equal to the heat gained by the water
bath. Therefore, we can write:
mAl ·CAl ·(T−Ti,Al) = mw·Cw·(T−Ti,w)
where Tis the final equilibrium temperature of the system.
Step 3: Solve for the final equilibrium temperature.
Substitute the given values and solve for T:
0.5·900 ·(T−100) = 1 ·4186 ·(T−20)
450(T−100) = 4186(T−20)
450T−45000 = 4186T−83720
3736T= 38720
T=38720
3736
T≈10.36◦C
Therefore, the final equilibrium temperature of the system is approximately
10.36◦C.
Question 26
Question
An insulated calorimeter contains 0.2 kg of water at 20
°
C. A 0.1 kg piece of cop-
per at 90
°
C is dropped into the water. Assuming no heat is lost to the surround-
ings, calculate the final temperature of the system. (Specific heat capacity of
water = 4.18×103J/kg
°
C, specific heat capacity of copper = 0.39×103J/kg
°
C).
Solution
Let’s denote the final temperature of the system as T. Since no heat is lost to
the surroundings, the total heat gained by copper must equal the total heat lost
by water. We’ll use the formula Q=mc∆T, where Qis the heat energy, mis
the mass, cis the specific heat capacity, and ∆Tis the temperature change.
21
Step 1: Calculate heat lost by water
The heat lost by water can be calculated as:
Qwater =mwater ×cwater ×(T−20)
Step 2: Calculate heat gained by copper
The heat gained by copper can be calculated as:
Qcopper =mcopper ×ccopper ×(90 −T)
Step 3: Equate the two heat values
Setting the heat lost by water equal to the heat gained by copper, we have:
mwater ×cwater ×(T−20) = mcopper ×ccopper ×(90 −T)
Step 4: Solve for T
Plugging in the values:
0.2×4.18 ×103×(T−20) = 0.1×0.39 ×103×(90 −T)
Solving the equation will give us the final temperature T.
Question 27
Question
A solid aluminum cube with a side length of 10 cm at 20◦C is heated until its
temperature reaches 70◦C. The coefficient of linear expansion for aluminum is
2.3×10−5◦C−1. Calculate the increase in volume of the cube as a result of the
temperature increase.
Solution
Step 1: Find the increase in length of the cube due to the temperature increase.
Given the coefficient of linear expansion, we can use the formula:
∆L=αLi∆T
where - ∆Lis the change in length, - αis the coefficient of linear expansion, -
Liis the initial length, - ∆Tis the change in temperature.
Substitute the values:
∆L= (2.3×10−5◦C−1)(10 cm)(70 −20)◦C
∆L= 0.03 cm
Step 2: Find the increase in volume of the cube. The volume of a cube is
given by V=L3. Let Vibe the initial volume and Vfbe the final volume. The
increase in volume can be calculated as:
∆V=Vf−Vi= (Li+ ∆L)3−L3
i
22
∆V= (10 cm + 0.03 cm)3−(10 cm)3
∆V≈0.0927 cm3
Therefore, the increase in volume of the aluminum cube as a result of the
temperature increase is approximately 0.0927 cm3.
Question 28
Question
A 500 g aluminum block is initially at a temperature of 100
°
C. The block is
placed in a container of water at 20
°
C. If the final equilibrium temperature
of the block and water is 25
°
C, calculate the mass of water in the container.
Assume no heat is lost to the surroundings.
Solution
Step 1: Find the heat lost by the aluminum block. The specific heat capacity of
aluminum is cAl = 0.904 J/g
°
C. The change in temperature (∆TAl) of the block
is 100 −25 = 75
°
C.
The heat lost by the block can be calculated using the formula:
QAl =mcAl∆TAl
Substitute the given values:
QAl = 500 g ×0.904 J/g
°
C×75
°
C
Calculating:
QAl = 33900 J
Step 2: Find the heat gained by the water. Let the mass of water in the
container be mwater grams. The specific heat capacity of water is cwater = 4.18
J/g
°
C. The change in temperature (∆Twater) of the water is 25 −20 = 5
°
C.
The heat gained by the water can be calculated using the formula:
Qwater =mcwater∆Twater
Substitute the given values:
Qwater =mwater g×4.18 J/g
°
C×5
°
C
Step 3: Set up the heat balance equation. Since no heat is lost to the
surroundings, we can set up the heat balance equation:
QAl =Qwater
Substitute the expressions for QAl and Qwater:
33900 J = mwater ×4.18 J/g
°
C×5
°
C
23
Solving for mwater:
mwater =33900 J
4.18 J/g
°
C×5
°
C
Calculating:
mwater = 1618.71 g
Therefore, the mass of water in the container is 1618.71 grams.
Question 29
Question
A copper rod of length 2 m has a cross-sectional area of 4 cm2. If the rod is
heated from 20
°
C to 90
°
C, calculate the increase in length of the rod. The linear
expansion coefficient of copper is 1.7×10−5K−1.
Solution
Step 1: Calculate the original length of the rod.
The cross-sectional area of the rod is 4 cm2= 4 ×10−4m2. Since the length of
the rod is 2 m, the volume V1of the rod is given by:
V1= Area ×Length = 4 ×10−4m2×2 m = 8 ×10−4m3
Step 2: Calculate the final volume of the rod.
The final length of the rod after heating is given by:
L2=L1(1 + α∆T)
where L1= 2 m is the original length, α= 1.7×10−5K−1is the linear expansion
coefficient, and ∆T= 90◦C−20◦C = 70◦C is the increase in temperature.
Therefore, the final volume V2is given by:
V2= Area×Length2= 4×10−4m2×2 m×(1+1.7×10−5K−1×70) = 8.028×10−4m3
Step 3: Calculate the increase in length.
The increase in volume ∆Vis given by:
∆V=V2−V1= 8.028 ×10−4m3−8×10−4m3= 0.028 ×10−4m3
Step 4: Calculate the increase in length ∆L.
Since the cross-sectional area remains constant, the increase in length is pro-
portional to the increase in volume. Therefore,
∆L
L1
=∆V
V1
∆L=L1×∆V
V1
= 2 m ×0.028 ×10−4m3
8×10−4m3= 0.007 m
Therefore, the increase in length of the copper rod is 0.007 meters.
24
Question 30
Question
A 2.0 kg piece of copper at 90◦C is placed in a container with 1.0 kg of water
at 20◦C. If the container is perfectly insulated and the specific heat capacity of
copper is 387 J/kg·K, specific heat capacity of water is 4186 J/kg·K, and the
latent heat of fusion of water is 334,000 J/kg, determine the final equilibrium
temperature of the system.
Solution
Step 1: Calculate the energy transferred from the copper to the water due to
heat conduction. The heat lost by the copper is equal to the heat gained by the
water:
m1c1(Tf−T1) = m2c2(Tf−T2) + m2L
where: - m1and T1are the mass and initial temperature of the copper, - m2
and T2are the mass and initial temperature of the water, - c1and c2are the
specific heat capacities of the copper and water, and - Lis the latent heat of
fusion of water.
Given: - m1= 2.0 kg, T1= 90 ◦C = 363 K, c1= 387 J/kg·K, - m2= 1.0
kg, T2= 20 ◦C = 293 K, c2= 4186 J/kg·K, - L= 334,000 J/kg.
Substitute the values:
2.0×387 ×(Tf−363) = 1.0×4186 ×(Tf−293) + 1.0×334,000
Step 2: Solve for the final equilibrium temperature Tf. Expand and simplify
the equation to solve for Tf:
774(Tf−363) = 4186(Tf−293) + 334,000
774Tf−281,262 = 4186Tf−1,229,198 + 334,000
774Tf= 4186Tf−1,125,460
3445Tf= 1,125,460
Tf=1,125,460
3445 ≈326.91
Therefore, the final equilibrium temperature of the system is approximately
326.91 K.
Question 31
Question
A copper cup has a mass of 250 g and is filled with 500 g of water at 25
°
C. If the
cup and water are heated until they reach thermal equilibrium, determine the
final temperature of the system. Assume the specific heat capacity of copper is
0.385 J/g
°
C and that of water is 4.18 J/g
°
C.
25
Solution
Step 1: Calculate the heat absorbed by the copper cup to reach the final tem-
perature.
Given: Mass of copper cup, mcopper = 250 g Specific heat capacity of copper,
ccopper = 0.385 J/g
°
C Initial temperature of copper cup, Tcopper initial = 25C
Final temperature of the system, Tfinal
The heat absorbed by the copper cup is given by the formula:
Qcopper =mcopper ·ccopper ·∆Tcopper
where ∆Tcopper =Tfinal −Tcopper initial
Step 2: Calculate the heat absorbed by the water to reach the final temper-
ature.
Given: Mass of water, mwater = 500 g Specific heat capacity of water,
cwater = 4.18 J/g
°
C Initial temperature of water, Twater initial = 25C
Since the copper cup and the water reach thermal equilibrium, the heat
absorbed by the water must be equal to the heat absorbed by the copper cup:
Qcopper =Qwater
Step 3: Set up and solve the equation to find the final temperature of the
system.
Since the heat absorbed by the water is given by:
Qwater =mwater ·cwater ·∆Twater
where ∆Twater =Tfinal −Twater initial
We can set up the equation:
mcopper ·ccopper ·∆Tcopper =mwater ·cwater ·∆Twater
Substitute the expressions for ∆Tcopper and ∆Twater into the equation:
mcopper ·ccopper ·(Tfinal −Tcopper initial) = mwater ·cwater ·(Tfinal −Twater initial)
Solve for Tfinal to find the final temperature of the system.
Question 32
Question
A copper vessel of mass 0.5 kg contains 2 kg of water at 25◦C. How much
heat must be supplied to the vessel to bring the water to boiling point 100◦C,
assuming no heat is lost to the surrounding?
26
Solution
Step 1: Calculate the heat required to raise the temperature of water from
25◦C to 100◦C. Given: Mass of water, mw= 2 kg Initial temperature of water,
Tinitial = 25◦C Final temperature of water, Tfinal = 100◦C Specific heat capacity
of water, cw= 4186 J/kg◦C
The heat required to raise the temperature of water can be calculated using
the formula:
Q=mw·cw·∆T
where ∆T=Tfinal −Tinitial.
Calculating the heat required:
∆T= 100◦C−25◦C = 75◦C
Q= 2 ·4186 ·75 = 627900 J = 627.9 kJ
Step 2: Calculate the heat required to raise the temperature of the copper
vessel. Given: Mass of copper vessel, mc= 0.5 kg Specific heat capacity of
copper, cc= 386 J/kg◦C Temperature change of copper vessel, ∆Tc= 100◦−
25◦= 75◦C
The heat required to raise the temperature of the copper vessel can be cal-
culated using the formula:
Qc=mc·cc·∆Tc
Calculating the heat required:
Qc= 0.5·386 ·75 = 14475 J = 14.5 kJ
Step 3: Add the heat required for the water and the copper vessel to find
the total heat required.
Qtotal =Q+Qc= 627.9 kJ + 14.5 kJ = 642.4 kJ
Therefore, 642.4 kJ of heat must be supplied to the vessel to bring the water
to boiling point 100◦C.
Question 33
Question
A copper rod of length 2 m and diameter 2 cm is initially at a temperature of
100◦C. It is immersed completely in a water container at 25◦C. Given that the
thermal conductivity of copper is 400 W/mK and the specific heat capacity of
copper is 390 J/kgK, calculate the time it takes for the rod to reach thermal
equilibrium with the water. Assume no heat loss to the surroundings.
27
Solution
Step 1: Find the cross-sectional area of the rod. The cross-sectional area Aof
the rod can be calculated using the formula for the area of a circle: A=πd2
4,
where dis the diameter of the rod. Given that the diameter is 2 cm, we have
d= 0.02 m. Therefore,
A=π(0.02)2
4= 3.14 ×10−4m2
Step 2: Calculate the volume of the rod. The volume Vof the rod can be
calculated using the formula: V=A×L, where Lis the length of the rod.
Given that the length is 2 m, we have
V= 3.14 ×10−4m2×2 m = 6.28 ×10−4m3
Step 3: Find the mass of the rod. The mass of the rod can be calculated using
the formula: m= volume ×density. The density of copper is approximately
8900 kg/m3. Therefore,
m= 6.28 ×10−4m3×8900 kg/m3= 5.5932 kg
Step 4: Calculate the heat energy required for the rod to reach equilibrium.
The heat energy Qrequired can be calculated using the equation: Q=mc∆T,
where ∆Tis the temperature difference between the rod and water, and cis
the specific heat capacity of copper. Given that the initial temperature of the
rod is 100◦C and of the water is 25◦C, we have ∆T= 100◦C−25◦C = 75◦C.
Therefore,
Q= 5.5932 kg ×390 J/kgK ×75 K = 163833 J
Step 5: Calculate the rate of heat transfer through the rod. The rate of
heat transfer Pcan be calculated using Fourier’s law of heat conduction: P=
−kAdT
dx , where kis the thermal conductivity of copper and dT
dx is the temperature
gradient. Since the rod is of constant diameter, the change in temperature along
the rod is uniform. Thus, P=−kA ∆T
L. Given that the thermal conductivity
of copper is 400 W/mK, we have
P=−400 W/mK ×3.14 ×10−4m2×75
2K = −47.1 W
Step 6: Calculate the time taken for the rod to reach thermal equilibrium.
The time ttaken for the rod to reach thermal equilibrium can be calculated
using the equation: Q=P t. Thus,
t=Q
P=163833 J
−47.1 W =−3478.11 s ≈58 minutes
Therefore, it takes approximately 58 minutes for the copper rod to reach
thermal equilibrium with the water.
28
Question 34
Question
A copper cube initially at a temperature of 100◦C is placed in a large thermal
reservoir at 0◦C. The sides of the cube are 10 cm long. Assuming all heat
capacities are constant over the temperature range, calculate how long it will
take for the cube to cool down to 30◦C. The thermal conductivity of copper is
401 W/(m*K) and its density is 8.96 g/cm3.
Solution
Step 1: Find the initial temperature difference between the cube and the reser-
voir. Given: Initial temperature of the cube (Tcube,i) = 100◦C = 100 K, Tem-
perature of the reservoir (Treservoir)=0◦C = 0 K. Initial temperature difference
=Tcube,i −Treservoir = 100 K.
Step 2: Calculate the area of the cube. Given: Side length of the cube (s)
= 10 cm = 0.1 m. Area of one face of the cube = s2. Total surface area of the
cube = 6 times the area of one face. Total surface area = 6 ×0.12m2= 0.06
m2.
Step 3: Calculate the initial rate of energy transfer (heat flow) from the cube
to the reservoir. Using the formula for the rate of heat flow through a material:
Q=kA∆T/d, where: - Q= rate of heat transfer (in watts), - k= thermal
conductivity of copper (401 W/(m*K)), - A= surface area of the cube (0.06
m2), - ∆T= initial temperature difference (100 K), - d= thickness of the cube
(since it is a cube, we can assume the thickness is the same as the side length).
Substitute the values into the formula: Q= 401 ×0.06 ×100/0.1 = 2406 W.
Step 4: Calculate the heat capacity of the cube. Given: Density of copper
= 8.96 g/cm3. The mass of the cube can be calculated using the density and
volume formula: m= density ×volume. Volume of the cube = side length3=
0.13m3. Mass of the cube = 8.96 g/cm3×0.001 m3= 8.96 kg.
The specific heat capacity of copper is 385 J/(kg*K). Heat capacity of the
cube = mass ×specific heat capacity = 8.96 ×385 J/K = 3455.6 J/K.
Step 5: Determine the time taken for the cube to cool down to 30◦C. The
change in temperature (∆T) = 100 K - 30 K = 70 K.
Using the formula for change in temperature in terms of heat capacity and
heat flow: ∆T=Q×t/C, where: - t= time taken (in seconds), - C= heat
capacity of the cube (3455.6 J/K).
Rearranging the formula to solve for t:t=C×∆T/Q.
Substitute the values into the formula: t= 3455.6×70/2406 = 100.4 seconds
=1 minute and 40.4 seconds.
29
Question 35
Question
A copper rod of length 1 m and diameter 2 cm is heated from 20◦C to 120◦C.
Given that the linear expansion coefficient of copper is 1.7×10−5◦C−1and
the specific heat capacity of copper is 0.385 J/g◦C, calculate the heat energy
required for this process.
Solution
Step 1: Calculate the change in length of the copper rod due to heating. The
change in length (∆L) of the copper rod can be calculated using the formula
for linear expansion:
∆L=L·α·∆T
where Lis the original length of the rod, αis the linear expansion coefficient
of copper, and ∆Tis the change in temperature. Substitute L= 1 m, α=
1.7×10−5◦C−1, ∆T= 120◦C−20◦C = 100◦C into the formula:
∆L= 1 m ·1.7×10−5◦C−1·100◦C
∆L= 0.0017 m = 1.7 mm
Step 2: Calculate the volume change of the copper rod. The initial volume
of the rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where ris the radius of the rod and his the original length of the rod. Given
that the diameter of the rod is 2 cm, the radius r= 1 cm = 0.01 m. Substitute
r= 0.01 m and h= 1 m into the formula:
V=π×(0.01)2×1 = π×0.0001 m3
V= 0.000314 m3
The change in volume (∆V) due to the change in length can be calculated
as:
∆V=π×r2×∆L
Substitute r= 0.01 m and ∆L= 0.0017 m into the formula:
∆V=π×0.0001 ×0.0017 = 5.366 ×10−7m3
Step 3: Calculate the mass of the copper rod. The density of copper is
approximately 8900 kg/m3. The mass of the rod can be calculated using the
formula:
m=ρ×V
30
Step 3: Integrate the differential equation. Integrating the equation ob-
tained in Step 1 gives:
ZQx
0
dQ =−kA Zx
0
d
dx(T)dx
Qx=−kA(Tx−T1)
Step 4: Solve the differential equation. Solving the differential equation
and applying the boundary conditions T(0) = 100◦C and T(L) = 0◦C:
T(x) = T1+ (T2−T1)1−x
L
Hence, the temperature as a function of distance xalong the rod in the
steady state is given by T(x) = T1+ (100◦C−T1)1−x
L.
Question 2
Question
A copper rod of length 2.0 m and diameter 1.0 cm is initially at a temperature
of 100◦C. The rod is placed in an ice-water mixture at 0◦C. If the rod loses
800 J of heat to the surroundings, calculate the final temperature of the rod.
Assume that the only significant heat transfer is along the length of the rod.
Solution
Step 1: Calculate the initial temperature difference between the rod and the
ice-water mixture. Given that the initial temperature of the rod is 100◦C and
the ice-water mixture is at 0◦C, the initial temperature difference (∆Ti) is:
∆Ti= 100 −0 = 100◦C
Step 2: Calculate the cross-sectional area of the rod. The radius of the rod
is r=1.0 cm
2= 0.005 m. The cross-sectional area (A) of the rod is:
A=πr2=π(0.005)2= 7.85 ×10−5m2
Step 3: Calculate the volume of the rod. The volume of the rod (V) is:
V=A×L= 7.85 ×10−5×2.0=1.57 ×10−4m3
Step 4: Calculate the mass of the rod. The density of copper is ρ=
8930 kg/m3. The mass of the rod (m) is:
m=ρ×V= 8930 ×1.57 ×10−4= 1.40 kg
Step 5: Calculate the specific heat capacity of copper. The specific heat
capacity of copper (c) is 390 J/(kg·K).
2
Step 6: Calculate the initial heat content of the rod. The initial heat content
of the rod (Qi) is given by:
Qi=mc∆Ti= 1.40 ×390 ×100 = 54,600 J
Step 7: Calculate the final temperature of the rod. Since the rod loses 800
J of heat to the surroundings and no work is done, the final heat content of the
rod (Qf) is Qi−800.
Qf= 54,600 −800 = 53,800 J
Step 8: Calculate the final temperature of the rod. The final temperature
difference (∆Tf) can be calculated using:
Qf=mc∆Tf
∆Tf=Qf
mc =53,800
1.40 ×390100.9◦C
Therefore, the final temperature of the rod is approximately 100.9◦C.
Question 3
Question
A piece of copper of mass 150 g at 150
°
C is placed in 500 g of water at 20
°
C.
Assuming no heat is lost to the surroundings, what is the final temperature of
the system when thermal equilibrium is reached? The specific heat capacity of
copper is 0.387 J/g
°
C and that of water is 4.18 J/g
°
C.
Solution
Step 1: Calculate the heat gained by the copper as it cools down to the final
temperature. The heat gained is equal to the heat lost by the water.
Heat lost by copper = Heat gained by water
mcopper ×ccopper ×(Tfinal −150) = mwater ×cwater ×(Tfinal −20)
Step 2: Substitute the known values into the equation.
150 ×0.387 ×(Tfinal −150) = 500 ×4.18 ×(Tfinal −20)
Step 3: Simplify the equation and solve for Tfinal.
58.05Tfinal −58.05 ×150 = 2090Tfinal −2090 ×20
58.05Tfinal −8707.5 = 2090Tfinal −41800
2090Tfinal −58.05Tfinal = 41800 −8707.5
2031.95Tfinal = 33092.5
Tfinal =33092.5
2031.95 ≈16.28
°
C
Therefore, the final temperature of the system when thermal equilibrium is
reached is approximately 16.28
°
C.
3
Question 4
Question
A piece of copper weighing 300 g at a temperature of 100
°
C is placed in a vessel
containing 200 g of water at 20
°
C. If the final temperature of the mixture is
25
°
C, calculate the specific heat capacity of copper. (Specific heat capacity of
water = 4200 J/kg
°
C)
Solution
Step 1: Calculate the heat lost by the copper piece when it cools from 100
°
C to
25
°
C. The formula for heat lost or gained is given by:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the substance (in kg), -
cis the specific heat capacity of the substance (in J/kg
°
C), - ∆Tis the change
in temperature (in
°
C).
Given that the mass of the copper, m= 0.3 kg, initial temperature, Ti=
100C, final temperature, Tf= 25C, and specific heat capacity of water, cwater =
4200 J/kg
°
C.
Substitute these values into the formula to find the heat lost by the copper:
Qcopper = (0.3)(ccopper)(100 −25)
Step 2: Calculate the heat gained by the water when it heats up from 20
°
C
to 25
°
C. The heat gained by the water is equal to the heat lost by the copper
(assuming no heat is lost to the surroundings). Therefore, we can write:
Qwater = (0.2)(4200)(25 −20)
Step 3: Set the heat lost by the copper equal to the heat gained by the water
and solve for ccopper:
Qcopper =Qwater
Step 4: Substitute the expressions for Qcopper and Qwater into the equation
from Step 3 and solve for ccopper. This will give the specific heat capacity of
copper.
Question 5
Question
A copper rod of length 2 m and cross-sectional area 4 cm2is initially at a uniform
temperature of 200◦C. The rod is then placed in a water bath at 20◦C. If the
thermal conductivity of copper is 400 W/(m·K) and the heat transfer coefficient
between copper and water is 1000 W/(m2·K), calculate the time required for the
rod to reach a temperature of 50◦C. Assume one-dimensional heat conduction
along the rod.
4
Solution
Step 1: First, we calculate the thermal resistance of the rod using the formula:
R=L
k·A
where: - Ris the thermal resistance, - Lis the length of the rod, - kis the thermal
conductivity of the material (copper in this case), - Ais the cross-sectional area
of the rod.
Plugging in the values:
R=2
400 ×4×10−4
R=2
0.16 = 12.5 K/W
Step 2: Next, we calculate the overall heat transfer coefficient (U) using the
formula:
U=1
hi
+A
k+1
ho−1
where: - hiis the heat transfer coefficient inside the rod, - hois the heat transfer
coefficient at the outer surface of the rod, - Ais the cross-sectional area of the
rod.
Given that hiand hoare the same due to one-dimensional heat conduction,
we get:
U=A
k+A
k−1
=2
400 +2
400−1
=1
0.01 = 100 W/(m2·K)
Step 3: Using the thermal resistance-capacitance method, the temperature
at any time tcan be calculated using the following equation:
T(t) = T∞+ (T0−T∞)·e−t
τ
where: - T(t) is the temperature at time t, - T∞is the ambient temperature, -
T0is the initial temperature, - τ=C·ρ·V
A·Uis the time constant derived from the
thermal properties of the object.
Step 4: Substituting the given values into the equation, we have:
50 = 20 + (200 −20) ·e
−t
C·ρ·V
A·U
30 = 180 ·e
−t
C·ρ·V
A·U
Step 5: Solving for the time constant τ:
t
τ= ln 180
30 = ln(6)
t= ln(6) ·τ
Thus, the time required for the rod to reach a temperature of 50◦C is ln(6) ·
C·ρ·V
A·U.
5
Question 6
Question
A sample of gas is heated at constant volume until the pressure doubles. If the
initial pressure was 1 atm, what is the final pressure of the gas in atm?
Solution
Step 1: Use the ideal gas law to relate the initial and final pressures of the gas.
- The ideal gas law is given by: P V =nRT , where Pis the pressure, Vis the
volume, nis the number of moles of gas, Ris the ideal gas constant, and Tis
the temperature in Kelvin.
Step 2: Since the volume is constant, the ideal gas law simplifies to P1=
nRT1and P2=nRT2, where P1and P2are the initial and final pressures, T1is
the initial temperature, and T2is the final temperature.
Step 3: Since the gas is heated at constant volume, the number of moles and
the gas constant are constant, so we have P2
P1=T2
T1.
Step 4: If the pressure doubles, then the final pressure P2= 2P1, and using
the relation from step 3, we have 2P1
1=T2
T1.
Step 5: Solving for the final temperature T2, we have T2= 2T1.
Step 6: Recall that temperature must be in Kelvin, so if the initial temper-
ature T1= 300 K, then the final temperature T2= 2 ×300 = 600 K.
Step 7: Recall the ideal gas law P V =nRT and use it to find the final pres-
sure P2. - Since P2=nRT2and T2= 600 K, and assuming n= 1 mole and R=
0.0821 atm L/(mol K), we have P2= (1 mol)(0.0821 atm L/(mol K))(600 K). -
Therefore, P2= 49.26 atm.
Final Answer
The final pressure of the gas is 49.26 atm.
Question 7
Question
A copper rod of length 0.5 m and diameter 0.02 m is used to conduct heat from
a furnace to a heat exchanger. The temperature of the hot end of the rod is 350
K and the temperature of the cold end is 300 K. The thermal conductivity of
copper is 400 W/(m K). Calculate the rate of heat conduction through the rod.
6
Solution
Step 1: Calculate the cross-sectional area of the rod. The cross-sectional area
Aof the rod can be calculated using the formula for the area of a circle:
A=πd2
4
where dis the diameter of the rod. Substituting d= 0.02 m into the formula
gives:
A=π×(0.02)2
4= 3.14 ×10−4m2
Step 2: Calculate the temperature difference across the rod. The tempera-
ture difference ∆Tacross the rod is given by:
∆T= 350 K −300 K = 50 K
Step 3: Calculate the rate of heat conduction. The rate of heat conduction
Qthrough 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
of the rod, ∆Tis the temperature difference, and Lis the length of the rod.
Substituting the given values gives:
Q=400 ×3.14 ×10−4×50
0.5= 40 W
Therefore, the rate of heat conduction through the rod is 40 W.
Question 8
Question
A 2 kg block of aluminum is initially at a temperature of 20
°
C. How much heat
is needed to raise the temperature of the block to 150
°
C? The specific heat
capacity of aluminum is 900 J/kg◦C.
Solution
Step 1: Determine the change in temperature. Given: Initial temperature,
Ti= 20◦C Final temperature, Tf= 150◦C
The change in temperature, ∆T=Tf−Ti= 150◦C−20◦C = 130◦C
Step 2: Calculate the heat required. The formula for heat (Q) is given by:
Q=mc∆T
7
where: m= mass of the aluminum block = 2 kg c= specific heat capacity of
aluminum = 900 J/kg◦C ∆T= change in temperature = 130
°
C
Substitute the values into the formula:
Q= 2 kg ×900 J/kg◦C×130◦C
Q= 2 ×900 ×130 J
Q= 234000 J
Therefore, the amount of heat needed to raise the temperature of the block
to 150
°
C is 234,000 J.
Question 9
Question
A copper sphere of radius 5 cm is heated from 20
°
C to 200
°
C. Calculate the heat
energy required to achieve this temperature change. The specific heat capacity
of copper is 0.385 J/g
°
C.
Solution
Step 1: Calculate the mass of the copper sphere using its volume and density.
The volume of a sphere is given by the formula V=4
3πr3, where ris the radius.
Given r= 5 cm, we have V=4
3π(5)3=500π
3cm3. The density of copper is
approximately 8.96 g/cm
³
. Using the formula Density = Mass
Volume , we can find
the mass of the sphere: 8.96 = m
500π
3
⇒m= 8.96 ×500π
3=4478π
75 g.
Step 2: Calculate the increase in temperature. The change in temperature
is given by ∆T=Tfinal −Tinitial = 200 −20 = 180
°
C.
Step 3: Calculate the heat energy using the formula Q=mc∆T, where mis
the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Substitute the values m=4478π
75 g, c= 0.385 J/g
°
C, and ∆T= 180
°
C into
the formula: Q=4478π
75 (0.385)(180) Q=4478π×0.385×180
75 Q=4478π×69
75 Q=
308922π
75 Q≈4104.3 J.
Therefore, the heat energy required to achieve the temperature change is
approximately 4104.3 J.
Question 10
Question
A 1 kg block of copper is initially at a temperature of 100◦C. The block is
then placed in contact with a large ice bath at 0◦C. Assuming no heat is lost
to the surroundings, calculate the final temperature of the block when thermal
equilibrium is reached. The specific heat capacity of copper is c= 385 J/kg◦C.
8
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100◦C
to the final temperature: The heat lost by the copper block is given by the
formula:
Q=mc∆T
where: - mis the mass of the block (1 kg), - cis the specific heat capacity of
copper (385 J/kg◦C), - ∆Tis the change in temperature.
The change in temperature is 100◦C - Tf(Tfis the final temperature).
Therefore, the heat lost by the copper block is:
Q= 1 kg ×385 J/kg◦C×(100◦C−Tf)
Step 2: Calculate the heat gained by the copper block as it warms up in the
ice bath: The heat gained by the copper block is given by the formula:
Q=mc∆T
The change in temperature is Tf−0◦C. Therefore, the heat gained by the
copper block is:
Q= 1 kg ×385 J/kg◦C×(Tf−0◦C)
Step 3: Since no heat is lost to the surroundings, the heat lost by the block
is equal to the heat gained by the block. Equating the two expressions for heat
and solving for Tf:
1×385 ×(100 −Tf)=1×385 ×(Tf−0)
38500 −385Tf= 385Tf
385Tf+ 385Tf= 38500
770Tf= 38500
Tf=38500
770 = 50◦C
Therefore, the final temperature of the block when thermal equilibrium is
reached is 50◦C.
Question 11
Question
A copper block of mass 0.5 kg at a temperature of 100
°
C is placed in a calorime-
ter containing 1 kg of water at 20
°
C. The final equilibrium temperature of the
system is 30
°
C. Assuming no heat is lost to the surroundings, calculate the
specific heat capacity of the copper block.
9
Solution
Step 1: Calculate the heat lost by the copper block when it cools down from
100
°
C to 30
°
C. The heat lost by the copper block is given by the equation:
Qlost =mc∆T
where: m= mass of the copper block = 0.5 kg c= specific heat capacity of
copper ∆T= change in temperature = (100
°
C - 30
°
C) = 70
°
C
Substitute the values into the equation:
Qlost = 0.5×c×70
Step 2: Calculate the heat gained by the water when it warms up from 20
°
C
to 30
°
C. The heat gained by the water is given by the equation:
Qgained =mc∆T
where: m= mass of the water = 1 kg c= specific heat capacity of water (known,
c= 4186 J/kg
°
C) ∆T= change in temperature = (30
°
C - 20
°
C) = 10
°
C
Substitute the values into the equation:
Qgained = 1 ×4186 ×10
Step 3: Since heat lost by the copper block = heat gained by the water, we
have:
0.5×c×70 = 1 ×4186 ×10
Solve for cto find the specific heat capacity of the copper block.
Question 12
Question
A piece of metal of mass 0.5 kg is heated to 100◦C and then placed in 2 kg
of water at 20◦C. If the final temperature of the system is 25◦C, calculate the
specific heat capacity of the metal. Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat gained by the water when the metal is placed in it.
The heat gained by the water can be calculated using the formula:
Qwater =mwatercwater∆T
where: - mwater = 2 kg is the mass of water - cwater = 4186 J/kg◦C is the
specific heat capacity of water - ∆T=Tfinal −Tinitial = 25 −20 = 5 ◦C is the
change in temperature
Substitute these values into the formula to find Qwater.
10
Step 2: Calculate the heat lost by the metal. The heat lost by the metal can
be calculated using the formula:
Qmetal =mmetalcmetal∆T
where: - mmetal = 0.5 kg is the mass of the metal (given) - cmetal is the specific
heat capacity of the metal (to be found) - ∆T=Tfinal −Tinitial = 100 −25 = 75
◦C is the change in temperature
Substitute these values into the formula to find Qmetal.
Step 3: Equate the heat gained by the water to the heat lost by the metal.
Since we are assuming no heat is lost to the surroundings, we can say:
Qwater =Qmetal
Step 4: Solve for the specific heat capacity of the metal, cmetal. Set up an
equation using the values calculated in Steps 1 and 2, then solve for cmetal.
Question 13
Question
A copper container contains 200 g of water at 20
°
C. If 100 g of ice at -10
°
C
is added to the water, what will be the final temperature of the mixture, as-
suming no heat is lost to the surroundings? (Specific heat capacity of water =
4.18 J/g
°
C, specific heat capacity of copper = 0.385 J/g
°
C, heat of fusion for ice
= 334 J/g)
Solution
Step 1: Calculate the heat absorbed by the water to reach its final temperature.
Let the final temperature of the mixture be T
°
C. The heat lost by the ice is
equal to the heat gained by the water:
mice ·cice ·(0 −T) = mwater ·cwater ·(T−20)
Substitute the given values:
100 g ·334 J/g ·(0 −T) = 200 g ·4.18 J/g
°
C·(T−20)
Step 2: Solve the equation to find the final temperature, T.
−33400 ·T= 836 ·(T−20)
−33400 ·T= 836T−16720
−34236T=−16720
T=−16720
−34236 ≈0.49
°
C
Thus, the final temperature of the mixture will be approximately 0.49
°
C.
11
Question 14
Question
A 200 g block of copper at 150
°
C is placed in 500 g of water at 25
°
C in a perfectly
insulated container. Assuming no heat is lost to the surroundings, what is the
final temperature of the system? (Specific heat capacity of copper = 0.39 J/g
°
C,
specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the copper block and the water.
The heat gained or lost (Q) by an object can be calculated using the formula:
Q=mc∆T
where mis the mass of the object, cis the specific heat capacity, and ∆Tis
the change in temperature.
For the copper block:
Qcopper = (200 g)(0.39 J/g
°
C)(Tfinal −150)
For the water:
Qwater = (500 g)(4.18 J/g
°
C)(Tfinal −25)
Since no heat is lost to the surroundings, the heat gained by the copper
block must be equal to the heat lost by the water:
Qcopper =−Qwater
Step 2: Set up the equation based on the principle of conservation of energy.
(200)(0.39)(Tfinal −150) = −(500)(4.18)(Tfinal −25)
Step 3: Solve for the final temperature (Tfinal).
78(Tfinal −150) = −2090(Tfinal −25)
78Tfinal −11700 = −2090Tfinal + 52250
2168Tfinal = 63950
Tfinal = 29.5C
Therefore, the final temperature of the system is 29.5
°
C.
12
Question 15
Question
An aluminum cube with a side length of 10 cm is heated from an initial tem-
perature of 20
°
C to a final temperature of 80
°
C. Calculate the amount of heat
transferred to the cube, given that the specific heat capacity of aluminum is
0.903 J/g
°
C.
Solution
Step 1: Calculate the mass of the aluminum cube. Given that the side length
of the cube is 10 cm, the volume of the cube is V= (10 cm)3= 1000 cm3=
1000 mL = 1000 g (since 1 mL of water = 1 g). The density of aluminum is
approximately 2.70 g/cm3, therefore the mass of the cube is 1000 g.
Step 2: Calculate the change in temperature. The change in temperature is
∆T=Tf−Ti= 80C−20C= 60C.
Step 3: Calculate the amount of heat transferred. The amount of heat
transferred can be calculated using the formula:
Q=mc∆T
where: - Qis the amount of heat transferred, - mis the mass of the aluminum
cube, - cis the specific heat capacity of aluminum, - ∆Tis the change in
temperature.
Plugging in the values:
Q= (1000 g)(0.903 J/g
°
C)(60C)
Q= 54,180 J
Therefore, the amount of heat transferred to the aluminum cube is 54,180
J.
Question 16
Question
A copper rod of length 2 m and cross-sectional area 0.1 m2is heated from 20◦C
to 80◦C. If the thermal conductivity of copper is 400 W/(m*K), determine the
rate at which heat flows through the rod.
Solution
Step 1: Calculate the temperature difference, ∆T. Given: Initial temperature,
T1= 20◦C Final temperature, T2= 80◦C Temperature difference:
∆T=T2−T1= 80 −20 = 60◦C
13
Step 2: Calculate the rate of heat flow, ˙
Q. The rate of heat flow is given by
Fourier’s law of heat conduction:
˙
Q=k·A·∆T
L
where: k= thermal conductivity of copper = 400 W/(m*K) A= cross-sectional
area of the rod = 0.1 m2L= length of the rod = 2 m Plugging in the values:
˙
Q=400 ×0.1×60
2
˙
Q= 1200 W
Therefore, the rate at which heat flows through the rod is 1200 W.
Question 17
Question
A copper cube with sides of length 10 cm is initially at a temperature of 200
°
C.
It is then heated until its temperature rises to 250
°
C. If the specific heat of
copper is 0.385 J/g
°
C, calculate the amount of heat energy required to heat the
cube.
Solution
Step 1: Calculate the mass of the copper cube. Given that the density of copper
is 8.96 g/cm3, we can find the mass of the cube using the formula:
mass = density ×volume
mass = 8.96 g/cm3×(10 cm)3
mass = 8.96 g/cm3×1000 cm3
mass = 8960 g = 8.96 kg
Step 2: Calculate the change in temperature. The change in temperature of
the cube is:
∆T=Tfinal −Tinitial
∆T= 250◦C−200◦C = 50◦C
Step 3: Calculate the amount of heat energy required. The formula to
calculate the heat energy is:
Q=mc∆T
where: Q= heat energy, m= mass of the copper cube, c= specific heat of
copper, ∆T= change in temperature.
14
Plugging in the values:
Q= 8.96 kg ×0.385 J/g
°
C×50
°
C
Q= 8.96 ×385 ×50 J = 17304 J
Therefore, the amount of heat energy required to heat the copper cube is
17304 J.
Question 18
Question
An aluminum block with a mass of 2 kg and a specific heat capacity of 900 J/kg◦C
is initially at a temperature of 100◦C. It is submerged in 3 kg of water at 20◦C.
Assuming no heat is exchanged with the surroundings, calculate the final equi-
librium temperature of the system.
Solution
Step 1: Calculate the heat transfer from the aluminum block to the water using
the formula:
Q=mc∆T
where: - Qis the heat transfer - mis the mass - cis the specific heat capacity
- ∆Tis the change in temperature
For the aluminum block:
∆Taluminum =Tfinal −Tinitial =Tfinal −100
Qaluminum = 2 ×900 ×(Tfinal −100)
Step 2: Calculate the heat transfer to the water using the same formula:
∆Twater =Tfinal −Tinitial =Tfinal −20
Qwater = 3 ×4186 ×(Tfinal −20)
Step 3: Since heat is conserved (no heat exchange with surroundings), the
heat lost by the aluminum block is equal to the heat gained by the water:
2×900 ×(Tfinal −100) = 3 ×4186 ×(Tfinal −20)
Step 4: Solve for Tfinal by simplifying the equation from Step 3. This involves
expanding and rearranging terms to isolate Tfinal. Once done, solve for Tfinal to
find the equilibrium temperature of the system.
15
Question 19
Question
A metal rod of length 1 m and uniform cross-sectional area of 0.01 m
²
is initially
at a temperature of 300 K. The rod is heated until it reaches a final temperature
of 400 K. If the specific heat capacity of the metal is 500 J/kg
·
K and the density
is 8000 kg/m
³
, calculate the amount of heat energy required to increase the
temperature of the rod.
Solution
Step 1: Firstly, we need to calculate the mass of the metal rod. Given that the
density of the metal is 8000 kg/m
³
and the volume of the rod is 0.01 m
²
×1 m =
0.01 m
³
, we have:
Mass = Density ×Volume = 8000 kg/m
³
×0.01 m
³
= 80 kg
Step 2: Next, we can calculate the change in temperature of the rod.
∆T=Tf−Ti= 400 K −300 K = 100 K
Step 3: Now, we can calculate the amount of heat energy required to increase
the temperature of the rod using the formula:
Q=mc∆T
where mis the mass of the rod, cis the specific heat capacity, and ∆Tis the
change in temperature. Plugging in the values, we get:
Q= 80 kg ×500 J/kg
·
K×100 K = 400,000 J
Therefore, the amount of heat energy required to increase the temperature
of the rod is 400,000 J.
Question 20
Question
A copper ball of mass 500 g at a temperature of 100
°
C is dropped into a container
of water at 20
°
C. If the final temperature of the system is 25
°
C and the specific
heat capacity of copper is 0.385 J/g
°
C, determine the mass of water in the
container. Assume no heat is lost to the surroundings.
16
Solution
Step 1: Use the formula for thermal energy to find the heat lost by the copper
ball and heat gained by the water: The thermal energy formula for a substance
is q=mc∆T, where - qis the heat energy in Joules (J), - mis the mass in grams
(g), - cis the specific heat capacity in J/g
°
C, - ∆Tis the change in temperature
in
°
C.
For the copper ball: qcopper =mcopperccopper∆Tcopper
For the water: qwater =mwatercwater∆Twater
Given that the final temperature is 25
°
C (which means ∆Tcopper = 75C
and ∆Twater = 5C), we can rewrite the thermal energy equation as follows:
qcopper = 500 ×0.385 ×75 qwater =mwater ×4.18 ×5
Step 2: Since the system is isolated and no heat is lost to the surroundings,
the heat lost by the copper ball must be equal to the heat gained by the water:
mcopperccopper∆Tcopper =mwatercwater∆Twater
Substitute the values and solve for mwater: 500×0.385×75 = mwater×4.18×5
1443.75 = 20.9mwater mwater =1443.75
20.9
Therefore, the mass of water in the container is approximately 69.11 g .
Question 21
Question
A 2 kg aluminum block initially at 100
°
C is dropped into a tank containing 10 kg
of water at 20
°
C. Assuming no heat is lost to the surroundings, calculate the final
temperature of the system when thermal equilibrium is reached. The specific
heat capacity of aluminum is 900 J/kg◦C and that of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the aluminum block as it cools down to the
final temperature. The heat lost by the aluminum block can be calculated using
the formula:
Q=mc∆T
where: - m= 2 kg is the mass of the aluminum block, - c= 900 J/kg◦C is the
specific heat capacity of aluminum, and - ∆Tis the temperature change of the
aluminum block.
Given that the initial temperature of the aluminum block is 100
°
C and the
final temperature is T
°
C, we have ∆T= 100 −T= 100 −T
°
C.
Therefore, the heat lost by the aluminum block is:
Qaluminum = 2 ×900 ×(100 −T)
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. The heat gained by the water can be calculated using the same
17
formula as above:
Q=mc∆T
where: - m= 10 kg is the mass of the water, - c= 4200 J/kg◦C is the specific
heat capacity of water, and - ∆Tis the temperature change of the water.
Given that the initial temperature of the water is 20
°
C and the final tem-
perature is T
°
C, we have ∆T=T−20
°
C.
Therefore, the heat gained by the water is:
Qwater = 10 ×4200 ×(T−20)
Step 3: Set up the equation for heat transfer at thermal equilibrium. Since
the total heat lost by the aluminum block must be equal to the total heat gained
by the water at thermal equilibrium, we can set up the equation:
Qaluminum =Qwater
2×900 ×(100 −T) = 10 ×4200 ×(T−20)
Step 4: Solve for the final temperature T.
1800 ×(100 −T) = 42000 ×(T−20)
180000 −1800T= 42000T−840000
222000 = 6000T
T= 37C
Therefore, the final temperature of the system when thermal equilibrium is
reached is 37C.
Question 22
Question
A copper rod of length 1.5 m and cross-sectional area 4 cm2has one end kept
at 100◦C and the other end at 50◦C. If the thermal conductivity of copper is
400 W/(m·K), determine the rate at which heat is conducted along the rod.
Solution
Step 1: Calculate the temperature difference along the rod. Given that the
temperature at one end of the rod is 100◦C and at the other end is 50◦C, the
temperature difference (∆T) is:
∆T= (100 −50)◦C = 50◦C
Step 2: Calculate the rate of heat conduction using Fourier’s law of heat
conduction which states: q=−kA ∆T
L. Given that the thermal conductivity k
18
of copper is 400 W/(m·K), the cross-sectional area Ais 4 cm2= 4 ×10−4m2,
and the length Lof the rod is 1.5 m, we can substitute these values into the
formula:
q=−400 ×4×10−4×50
1.5
Step 3: Calculate the rate of heat conduction.
q=−0.4×4×10−4×33.3 = −0.05328 W
Therefore, the rate at which heat is conducted along the rod is 0.05328 W.
Question 23
Question
A copper ball of mass 400 g is heated to 150◦C and then dropped into a vessel
containing 800 g of water at 20◦C. If the final temperature of the mixture is
30◦C, determine the specific heat capacity of the copper ball. Assume no heat
is lost to the surroundings.
Solution
Step 1: Find the heat absorbed by the copper ball from the initial temperature
to the final temperature.
Q1=mc∆T
Q1= (0.4 kg)(386 J/kg◦C)(30◦C−150◦C)
Q1=−46480 J
Step 2: Find the heat released by the copper ball to raise the temperature
of the water from 20◦C to 30◦C.
Q2=mc∆T
Q2= (0.8 kg)(4186 J/kg◦C)(30◦C−20◦C)
Q2= 33488 J
Step 3: As no heat is lost to the surroundings, the heat absorbed by the
copper ball equals the heat released by it, so we have:
Q1=−Q2
Step 4: Equating Q1and Q2and solving for the specific heat capacity of the
copper ball, c:
−46480 = 33488
c=46480
33488
c≈1.387 J/kg◦C
19
Therefore, the specific heat capacity of the copper ball is approximately
1.387 J/kg◦C.
Question 24
Question
A copper ball at 100◦C is dropped into a large vat of water at 20◦C. If the mass
of the copper ball is 0.5 kg and the heat capacity of copper is 390 J/kg·K, and
the specific heat capacity of water is 4186 J/kg·K, what is the final temperature
of the copper ball and water when they reach thermal equilibrium?
Solution
Step 1: Calculate the heat lost by the copper ball as it cools down to the final
temperature. The heat lost is given by the equation:
Qlost =−mc∆T,
where: - mis the mass of the copper ball (0.5 kg), - cis the heat capacity of
copper (390 J/kg·K), - ∆Tis the change in temperature of the copper ball.
The change in temperature is given by:
∆T=Tfinal −Tinitial.
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature. The heat gained is given by the equation:
Qgained =mc∆T,
where: - mis the mass of the water (equal to the mass of the copper ball, 0.5
kg), - cis the specific heat capacity of water (4186 J/kg·K), - ∆Tis the change
in temperature of the water.
Step 3: Set the heat lost equal to the heat gained to find the final tempera-
ture.
Qlost =Qgained.
Do you want to proceed with calculating the final temperature using the
steps above?
Question 25
Question
A 0.5 kg aluminum block initially at 100◦C is placed in a 1 kg water bath at
20◦C. Assume no heat is lost to the surroundings. Given that the specific heat
capacity of aluminum is 900 J/kg◦C and the specific heat capacity of water is
4186 J/kg◦C, determine the final equilibrium temperature of the system.
20
Solution
Step 1: Identify the information given in the question.
In this problem, we are given: - Mass of the aluminum block, mAl = 0.5 kg
- Initial temperature of the aluminum block, Ti,Al = 100◦C - Specific heat
capacity of aluminum, CAl = 900 J/kg◦C - Mass of the water bath, mw= 1 kg
- Initial temperature of the water bath, Ti,w= 20◦C - Specific heat capacity of
water, Cw= 4186 J/kg◦C
Step 2: Determine the heat lost by the aluminum block and gained by the
water bath.
The heat lost by the aluminum block is equal to the heat gained by the water
bath. Therefore, we can write:
mAl ·CAl ·(T−Ti,Al) = mw·Cw·(T−Ti,w)
where Tis the final equilibrium temperature of the system.
Step 3: Solve for the final equilibrium temperature.
Substitute the given values and solve for T:
0.5·900 ·(T−100) = 1 ·4186 ·(T−20)
450(T−100) = 4186(T−20)
450T−45000 = 4186T−83720
3736T= 38720
T=38720
3736
T≈10.36◦C
Therefore, the final equilibrium temperature of the system is approximately
10.36◦C.
Question 26
Question
An insulated calorimeter contains 0.2 kg of water at 20
°
C. A 0.1 kg piece of cop-
per at 90
°
C is dropped into the water. Assuming no heat is lost to the surround-
ings, calculate the final temperature of the system. (Specific heat capacity of
water = 4.18×103J/kg
°
C, specific heat capacity of copper = 0.39×103J/kg
°
C).
Solution
Let’s denote the final temperature of the system as T. Since no heat is lost to
the surroundings, the total heat gained by copper must equal the total heat lost
by water. We’ll use the formula Q=mc∆T, where Qis the heat energy, mis
the mass, cis the specific heat capacity, and ∆Tis the temperature change.
21
Step 1: Calculate heat lost by water
The heat lost by water can be calculated as:
Qwater =mwater ×cwater ×(T−20)
Step 2: Calculate heat gained by copper
The heat gained by copper can be calculated as:
Qcopper =mcopper ×ccopper ×(90 −T)
Step 3: Equate the two heat values
Setting the heat lost by water equal to the heat gained by copper, we have:
mwater ×cwater ×(T−20) = mcopper ×ccopper ×(90 −T)
Step 4: Solve for T
Plugging in the values:
0.2×4.18 ×103×(T−20) = 0.1×0.39 ×103×(90 −T)
Solving the equation will give us the final temperature T.
Question 27
Question
A solid aluminum cube with a side length of 10 cm at 20◦C is heated until its
temperature reaches 70◦C. The coefficient of linear expansion for aluminum is
2.3×10−5◦C−1. Calculate the increase in volume of the cube as a result of the
temperature increase.
Solution
Step 1: Find the increase in length of the cube due to the temperature increase.
Given the coefficient of linear expansion, we can use the formula:
∆L=αLi∆T
where - ∆Lis the change in length, - αis the coefficient of linear expansion, -
Liis the initial length, - ∆Tis the change in temperature.
Substitute the values:
∆L= (2.3×10−5◦C−1)(10 cm)(70 −20)◦C
∆L= 0.03 cm
Step 2: Find the increase in volume of the cube. The volume of a cube is
given by V=L3. Let Vibe the initial volume and Vfbe the final volume. The
increase in volume can be calculated as:
∆V=Vf−Vi= (Li+ ∆L)3−L3
i
22
∆V= (10 cm + 0.03 cm)3−(10 cm)3
∆V≈0.0927 cm3
Therefore, the increase in volume of the aluminum cube as a result of the
temperature increase is approximately 0.0927 cm3.
Question 28
Question
A 500 g aluminum block is initially at a temperature of 100
°
C. The block is
placed in a container of water at 20
°
C. If the final equilibrium temperature
of the block and water is 25
°
C, calculate the mass of water in the container.
Assume no heat is lost to the surroundings.
Solution
Step 1: Find the heat lost by the aluminum block. The specific heat capacity of
aluminum is cAl = 0.904 J/g
°
C. The change in temperature (∆TAl) of the block
is 100 −25 = 75
°
C.
The heat lost by the block can be calculated using the formula:
QAl =mcAl∆TAl
Substitute the given values:
QAl = 500 g ×0.904 J/g
°
C×75
°
C
Calculating:
QAl = 33900 J
Step 2: Find the heat gained by the water. Let the mass of water in the
container be mwater grams. The specific heat capacity of water is cwater = 4.18
J/g
°
C. The change in temperature (∆Twater) of the water is 25 −20 = 5
°
C.
The heat gained by the water can be calculated using the formula:
Qwater =mcwater∆Twater
Substitute the given values:
Qwater =mwater g×4.18 J/g
°
C×5
°
C
Step 3: Set up the heat balance equation. Since no heat is lost to the
surroundings, we can set up the heat balance equation:
QAl =Qwater
Substitute the expressions for QAl and Qwater:
33900 J = mwater ×4.18 J/g
°
C×5
°
C
23
Solving for mwater:
mwater =33900 J
4.18 J/g
°
C×5
°
C
Calculating:
mwater = 1618.71 g
Therefore, the mass of water in the container is 1618.71 grams.
Question 29
Question
A copper rod of length 2 m has a cross-sectional area of 4 cm2. If the rod is
heated from 20
°
C to 90
°
C, calculate the increase in length of the rod. The linear
expansion coefficient of copper is 1.7×10−5K−1.
Solution
Step 1: Calculate the original length of the rod.
The cross-sectional area of the rod is 4 cm2= 4 ×10−4m2. Since the length of
the rod is 2 m, the volume V1of the rod is given by:
V1= Area ×Length = 4 ×10−4m2×2 m = 8 ×10−4m3
Step 2: Calculate the final volume of the rod.
The final length of the rod after heating is given by:
L2=L1(1 + α∆T)
where L1= 2 m is the original length, α= 1.7×10−5K−1is the linear expansion
coefficient, and ∆T= 90◦C−20◦C = 70◦C is the increase in temperature.
Therefore, the final volume V2is given by:
V2= Area×Length2= 4×10−4m2×2 m×(1+1.7×10−5K−1×70) = 8.028×10−4m3
Step 3: Calculate the increase in length.
The increase in volume ∆Vis given by:
∆V=V2−V1= 8.028 ×10−4m3−8×10−4m3= 0.028 ×10−4m3
Step 4: Calculate the increase in length ∆L.
Since the cross-sectional area remains constant, the increase in length is pro-
portional to the increase in volume. Therefore,
∆L
L1
=∆V
V1
∆L=L1×∆V
V1
= 2 m ×0.028 ×10−4m3
8×10−4m3= 0.007 m
Therefore, the increase in length of the copper rod is 0.007 meters.
24
Question 30
Question
A 2.0 kg piece of copper at 90◦C is placed in a container with 1.0 kg of water
at 20◦C. If the container is perfectly insulated and the specific heat capacity of
copper is 387 J/kg·K, specific heat capacity of water is 4186 J/kg·K, and the
latent heat of fusion of water is 334,000 J/kg, determine the final equilibrium
temperature of the system.
Solution
Step 1: Calculate the energy transferred from the copper to the water due to
heat conduction. The heat lost by the copper is equal to the heat gained by the
water:
m1c1(Tf−T1) = m2c2(Tf−T2) + m2L
where: - m1and T1are the mass and initial temperature of the copper, - m2
and T2are the mass and initial temperature of the water, - c1and c2are the
specific heat capacities of the copper and water, and - Lis the latent heat of
fusion of water.
Given: - m1= 2.0 kg, T1= 90 ◦C = 363 K, c1= 387 J/kg·K, - m2= 1.0
kg, T2= 20 ◦C = 293 K, c2= 4186 J/kg·K, - L= 334,000 J/kg.
Substitute the values:
2.0×387 ×(Tf−363) = 1.0×4186 ×(Tf−293) + 1.0×334,000
Step 2: Solve for the final equilibrium temperature Tf. Expand and simplify
the equation to solve for Tf:
774(Tf−363) = 4186(Tf−293) + 334,000
774Tf−281,262 = 4186Tf−1,229,198 + 334,000
774Tf= 4186Tf−1,125,460
3445Tf= 1,125,460
Tf=1,125,460
3445 ≈326.91
Therefore, the final equilibrium temperature of the system is approximately
326.91 K.
Question 31
Question
A copper cup has a mass of 250 g and is filled with 500 g of water at 25
°
C. If the
cup and water are heated until they reach thermal equilibrium, determine the
final temperature of the system. Assume the specific heat capacity of copper is
0.385 J/g
°
C and that of water is 4.18 J/g
°
C.
25
Solution
Step 1: Calculate the heat absorbed by the copper cup to reach the final tem-
perature.
Given: Mass of copper cup, mcopper = 250 g Specific heat capacity of copper,
ccopper = 0.385 J/g
°
C Initial temperature of copper cup, Tcopper initial = 25C
Final temperature of the system, Tfinal
The heat absorbed by the copper cup is given by the formula:
Qcopper =mcopper ·ccopper ·∆Tcopper
where ∆Tcopper =Tfinal −Tcopper initial
Step 2: Calculate the heat absorbed by the water to reach the final temper-
ature.
Given: Mass of water, mwater = 500 g Specific heat capacity of water,
cwater = 4.18 J/g
°
C Initial temperature of water, Twater initial = 25C
Since the copper cup and the water reach thermal equilibrium, the heat
absorbed by the water must be equal to the heat absorbed by the copper cup:
Qcopper =Qwater
Step 3: Set up and solve the equation to find the final temperature of the
system.
Since the heat absorbed by the water is given by:
Qwater =mwater ·cwater ·∆Twater
where ∆Twater =Tfinal −Twater initial
We can set up the equation:
mcopper ·ccopper ·∆Tcopper =mwater ·cwater ·∆Twater
Substitute the expressions for ∆Tcopper and ∆Twater into the equation:
mcopper ·ccopper ·(Tfinal −Tcopper initial) = mwater ·cwater ·(Tfinal −Twater initial)
Solve for Tfinal to find the final temperature of the system.
Question 32
Question
A copper vessel of mass 0.5 kg contains 2 kg of water at 25◦C. How much
heat must be supplied to the vessel to bring the water to boiling point 100◦C,
assuming no heat is lost to the surrounding?
26
Solution
Step 1: Calculate the heat required to raise the temperature of water from
25◦C to 100◦C. Given: Mass of water, mw= 2 kg Initial temperature of water,
Tinitial = 25◦C Final temperature of water, Tfinal = 100◦C Specific heat capacity
of water, cw= 4186 J/kg◦C
The heat required to raise the temperature of water can be calculated using
the formula:
Q=mw·cw·∆T
where ∆T=Tfinal −Tinitial.
Calculating the heat required:
∆T= 100◦C−25◦C = 75◦C
Q= 2 ·4186 ·75 = 627900 J = 627.9 kJ
Step 2: Calculate the heat required to raise the temperature of the copper
vessel. Given: Mass of copper vessel, mc= 0.5 kg Specific heat capacity of
copper, cc= 386 J/kg◦C Temperature change of copper vessel, ∆Tc= 100◦−
25◦= 75◦C
The heat required to raise the temperature of the copper vessel can be cal-
culated using the formula:
Qc=mc·cc·∆Tc
Calculating the heat required:
Qc= 0.5·386 ·75 = 14475 J = 14.5 kJ
Step 3: Add the heat required for the water and the copper vessel to find
the total heat required.
Qtotal =Q+Qc= 627.9 kJ + 14.5 kJ = 642.4 kJ
Therefore, 642.4 kJ of heat must be supplied to the vessel to bring the water
to boiling point 100◦C.
Question 33
Question
A copper rod of length 2 m and diameter 2 cm is initially at a temperature of
100◦C. It is immersed completely in a water container at 25◦C. Given that the
thermal conductivity of copper is 400 W/mK and the specific heat capacity of
copper is 390 J/kgK, calculate the time it takes for the rod to reach thermal
equilibrium with the water. Assume no heat loss to the surroundings.
27
Solution
Step 1: Find the cross-sectional area of the rod. The cross-sectional area Aof
the rod can be calculated using the formula for the area of a circle: A=πd2
4,
where dis the diameter of the rod. Given that the diameter is 2 cm, we have
d= 0.02 m. Therefore,
A=π(0.02)2
4= 3.14 ×10−4m2
Step 2: Calculate the volume of the rod. The volume Vof the rod can be
calculated using the formula: V=A×L, where Lis the length of the rod.
Given that the length is 2 m, we have
V= 3.14 ×10−4m2×2 m = 6.28 ×10−4m3
Step 3: Find the mass of the rod. The mass of the rod can be calculated using
the formula: m= volume ×density. The density of copper is approximately
8900 kg/m3. Therefore,
m= 6.28 ×10−4m3×8900 kg/m3= 5.5932 kg
Step 4: Calculate the heat energy required for the rod to reach equilibrium.
The heat energy Qrequired can be calculated using the equation: Q=mc∆T,
where ∆Tis the temperature difference between the rod and water, and cis
the specific heat capacity of copper. Given that the initial temperature of the
rod is 100◦C and of the water is 25◦C, we have ∆T= 100◦C−25◦C = 75◦C.
Therefore,
Q= 5.5932 kg ×390 J/kgK ×75 K = 163833 J
Step 5: Calculate the rate of heat transfer through the rod. The rate of
heat transfer Pcan be calculated using Fourier’s law of heat conduction: P=
−kAdT
dx , where kis the thermal conductivity of copper and dT
dx is the temperature
gradient. Since the rod is of constant diameter, the change in temperature along
the rod is uniform. Thus, P=−kA ∆T
L. Given that the thermal conductivity
of copper is 400 W/mK, we have
P=−400 W/mK ×3.14 ×10−4m2×75
2K = −47.1 W
Step 6: Calculate the time taken for the rod to reach thermal equilibrium.
The time ttaken for the rod to reach thermal equilibrium can be calculated
using the equation: Q=P t. Thus,
t=Q
P=163833 J
−47.1 W =−3478.11 s ≈58 minutes
Therefore, it takes approximately 58 minutes for the copper rod to reach
thermal equilibrium with the water.
28
Question 34
Question
A copper cube initially at a temperature of 100◦C is placed in a large thermal
reservoir at 0◦C. The sides of the cube are 10 cm long. Assuming all heat
capacities are constant over the temperature range, calculate how long it will
take for the cube to cool down to 30◦C. The thermal conductivity of copper is
401 W/(m*K) and its density is 8.96 g/cm3.
Solution
Step 1: Find the initial temperature difference between the cube and the reser-
voir. Given: Initial temperature of the cube (Tcube,i) = 100◦C = 100 K, Tem-
perature of the reservoir (Treservoir)=0◦C = 0 K. Initial temperature difference
=Tcube,i −Treservoir = 100 K.
Step 2: Calculate the area of the cube. Given: Side length of the cube (s)
= 10 cm = 0.1 m. Area of one face of the cube = s2. Total surface area of the
cube = 6 times the area of one face. Total surface area = 6 ×0.12m2= 0.06
m2.
Step 3: Calculate the initial rate of energy transfer (heat flow) from the cube
to the reservoir. Using the formula for the rate of heat flow through a material:
Q=kA∆T/d, where: - Q= rate of heat transfer (in watts), - k= thermal
conductivity of copper (401 W/(m*K)), - A= surface area of the cube (0.06
m2), - ∆T= initial temperature difference (100 K), - d= thickness of the cube
(since it is a cube, we can assume the thickness is the same as the side length).
Substitute the values into the formula: Q= 401 ×0.06 ×100/0.1 = 2406 W.
Step 4: Calculate the heat capacity of the cube. Given: Density of copper
= 8.96 g/cm3. The mass of the cube can be calculated using the density and
volume formula: m= density ×volume. Volume of the cube = side length3=
0.13m3. Mass of the cube = 8.96 g/cm3×0.001 m3= 8.96 kg.
The specific heat capacity of copper is 385 J/(kg*K). Heat capacity of the
cube = mass ×specific heat capacity = 8.96 ×385 J/K = 3455.6 J/K.
Step 5: Determine the time taken for the cube to cool down to 30◦C. The
change in temperature (∆T) = 100 K - 30 K = 70 K.
Using the formula for change in temperature in terms of heat capacity and
heat flow: ∆T=Q×t/C, where: - t= time taken (in seconds), - C= heat
capacity of the cube (3455.6 J/K).
Rearranging the formula to solve for t:t=C×∆T/Q.
Substitute the values into the formula: t= 3455.6×70/2406 = 100.4 seconds
=1 minute and 40.4 seconds.
29
Question 35
Question
A copper rod of length 1 m and diameter 2 cm is heated from 20◦C to 120◦C.
Given that the linear expansion coefficient of copper is 1.7×10−5◦C−1and
the specific heat capacity of copper is 0.385 J/g◦C, calculate the heat energy
required for this process.
Solution
Step 1: Calculate the change in length of the copper rod due to heating. The
change in length (∆L) of the copper rod can be calculated using the formula
for linear expansion:
∆L=L·α·∆T
where Lis the original length of the rod, αis the linear expansion coefficient
of copper, and ∆Tis the change in temperature. Substitute L= 1 m, α=
1.7×10−5◦C−1, ∆T= 120◦C−20◦C = 100◦C into the formula:
∆L= 1 m ·1.7×10−5◦C−1·100◦C
∆L= 0.0017 m = 1.7 mm
Step 2: Calculate the volume change of the copper rod. The initial volume
of the rod can be calculated using the formula for the volume of a cylinder:
V=πr2h
where ris the radius of the rod and his the original length of the rod. Given
that the diameter of the rod is 2 cm, the radius r= 1 cm = 0.01 m. Substitute
r= 0.01 m and h= 1 m into the formula:
V=π×(0.01)2×1 = π×0.0001 m3
V= 0.000314 m3
The change in volume (∆V) due to the change in length can be calculated
as:
∆V=π×r2×∆L
Substitute r= 0.01 m and ∆L= 0.0017 m into the formula:
∆V=π×0.0001 ×0.0017 = 5.366 ×10−7m3
Step 3: Calculate the mass of the copper rod. The density of copper is
approximately 8900 kg/m3. The mass of the rod can be calculated using the
formula:
m=ρ×V
30
where ρis the density of copper and Vis the initial volume of the rod. Substitute
ρ= 8900 kg/m3and V= 0.000314 m3into the formula:
m= 8900 ×0.000314 = 2.7956 kg
Step 4: Calculate the heat energy required. The heat energy (Q) required
to raise the temperature of the copper rod can be calculated using the formula:
Q=m×c×∆T
where mis the mass of the rod, cis the specific heat capacity of copper, and
∆Tis the change in temperature. Substitute m= 2.7956 kg, c= 0.385 J/g◦C,
and ∆T= 120◦C−20◦C = 100◦C into the formula:
31
