PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 7
Liberty University
Question 1
Question
A copper block initially at a temperature of 100
°
C is placed in a large container
of water at 20
°
C. The block has a mass of 2 kg and the specific heat capacity of
copper is 390 J/kg
°
C. Assuming no heat is lost to the surroundings, calculate
the final equilibrium temperature of the system when the block and water reach
thermal equilibrium.
Given: Specific heat capacity of water = 4186J/kgC Density of water =
1000kg/m3
Solution
Step 1: Calculate the heat lost by the copper block until it reaches thermal
equilibrium with the water: The heat lost by the copper block (Qcopper) is equal
to the heat gained by the water. The formula for heat transfer is Q=mc∆T,
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
The initial temperature of the copper block is 100
°
C and the final equilibrium
temperature is T, so ∆T=T−100. Therefore,
Qcopper =mc(∆T) = 2 ×390 ×(T−100)
Step 2: Calculate the heat gained by the water: The heat gained by the
water (Qwater ) is also given by Qwater =mc∆T, where mis the mass, cis the
specific heat capacity, and ∆Tis the change in temperature.
The mass of water is not given, but we can calculate it using the density
formula m=ρV , where ρis the density and Vis the volume. Let the volume
of water be V.
m=ρV = 1000 ×V
Step 3: Setting up an equation for the heat transfer to the water: As no
heat is lost to the surroundings, the heat lost by the copper block is equal to
the heat gained by the water. This allows us to write the equation:
2×390 ×(T−100) = 1000 ×4186 ×(T−20)
Step 4: Solve the equation to find the equilibrium temperature T: Solving
the equation from step 3:
780(T−100) = 4186000(T−20)
Solving for Tgives T= 24.6
°
C.
Therefore, the final equilibrium temperature when the copper block and
water reach thermal equilibrium is 24.6
°
C.
Question 2
Question
A piece of copper of mass 200 g at 200
°
C is placed in a container with 800
g of water at 20
°
C. If no heat is lost to the surroundings, what is the final
temperature of the system? Assume the specific heat capacities of copper and
water are cCu = 0.385 J/g
°
C and cH2O = 4.18 J/g
°
C, respectively, and the latent
heat of fusion for water is 334 J/g.
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
The heat lost by the copper (QCu) can be calculated using the formula:
QCu =mCu ·cCu ·∆TCu
where: mCu = mass of copper = 200 g, cCu = specific heat capacity of copper
= 0.385 J/g
°
C, ∆TCu = change in temperature of copper = final temperature -
initial temperature.
Similarly, the heat gained by the water (QH2O) can be calculated using the
formula:
QH2O =mH2O ·cH2O ·∆TH2O
where: mH2O = mass of water = 800 g, cH2O = specific heat capacity of water
= 4.18 J/g
°
C, ∆TH2O = change in temperature of water = final temperature -
initial temperature.
Since the system is isolated and no heat is lost to the surroundings, QCu =
−QH2O. Let the final temperature be Tf.
Step 2: Set up the equation using the heat lost and gained expressions. Since
QCu =−QH2O, we have:
mCu ·cCu ·(Tf−200) = −mH2O ·cH2O ·(Tf−20)
2
Step 3: Solve for the final temperature. Plug in the given values:
200 ·0.385 ·(Tf−200) = −800 ·4.18 ·(Tf−20)
Solving this equation will give the final temperature Tf.
Question 3
Question
A piece of iron has a mass of 500 g and is initially at a temperature of 100
°
C.
It is dropped into a bucket containing 2 kg of water at 20
°
C. Assuming no heat
is lost to the surroundings, calculate the final equilibrium temperature of the
system. The specific heat capacity of iron is 450 J/kg◦C and that of water is
4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the iron.
The heat lost by the iron can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the iron (in kg), - cis the specific heat capacity of
iron (in J/kg◦C), and - ∆Tis the change in temperature of the iron (initial
temperature - final temperature).
Converting the mass of the iron to kilograms:
m=500 g
1000 = 0.5 kg
Plugging in the values:
Qiron = 0.5×450 ×(100 −Tf)
Qiron = 225 ×(100 −Tf)
Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the water (in kg), - cis the specific heat capacity of
water (in J/kg◦C), and - ∆Tis the change in temperature of the water (final
temperature - initial temperature).
Converting the mass of the water to kilograms:
m= 2 kg
3
Plugging in the values:
Qwater = 2 ×4186 ×(Tf−20)
Qwater = 8372 ×(Tf−20)
Step 3: Setting up the energy balance equation.
Since there is no heat lost to the surroundings, the heat lost by the iron is equal
to the heat gained by the water:
Qiron =Qwater
225 ×(100 −Tf) = 8372 ×(Tf−20)
Step 4: Solve for Tf.
Expanding and simplifying the equation:
22500 −225Tf= 8372Tf−167440
225Tf+ 8372Tf= 167440 + 22500
8597Tf= 190940
Tf=190940
8597
Tf≈22.23◦C
Therefore, the final equilibrium temperature of the system is approximately
22.23◦C.
Question 4
Question
A copper rod of length 2 meters and diameter 1 centimeter is heated from 20
°
C
to 80
°
C. If the thermal conductivity of copper is 390 W/mK, calculate the rate
at which heat flows through the rod.
Solution
Step 1: Calculate the cross-sectional area of the copper rod. The cross-sectional
area of the rod can be calculated using the formula for the area of a circle:
A=πr2, where ris the radius of the rod. Given that the diameter of the rod
is 1 centimeter, the radius ris 0.5 cm (or 0.005 m). The cross-sectional area is
then:
A=π(0.005)2= 0.00007854 m2
Step 2: Calculate the temperature difference. The temperature difference
∆Tis given by:
∆T=T2−T1= 80C−20C= 60C
4
Step 3: Calculate the rate of heat flow. The rate of heat flow (Q) through
the rod is given by Fourier’s Law:
Q=k·A·∆T
L
where: k= thermal conductivity of copper (390 W/mK), A= cross-sectional
area of the rod (0.00007854 m2), ∆T= temperature difference (60C), L= length
of the rod (2 m).
Substitute the values into the formula:
Q= 390 ×0.00007854 ×60
2= 0.921 W
Therefore, the rate at which heat flows through the rod is 0.921 W.
Question 5
Question
A copper rod of length 2 meters and cross-sectional area 2 ×10−4m2is initially
at a temperature of 200◦C. If 5000 J of heat is added to the rod, what is the
final temperature of the rod? (Note: The specific heat capacity of copper is
390 J/kg ·K and the density of copper is 8900 kg/m3.)
Solution
Step 1: Find the mass of the copper rod. The mass of the copper rod can be
calculated using the formula:
mass = density ×volume = ρ×A×L
where: ρ= density of copper = 8900 kg/m3,A= cross-sectional area = 2 ×
10−4m2,L= length of the rod = 2 meters.
Substitute the given values into the formula:
mass = 8900 ×2×10−4×2=3.56 kg
Step 2: Calculate the change in temperature. The heat added to the rod
can be expressed as:
Q=mc∆T
where: Q= heat added = 5000 J, m= mass of the rod = 3.56 kg, c= specific
heat capacity of copper = 390 J/kg ·K, ∆T= change in temperature.
Solve for ∆T:
∆T=Q
mc =5000
3.56 ×390 ≈3.24 K
Step 3: Find the final temperature of the rod. The final temperature Tfcan
be calculated as:
Tf=Ti+ ∆T
5
where: Ti= initial temperature of the rod = 200◦C⇒200 + 273 = 473 K.
Substitute the values into the formula:
Tf= 473 + 3.24 = 476.24 K
Convert the final temperature from Kelvin back to degrees Celsius:
Tf= 476.24 −273 ≈203.24 ◦C
Therefore, the final temperature of the copper rod is approximately 203.24◦C.
Question 6
Question
A metal block of mass 2 kg is heated to a temperature of 80
°
C and then placed in
a large container of water at 20
°
C. The initial temperature of the water is 25
°
C
and its mass is 5 kg. Assuming no heat loss to the surroundings, calculate the
final temperature of the system (metal block + water) when thermal equilibrium
is reached. The specific heat capacity of the metal is 500 J/kg
°
C and that of
water is 4186 J/kg
°
C.
Solution
Let the final temperature of the system be T
°
C. The heat lost by the metal
block must equal the heat gained by the water.
Step 1: Calculate the heat lost by the metal block. The heat lost by the
metal block can be calculated using the formula:
Qlost =mc∆T
where: - mis the mass of the metal block, - cis the specific heat capacity of
the metal, and - ∆Tis the change in temperature of the metal block.
Substitute the given values:
Qlost = 2 kg ×500 J/kg
°
C×(80 −T)
°
C
Step 2: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the same formula:
Qgained =mc∆T
where: - mis the mass of the water, - cis the specific heat capacity of water,
and - ∆Tis the change in temperature of the water.
Substitute the given values:
Qgained = 5 kg ×4186 J/kg
°
C×(T−25)
°
C
6
Step 3: Set up the equation for heat equilibrium. Since there is no heat
loss to the surroundings, the heat lost by the metal block must equal the heat
gained by the water:
2×500 ×(80 −T)=5×4186 ×(T−25)
Step 4: Solve the equation for T. Simplify and solve the equation to find
the final temperature Tof the system. This involves expanding and rearranging
terms before solving for T.
Question 7
Question
A sample of aluminum initially at a temperature of 100
°
C is placed in contact
with a reservoir at 0
°
C. The sample exchanges heat with the reservoir until it
reaches thermal equilibrium. Given that the mass of the aluminum sample is
500g and the specific heat capacity of aluminum is 0.9 J/g
°
C, calculate the final
temperature of the aluminum sample.
Solution
Step 1: Calculate the heat transferred from the aluminum sample to the reser-
voir using the formula:
Q=mc∆T
where: Q= heat transferred, m= mass of the aluminum sample = 500g, c=
specific heat capacity of aluminum = 0.9 J/g
°
C, ∆T= change in temperature.
The initial temperature of the aluminum sample is 100
°
C and the final tem-
perature is θ
°
C. Therefore, the change in temperature, ∆T=θ−100. Substitute
the given values into the formula:
Q= 500 ×0.9×(θ−100)
Step 2: Calculate the heat absorbed by the reservoir, which is equal in
magnitude to the heat lost by the aluminum sample.
Q=−mc∆T
Substitute the given values into the formula:
Q=−500 ×0.9×θ
Step 3: Set the two equations equal to each other and solve for θ.
500 ×0.9×(θ−100) = −500 ×0.9×θ
Step 4: Solve the equation for θ.
450(θ−100) = −450θ
7
450θ−45000 = −450θ
450θ+ 450θ= 45000
900θ= 45000
θ=45000
900
θ= 50
Therefore, the final temperature of the aluminum sample is 50
°
C.
Question 8
Question
A copper bar of length 2 m is initially at a uniform temperature of 400 K. One
end of the bar is then heated until its temperature is 600 K. If the specific heat
capacity of copper is 390 J/kg·K and its density is 8930 kg/m3, find the amount
of heat energy transferred to the bar during this process. Assume no heat is
lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper bar. Given that the density of copper
is 8930 kg/m3, the mass of the copper bar can be calculated using the formula:
mass = density ×volume
The volume of the copper bar is V= 2 m ×A, where Ais the cross-sectional
area of the bar. Since the bar is uniform, we can assume the cross-sectional area
is constant.
Step 2: Calculate the amount of heat energy transferred. The amount of
heat energy transferred can be calculated using the formula:
Q=mc∆T
where mis the mass of the copper bar, cis the specific heat capacity of copper,
and ∆Tis the change in temperature of the copper bar.
Step 3: Substitute the known values into the formula. We have already
calculated the mass of the copper bar in Step 1. The specific heat capacity of
copper is 390 J/kg·K. The initial temperature is 400 K and the final temperature
is 600 K, so the change in temperature is ∆T= 600 K −400 K = 200 K.
Step 4: Calculate the amount of heat energy transferred. Substitute the
values into the formula to find the amount of heat energy transferred:
Q=mc∆T
Q= mass ×c×∆T
8
Q= (8930 kg/m3×2 m ×A)×390 J/kg ·K×200 K
Q= 17,860 AJ
Therefore, the amount of heat energy transferred to the copper bar during
this process is 17,860 AJ.
Question 9
Question
A 2 kg block of copper at 100◦C is placed into a container filled with 1 kg of
water at 20◦C. Assuming no heat is lost to the surroundings, what will be the
final temperature of the system? (Specific heat capacity of copper = 390 J/kg◦C,
specific heat capacity of water = 4186 J/kg◦C)
Solution
Let the final temperature of the system be T◦C.
Step 1: Write the heat equation for the system. The heat gained by the
copper block (cooling down from 100◦C to T◦C) is equal to the heat lost by
the water (heating up from 20◦C to T◦C). This can be expressed as:
mcopper ·ccopper ·(T−100) = −mwater ·cwater ·(20 −T)
where: mcopper = 2 kg (mass of copper block), ccopper = 390 J/kg◦C (specific
heat capacity of copper), mwater = 1 kg (mass of water), cwater = 4186 J/kg◦C
(specific heat capacity of water).
Step 2: Solve the heat equation. Substitute the given values into the heat
equation and solve for T:
2×390 ×(T−100) = −1×4186 ×(20 −T)
780 ×(T−100) = −4186 ×(20 −T)
780T−78000 = −83720 + 4186T
3614T= 5720
T=5720
3614 ≈1.58
°
C
Step 3: Answer Therefore, the final temperature of the system will be
approximately 1.58◦C.
9
Question 10
Question
A metal bar of mass 0.5 kg at an initial temperature of 100
°
C is placed in a
large ice bath at 0
°
C. The specific heat of the metal is 450 J/kg
°
C. Assuming
no heat is lost to the surroundings, calculate the final temperature of the metal
bar when it reaches thermal equilibrium with the ice bath.
Solution
Step 1: Calculate the heat lost by the metal bar as it cools down from 100
°
C
to the final temperature. Given: - Initial temperature, Ti= 100C- Final
temperature, Tf=? - Specific heat, c= 450J/kgC - Mass of the metal bar,
m= 0.5kg - Heat lost, Q=mc∆T
From the formula,
Q=mc∆T
we have,
Q= 0.5×450 ×(100 −Tf)
Q= 225(100 −Tf)
Step 2: Calculate the heat gained by the metal bar as it absorbs heat from
the ice bath to raise its temperature to the final temperature. Given: - Initial
temperature of the ice bath, Tice = 0C- Final temperature, Tf- Heat gained,
Q=mc∆T
From the formula,
Q=mc∆T
we have,
Q= 0.5×450 ×(Tf−0)
Q= 225Tf
Step 3: Equate the heat lost and heat gained equations to find the final
temperature.
225(100 −Tf) = 225Tf
22500 −225Tf= 225Tf
225Tf+ 225Tf= 22500
450Tf= 22500
Tf= 50C
Therefore, the final temperature of the metal bar when it reaches thermal
equilibrium with the ice bath is 50
°
C.
10
Question 11
Question
A copper pot contains 0.5 kg of water at 20
°
C. How much heat is required to
raise the temperature of the water to 100
°
C? (Specific heat capacity of copper
= 390 J/kg◦C, specific heat capacity of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat required to raise the temperature of the water to
100
°
C.
The heat required to raise the temperature of a substance can be calculated
using the formula:
Q=mc∆T
where: Q= heat energy (Joules), m= mass of the substance (kg), c= spe-
cific heat capacity of the substance (J/kg◦C), and ∆T= change in temperature
(◦C).
Given: m= 0.5 kg, cwater = 4186 J/kg◦C, ∆T= 100 −20 = 80◦C.
Substitute these values into the formula to find the heat required for the
water:
Qwater = 0.5 kg ×4186 J/kg◦C×80◦C
Qwater = 167440 J
Therefore, the heat required to raise the temperature of the water to 100
°
C
is 167440 Joules.
Step 2: Calculate the heat capacity of the copper pot.
The heat capacity of the copper pot can be calculated using the formula:
Q=mc∆T
where: Q= heat energy (Joules), m= mass of the substance (kg), c= spe-
cific heat capacity of the substance (J/kg◦C), and ∆T= change in temperature
(◦C).
Given: mcopper ≈mwater,ccopper = 390 J/kg◦C, ∆Tcopper = 100−20 = 80◦C.
Substitute these values into the formula to find the heat capacity of the
copper pot:
Qcopper = 0.5 kg ×390 J/kg◦C×80◦C
Qcopper = 15600 J
Therefore, the heat capacity of the copper pot is 15600 Joules.
Step 3: The total heat required is the sum of the heat required for the water
and the copper pot.
11
Qtotal =Qwater +Qcopper
Qtotal = 167440 J + 15600 J
Qtotal = 183040 J
Therefore, the total heat required to raise the temperature of the water to
100
°
C is 183040 Joules.
Question 12
Question
A 0.5 kg block of copper is heated from 20
°
C to 50
°
C. Calculate the amount of
heat transferred to the block. (Specific heat capacity of copper is 390 J/kg ·K)
Solution
Step 1: Calculate the change in temperature of the block. Given: Initial tem-
perature (Ti) = 20
°
C Final temperature (Tf) = 50
°
C
The change in temperature (∆T) can be calculated as:
∆T=Tf−Ti= 50C−20C= 30C
Step 2: Use the formula for heat transfer to find the amount of heat trans-
ferred. The formula for heat transfer is:
Q=mc∆T
where: Q= amount of heat transferred m= mass of the block (0.5 kg) c=
specific heat capacity of copper (390 J/kg*K) ∆T= change in temperature
(30
°
C)
Substitute the given values into the formula:
Q= 0.5 kg ×390 J/kg ·K×30 K
Q= 5850 J
Therefore, the amount of heat transferred to the block is 5850 J.
Question 13
Question
A copper ball at a temperature of 200
°
C is placed into a container of water at
20
°
C. The mass of the copper ball is 500 g and the mass of the water is 2 kg.
Assuming no heat is lost to the surroundings and that the specific heat capacities
of copper and water are 0.385 J/g
°
C and 4.18 J/g
°
C respectively, calculate the
final temperature of the system after thermal equilibrium is reached.
12
Solution
Step 1: Calculate the heat lost by the copper ball as it cools down to the final
temperature. The formula to calculate heat is:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity, and - ∆Tis the change in temperature.
The heat lost by the copper ball can be calculated as:
Qcopper =mcopper ×ccopper ×(200 −Tf)
Qcopper = 500 ×0.385 ×(200 −Tf)
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. Using the same formula as above:
Qwater =mwater ×cwater ×(Tf−20)
Qwater = 2000 ×4.18 ×(Tf−20)
Step 3: Since energy is conserved in this closed system, the heat lost by the
copper ball is equal to the heat gained by the water. Therefore, we can set the
two equations equal to each other and solve for the final temperature (Tf).
500 ×0.385 ×(200 −Tf) = 2000 ×4.18 ×(Tf−20)
Step 4: Solve for Tf.
192.5×(200 −Tf) = 8360 ×(Tf−20)
38500 −192.5Tf= 8360Tf−167200
38500 + 167200 = 192.5Tf+ 8360Tf
205700 = 8552.5Tf
Tf=205700
8552.5
Tf≈24.06C
Therefore, the final temperature of the system after thermal equilibrium is
reached is approximately 24.06C.
Question 14
Question
A 0.5 kg block of copper is initially at a temperature of 50◦C. How much heat
needs to be added to the block to raise its temperature to 100◦C? The specific
heat capacity of copper is 390 J/kg ·K.
13
Solution
Step 1: Determine the change in temperature. The change in temperature (∆T)
can be calculated using the formula:
∆T=Tf−Ti
where Tfis the final temperature and Tiis the initial temperature. Given that
the initial temperature Ti= 50◦Cand the final temperature Tf= 100◦C, we
have:
∆T= 100◦C−50◦C= 50◦C
Step 2: Calculate the amount of heat using the specific heat capacity formula.
The amount of heat (Q) needed to change the temperature of a substance can
be calculated using the formula:
Q=mc∆T
where: m= 0.5 kg (mass of the copper block), c= 390 J/kg ·K (specific heat
capacity of copper), and ∆T= 50◦C(change in temperature).
Substitute the values into the formula:
Q= (0.5 kg)(390 J/kg ·K)(50 ◦C)
Step 3: Calculate the amount of heat.
Q= 0.5×390 ×50 = 9750 J
Therefore, 9750 J of heat needs to be added to the copper block to raise its
temperature to 100◦C.
Question 15
Question
A 2 kg block of copper at an initial temperature of 300 K is placed in a container
with 1 kg of water at 350 K. The thermal conductivity of copper is 400 W/mK
and that of water is 0.6 W/mK. Assuming no heat is lost to the surroundings,
calculate the final equilibrium temperature of the system.
Solution
Step 1: Calculate the heat gained or lost by each substance using the equation
Q=mc∆T, where mis mass, cis specific heat capacity, and ∆Tis the change
in temperature.
For copper: Qcopper =mcopperccopper∆Tcopper
Given: mcopper = 2 kg, ccopper = 390 J/kgK, ∆Tcopper =T−300 K.
Step 2: Calculate Qcopper.
14
Qcopper = 2 ×390 ×(T−300)
Step 3: For water: Qwater =mwatercwater∆Twater
Given: mwater = 1 kg, cwater = 4186 J/kgK, ∆Twater =T−350 K.
Step 4: Calculate Qwater.
Qwater = 1 ×4186 ×(T−350)
Step 5: Since there is no heat loss, Qcopper =−Qwater.
Step 6: Solve the equation for T.
2×390 ×(T−300) = −1×4186 ×(T−350)
Step 7: Solve for Tto find the final equilibrium temperature.
Question 16
Question
An iron bar of length 1.5 m and cross-sectional area 4 cm2is heated at one end
to 100◦C. If the coefficient of linear expansion of iron is 1.2×10−5◦C−1and
its Young’s modulus is 2 ×1011 N/m2, determine the increase in length of the
bar due to heating.
Solution
Step 1: Calculate the initial length change of the iron bar due to heating. Given
that the coefficient of linear expansion of iron is α= 1.2×10−5◦C−1, the change
in length ∆Lof the iron bar when heated from 0◦C to 100◦C can be calculated
using the formula:
∆L=L·α·∆T
where Lis the initial length, αis the coefficient of linear expansion, and ∆Tis
the change in temperature. Substitute the values L= 1.5 m, α= 1.2×10−5
◦C−1, and ∆T= 100 ◦C into the formula:
∆L= 1.5 m ·1.2×10−5◦C−1·100 ◦C
∆L= 1.8×10−3m=1.8 mm
Step 2: Calculate the increase in length of the iron bar due to heating. The
increase in length of the iron bar due to heating is related to the stress and
Young’s modulus by the formula:
Stress = F orce
Area =Y·∆L
L
where Yis the Young’s modulus, ∆Lis the change in length, and Lis the initial
length. Rearrange the formula to solve for the increase in length ∆Lincrease:
∆Lincrease =Stress ·L
Y
15
Substitute the values Y= 2 ×1011 N/m2,L= 1.5 m, and Stress = F orce
Area into
the formula:
∆Lincrease =
F
Area ·1.5
2×1011
Since F= 0 initially and Fonly arises due to the increase in length,
∆Lincrease =
F
Area ·1.5
2×1011 =
F
4×10−4·1.5
2×1011 =1.5·F
8×107
∆Lincrease =1.5·F
8×107
Therefore, the increase in length of the iron bar due to heating is 1.5·F
8×107
meters.
Question 17
Question
A 0.2 kg block of ice at -10
°
C is placed on a stove where it absorbs 5000 J of
heat. The specific heat capacity of ice is 2100 J/kg ·K, the specific heat capacity
of water is 4200 J/kg ·K, and the latent heat of fusion of ice is 3.34 ×105J/kg.
Determine the final temperature of the water.
Solution
Step 1: Calculate the heat required for the ice to reach 0
°
C. The specific heat
equation is given by:
Q=mc∆T
where Qis the heat absorbed, mis the mass of the substance, cis the specific
heat capacity, and ∆Tis the change in temperature.
Substitute the given values for ice:
5000 = 0.2×2100 ×(0 −(−10))
5000 = 0.2×2100 ×10
5000 = 4200 ×2
∆Q= 8400 J
Step 2: Calculate the heat required for the ice to melt at 0
°
C. The heat
required for a change of state is given by:
Q=mL
where Qis the heat absorbed, mis the mass of the substance, and Lis the
latent heat of fusion.
16
Substitute the given values for ice:
8400 = 0.2×3.34 ×105
8400 = 66800 J
Step 3: Calculate the heat required for the water to increase to the final
temperature. The specific heat equation is given by:
Q=mc∆T
Substitute the given values for water:
5000 = 0.2×4200 ×(Tfinal −0)
5000 = 840 ×Tfinal
Tfinal =5000
840
Tfinal ≈5.95
°
C
Therefore, the final temperature of the water is approximately 5.95
°
C.
Question 18
Question
A copper rod of length 2.0 m and cross-sectional area 4.0 cm2is initially at
a temperature of 100◦C. It is placed in a large bath of water at 10◦C until
it reaches thermal equilibrium. If the rod has a thermal conductivity of 400
W/(m*K) and a specific heat capacity of 385 J/(kg*K), determine the heat
transfer in the rod during this process.
Solution
Step 1: Calculate the volume of the copper rod. The volume Vof the copper
rod can be calculated as:
V= Area ×Length
V= 4.0×10−4m2×2.0 m
V= 8 ×10−4m3
Step 2: Calculate the mass of the copper rod. The mass mof the copper
rod can be calculated using the density of copper, which is 8900 kg/m3:
m= Density ×Volume
m= 8900 kg/m3×8×10−4m3
m= 7.12 kg
17
Step 3: Calculate the initial energy in the copper rod. The initial energy Qi
in the copper rod can be calculated using the specific heat capacity of copper:
Qi=m×c×∆T
Qi= 7.12 kg ×385 J/(kg*K) ×(100 −10) K
Qi= 2.34 ×104J
Step 4: Calculate the rate of heat flow out of the rod. The rate of heat flow
Pout of the rod can be calculated using the thermal conductivity of copper:
P=k×Area ×∆T
L
P=400 W/(m*K) ×4.0×10−4m2×(100 −10) K
2.0 m
P= 1400 W
Step 5: Determine 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 initial energy and the rate of heat flow:
Qi=P×t
2.34 ×104J = 1400 W ×t
t≈16.7 s
Step 6: Calculate the total heat transfer in the rod during this process. The
total heat transfer Qin the rod can be calculated as:
Q=P×t
Q= 1400 W ×16.7 s
Q= 2.34 ×104J
Therefore, the heat transfer in the rod during this process is 23,400 J.
Question 19
Question
A copper block of mass 500 g is initially at a temperature of 80◦C. It is placed
in a calorimeter containing 200 g of water at 20◦C. If the final temperature of
the system is 30◦C, calculate the specific heat capacity of copper. Assume no
heat is lost to the surroundings.
18
Solution
Step 1: Calculate the heat lost by the copper block.
Heat lost by copper = mass ×specific heat capacity ×∆temperature
Qcopper =mcopper ×ccopper ×∆T
Given:
mcopper = 500 g = 0.5 kg
cwater = 4200 J/kg ·◦C
Initial temperature of copper, Tinitial = 80◦C Final temperature of system,
Tfinal = 30◦C
∆T=Tfinal −Tinitial = 30 −80 = −50 C
Qcopper = 0.5×ccopper ×(−50)
Step 2: Calculate the heat gained by the water and calorimeter.
Qwater =mwater ×cwater ×∆T
Qcalorimeter =Qwater
Given:
mwater = 200 g = 0.2 kg
∆T=Tfinal −Tinitial = 30 −20 = 10 C
Qwater = 0.2×4200 ×10
Step 3: To find the specific heat capacity of copper, set the total heat lost
by copper equal to the total heat gained by water and calorimeter.
Qcopper =Qwater +Qcalorimeter
0.5×ccopper ×(−50) = 0.2×4200 ×10
ccopper =0.2×4200 ×10
0.5×(−50)
ccopper = 1680 J/kg·C
Therefore, the specific heat capacity of copper is 1680 J/kg·◦C.
Question 20
Question
A gas expands at constant pressure while absorbing 1500 J of heat. If the
internal energy of the gas increases by 2500 J, determine the work done by the
gas during the process.
19
Solution
Step 1: Recall that the first law of thermodynamics states that the change in
internal energy (∆U) of a system is equal to the heat added to the system minus
the work done by the system. Mathematically, this can be expressed as:
∆U=Q−W
where ∆Uis the change in internal energy, Qis the heat added to the system,
and Wis the work done by the system.
Step 2: From the information given in the question, we know that Q= 1500
J (heat absorbed by the gas) and ∆U= 2500 J (increase in internal energy).
Step 3: Substituting the known values into the first law of thermodynamics
equation, we have:
2500 J = 1500 J −W
Step 4: Rearranging the equation to solve for W, we get:
W= 1500 J −2500 J = −1000 J
Step 5: Therefore, the work done by the gas during the process is −1000 J.
This negative sign indicates that work is done on the gas (the gas is compressed)
rather than by the gas (the gas expands).
Question 21
Question
A copper sphere of radius 5 cm is heated to a temperature of 120◦C. If the
specific heat capacity of copper is 0.385 J/g◦C and the density of copper is
8.96 g/cm3, calculate the amount of heat required to raise the temperature of
the sphere to 120◦C.
Solution
Step 1: Calculate the mass of the copper sphere. Given that the density of
copper is 8.96 g/cm3, we can use the formula for the volume of a sphere to find
the mass of the sphere. The volume of a sphere is given by V=4
3πr3, where r
is the radius of the sphere. Substitute r= 5 cm into the formula to get:
V=4
3×π×(5)3cm3=500π
3cm3
Since density = mass
volume , we can rearrange the formula to find the mass:
mass = density ×volume = 8.96 g/cm3×500π
3cm3
20
Step 2: Calculate the amount of heat required. The specific heat formula is
Q=mc∆T, where: - Qis the heat energy, - mis the mass of the substance, -
cis the specific heat capacity, and - ∆Tis the change in temperature.
Given that the specific heat capacity (c) is 0.385 J/g◦C and the temperature
change (∆T) is 120◦C, we can substitute these values along with the mass we
calculated in Step 1 into the formula to find the amount of heat required.
Question 22
Question
A copper block initially at a temperature of 100
°
C is dropped into a calorimeter
containing 500 g of water at 20
°
C. If the final temperature of the system is
25
°
C, calculate the specific heat capacity of the copper block. The specific heat
capacity of water is 4.18 J/g
°
C and no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by
the water. Given: Mass of water, mw= 500 g Initial temperature of water,
Tinitial, water = 20◦C Final temperature of the system, Tfinal = 25◦C Specific
heat capacity of water, cwater = 4.18 J/g
°
C
Let the specific heat capacity of the copper block be denoted as ccopper.
The heat lost by the copper block is equal to the heat gained by the water:
mccopper(Tfinal −Tinitial, copper) = mcwater(Tfinal −Tinitial, water)
Step 2: Substitute the known values into the equation and solve for ccopper.
ccopper(25 −100) = 500 ×4.18 ×(25 −20)
ccopper(−75) = 500 ×4.18 ×5
ccopper =500 ×4.18 ×5
75
ccopper =1045 ×5
3
ccopper =5225
3
ccopper ≈1741.67 J/kg
°
C
Therefore, the specific heat capacity of the copper block is approximately
1741.67 J/kg
°
C.
21
Question 23
Question
A block of copper with a mass of 500 g is heated from 20
°
C to 100
°
C. Calculate
the amount of heat required to raise its temperature. The specific heat capacity
of copper is 0.385 J/g
°
C.
Solution
Step 1: Calculate the change in temperature. Step 2: Use the formula Q=
mc∆Tto find the amount of heat required.
Step 1: The change in temperature can be calculated as:
∆T=Tfinal −Tinitial = 100C−20C= 80C
Step 2: Using the formula Q=mc∆T, where: - mis the mass of the block
(500 g), - cis the specific heat capacity of copper (0.385 J/g
°
C), - ∆Tis the
change in temperature (80
°
C),
we can substitute the values and solve for Q:
Q= (500 g)(0.385 J/g
°
C)(80
°
C)
Q= 15400 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 100
°
C is 15400 J.
Question 24
Question
A 500 g aluminum pot at an initial temperature of 20
°
C contains 0.50 kg of
water at 10
°
C. How much heat is required to raise the temperature of the pot
and its contents to 100
°
C? Assume the specific heat capacity of aluminum is
900 J/kg
°
C, and the specific heat capacity of water is 4186 J/kg
°
C.
Solution
Step 1: Calculate the heat required to raise the temperature of the aluminum
pot: The formula to calculate heat is: Q=mc∆T, where Qis the heat, mis
the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Given mAl = 0.5 kg, cAl = 900 J/kg
°
C, Ti,Al = 20
°
C, and Tf= 100
°
C. We
can calculate ∆TAl as follows:
∆TAl =Tf−Ti,Al = 100 −20 = 80
°
C
22
Substitute the values into the formula:
QAl =mAlcAl∆TAl
QAl = 0.5×900 ×80
QAl = 36,000 J
Therefore, the heat required to raise the temperature of the aluminum pot
to 100
°
C is 36,000 J.
Step 2: Calculate the heat required to raise the temperature of the water:
Given mwater = 0.5 kg, cwater = 4186 J/kg
°
C, Ti,water = 10
°
C, and Tf= 100
°
C.
We can calculate ∆Twater as follows:
∆Twater =Tf−Ti,water = 100 −10 = 90
°
C
Substitute the values into the formula:
Qwater =mwatercwater∆Twater
Qwater = 0.5×4186 ×90
Qwater = 188,370 J
Therefore, the heat required to raise the temperature of the water to 100
°
C
is 188,370 J.
Step 3: Calculate the total heat required for both the pot and water:
Qtotal =QAl +Qwater
Qtotal = 36,000 + 188,370
Qtotal = 224,370 J
Therefore, the total heat required to raise the temperature of the pot and
its contents to 100
°
C is 224,370 J.
Question 25
Question
An aluminum bar of mass 0.5 kg and specific heat capacity 900 J/kg·K is initially
at a temperature of 200◦C. The bar is placed in a container of water at 20◦C.
The final equilibrium temperature of the system is 25◦C. What is the mass of
the water in the container? Assume no heat is lost to the surroundings.
23
Solution
Step 1: Calculate the heat lost by the aluminum bar as it cools down from
200◦C to 25◦C. The formula for heat transfer is given by Q=mc∆T, where
-Qis the heat transfer, - mis the mass of the object, - cis the specific heat
capacity of the object, and - ∆Tis the change in temperature of the object.
Given that m= 0.5 kg, c= 900 J/kg·K, and ∆T= 200 −25 = 175 K, we
can calculate the heat lost by the aluminum bar:
Qaluminum = (0.5)(900)(175)
Step 2: Calculate the heat gained by the water in the container as it warms
up from 20◦C to 25◦C. Let Mbe the mass of water in the container. The heat
gained by the water can be calculated using the same formula as above:
Qwater = (M)(4200)(5)
Step 3: The heat lost by the aluminum bar is equal to the heat gained by
the water.
Qaluminum =Qwater
(0.5)(900)(175) = (M)(4200)(5)
Step 4: Solve for the mass of water, M.
0.5×900 ×175 = M×4200 ×5
78750 = 21000M
M=78750
21000 = 3.75 kg
Therefore, the mass of water in the container is 3.75 kg.
Question 26
Question
A 1 kg block of copper at an initial temperature of 100◦C is placed in 2 kg of
water at 20◦C. The specific heat capacity of copper is 0.385 J/g◦C and that of
water is 4.18 J/g◦C. Assuming no heat is lost to the surroundings, calculate the
final temperature when the system reaches thermal equilibrium.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
water.
Heat lost by copper = mcopper ·ccopper ·∆T
Heat gained by water = mwater ·cwater ·∆T
24
where m= mass, c= specific heat capacity, ∆T= change in temperature.
Substitute the given values:
Heat lost by copper = 1 kg ×0.385 J/g◦C×(Tfinal −100◦C)
Heat gained by water = 2 kg ×4.18 J/g◦C×(Tfinal −20◦C)
Step 2: Set the heat lost equal to the heat gained to find the final tempera-
ture.
1×0.385 ×(Tfinal −100) = 2 ×4.18 ×(Tfinal −20)
Step 3: Solve for Tfinal.
⇒0.385Tfinal −38.5=8.36Tfinal −167.2
⇒8.36Tfinal −0.385Tfinal = 167.2−38.5
⇒7.975Tfinal = 128.7
⇒Tfinal =128.7
7.975
⇒Tfinal ≈16.1◦C
Therefore, the final temperature when the system reaches thermal equilib-
rium is approximately 16.1◦C.
Question 27
Question
A copper container of mass 0.5 kg contains 0.2 kg of water at 30
°
C. How much
heat must be supplied to the system in order to raise the temperature of the wa-
ter in the container to 90
°
C? The specific heat capacity of copper is 390 J/kg◦C
and the specific heat capacity of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
30
°
C to 90
°
C. The formula to calculate the heat required to change the temper-
ature of a substance is given by:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity of the substance, and - ∆Tis the change in temperature.
For the water:
Qwater = 0.2 kg ×4186 J/kg◦C×(90 −30)◦C
Qwater = 0.2×4186 ×60
25
Qwater = 50232 J
Step 2: Calculate the heat required to raise the temperature of the copper
container from 30
°
C to 90
°
C. Using the same formula, for the copper container:
Qcopper = 0.5 kg ×390 J/kg◦C×(90 −30)◦C
Qcopper = 0.5×390 ×60
Qcopper = 11700 J
Step 3: Calculate the total heat required for both the water and the copper
container.
Qtotal =Qwater +Qcopper
Qtotal = 50232 + 11700
Qtotal = 61932 J
Therefore, the total heat that must be supplied to the system is 61932 J.
Question 28
Question
A 200 g piece of aluminum at 80
°
C is placed in a calorimeter containing 800
g of water at 20
°
C. The specific heat capacity of aluminum is 0.897 J/g
°
C
and the specific heat capacity of water is 4.18 J/g
°
C. If no heat is lost to the
surroundings, what is the final equilibrium temperature of the system?
Solution
Step 1: Calculate the heat lost by the aluminum as it cools down to the final
equilibrium temperature. Using the formula for heat transfer: Q=mc∆T,
where Qis the heat, mis the mass, cis the specific heat capacity, and ∆Tis
the change in temperature. The heat lost by the aluminum can be calculated
as:
QAluminum = (200 g)(0.897 J/g
°
C)(80 −T)
°
C
where Tis the final equilibrium temperature of the system.
Step 2: Calculate the heat gained by the water as it warms up to the final
equilibrium temperature. Using the same formula as above, the heat gained by
the water can be calculated as:
QWater = (800 g)(4.18 J/g
°
C)(T−20)
°
C
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum must be equal to the heat gained by the water. Therefore, we set the
two equations equal to each other:
(200 g)(0.897 J/g
°
C)(80 −T) = (800 g)(4.18 J/g
°
C)(T−20)
26
Step 4: Solve the equation for the final equilibrium temperature T.
179.4(80 −T) = 3344(T−20)
14352 −179.4T= 3344T−66880
69816 = 513.4T
T=69816
513.4
T≈136
°
C
Therefore, the final equilibrium temperature of the system is approximately
136
°
C.
Question 29
Question
A copper rod of length 2.5 m and diameter 2 cm is initially at a temperature of
25
°
C. If 6.0 x 105Jofheatenergyissuppliedtotherod, calculatethef inaltemperatureoftherod, giventhatthespecificheatcapacityof copperis390
J/kg ·K and the density of copper is 8.96 g/cm3. Assume the rod is perfectly
insulated.
Solution
Step 1: Calculate the mass of the copper rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder:
V=πr2h,
where ris the radius and his the height (or length) of the rod. Given that
the diameter of the rod is 2 cm, the radius r= 1 cm = 0.01 m and the height
h= 2.5 m. Hence,
V=π×(0.01 m)2×2.5 m.
Calculating this gives us the volume of the rod.
Step 2: Calculate the mass of the copper rod. The density of copper is given
as 8.96 g/cm3. Converting this to kg/m3, we get 8960 kg/m3. The mass mof
the rod can be calculated as
m= density ×volume.
Step 3: Calculate the specific heat energy absorbed by the copper rod. The
specific heat energy Qabsorbed by the rod can be calculated using the formula
Q=mc∆T,
where mis the mass of the rod, cis the specific heat capacity of copper, and
∆Tis the change in temperature.
27
Step 4: Calculate the final temperature of the rod. Since the rod is perfectly
insulated, all the heat supplied to it will result in a temperature increase. Thus,
the final temperature can be calculated using the formula
Q=mc(Tf−Ti),
where Tfis the final temperature and Tiis the initial temperature. Rearranging
this formula gives us the final temperature Tf.
Now we can substitute the given values into the equations to find the final
temperature of the rod.
Question 30
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. It is then placed in an ice bath at 0◦C. If the rod loses heat energy at
a rate of 200 W, determine the time it takes for the temperature of the rod to
drop to 10◦C. Assume the specific heat capacity of copper is 390 J/kg ·K and
the density of copper is 8900 kg/m3.
Solution
Step 1: Calculate the mass of the copper rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder: V=πr2h, where
r= 0.005 m and h= 2 m.
V=π×(0.005 m)2×2 m = 0.000157 m3
The mass can be calculated using the formula m= density ×volume.
m= 8900 kg/m3×0.000157 m3= 1.3973 kg
Step 2: Calculate the heat energy required to change the temperature of the
copper rod from 100◦C to 10◦C. The formula for heat energy is Q=mc∆T,
where c= 390J/kg·K is the specific heat capacity of copper and ∆T= 100−10 =
90 K.
Q= 1.3973 kg ×390 J/kg ·K×90 K = 49257.3 J
Step 3: Calculate the time taken for the rod to cool from 100◦C to 10◦C.
Given that heat energy is lost at a rate of 200 W, the time taken can be calcu-
lated using the formula t=Q
P, where P= 200 W.
t=49257.3 J
200 W = 246.29 s
Therefore, it will take approximately 246.29 seconds for the temperature of
the rod to drop from 100◦C to 10◦C.
28
Question 31
Question
A copper rod of length 2 m and diameter 1 cm is initially at a uniform tempera-
ture of 100◦C. It is then placed in an ice-water mixture at a temperature of 0◦C.
Assuming the rod loses heat only through its lateral surface, calculate the time
it takes for the temperature at the center of the rod to drop to 10◦C. Given:
Thermal conductivity of copper = 400 W/mK, density of copper = 8900 kg/m3,
specific heat capacity of copper = 390 J/kgK.
Solution
Step 1: The initial temperature distribution in the rod will be given by the
one-dimensional heat conduction equation:
∂T
∂t =α∂2T
∂x2
where Tis the temperature, tis time, xis the distance along the rod, and αis
the thermal diffusivity given by α=k
ρc . Since the rod loses heat only through
its lateral surface, we consider heat conduction only along the length of the rod
(x-direction). The boundary conditions are:
At x=0: T(0, t) = 100◦C
At x=L/2 : T(L/2, t) = 10◦C
At t=0: T(x, 0) = 100◦C
Step 2: Let’s find the thermal diffusivity of copper using the given data:
α=k
ρc =400 W/mK
8900 kg/m3×390 J/kgK
α≈1.034 ×10−5m2/s
Step 3: The general solution to the one-dimensional heat conduction equa-
tion with the given boundary conditions is:
T(x, t) =
∞
X
n=1 Ancos nπx
Le−(nπ
L)2αt
where Lis the length of the rod.
Step 4: For T(L/2, t) = 10◦C, we have:
10 =
∞
X
n=1 Ancos nπ
4e−(nπ
2)2αt
Step 5: To find the time it takes for the temperature at the center of the rod
to drop to 10◦C, we would need to solve for tin the above equation to find the
sum of the series. This involves quite complex math and may require numerical
methods for the summation.
29
Question 32
Question
A copper rod of length 2 m is initially at a temperature of 100
°
C. One end of
the rod is then placed in an ice-water mixture at 0
°
C, while the other end is
placed in a furnace at 500
°
C. If the thermal conductivity of copper is 400 W/(m
K), determine the time taken for the whole rod to reach a uniform temperature.
Assume the rod is well-insulated along its length except at the ends.
Solution
Step 1: Calculate the rate of heat transfer per unit length for the rod.
The rate of heat transfer per unit length (Q) through the rod is given by
Fourier’s law as:
Q=kA∆T
L
where: - kis the thermal conductivity of copper (400 W/(m K)), - Ais the
cross-sectional area of the rod, - ∆Tis the temperature difference (500
°
C - 0
°
C
= 500 K), - Lis the length of the rod (2 m).
Step 2: Calculate the rate of heat transfer for the entire rod.
The rate of heat transfer for the entire rod is given by:
Qtotal =Q·Length of the rod
Step 3: Apply the principle of conservation of energy to find the time taken
for the rod to reach a uniform temperature.
The rate of heat transfer at the furnace end is equal to the rate of heat
transfer at the ice-water end. This implies that the total heat received by the
ice end is equal to the total heat loss at the furnace end.
The equation for this conservation of energy is:
Qtotal =mc∆T
where: - mis the mass of the rod, - cis the specific heat capacity of copper,
- ∆Tis the final temperature of the rod.
Solve for the time taken for the whole rod to reach a uniform temperature.
Question 33
Question
A steel rod of length 2.5 m and diameter 5 cm is initially at a temperature of
200
°
C. If the rod is heated until its temperature reaches 500
°
C, determine the
amount of heat transferred to the rod. Assume the thermal conductivity of steel
is 50 W/(m
·
K) and its specific heat capacity is 450 J/(kg
·
K).
30
Solution
Step 1: Calculate the mass of the steel rod. The volume of the steel rod can be
calculated using the formula for the volume of a cylinder:
V=πr2h,
where ris the radius of the rod and his the length of the rod. Given that
the diameter of the rod is 5 cm, the radius r=5
2cm = 0.025 m and the length
h= 2.5 m. Now, we can calculate the volume:
V=π(0.025)2×2.5.
Step 2: Calculate the mass of the steel rod. The density of steel is approxi-
mately 7850 kg/m
³
. Therefore, the mass of the steel rod can be calculated using
the formula:
mass = density ×volume.
Step 3: Calculate the heat energy absorbed by the steel rod. The heat energy
absorbed by the rod can be calculated using the formula:
Q=mc∆T,
where mis the mass of the rod, cis the specific heat capacity of steel, and ∆T
is the change in temperature.
Step 4: Calculate the change in temperature. The change in temperature,
∆T, is equal to the final temperature minus the initial temperature:
∆T= 500 −200.
Step 5: Find the amount of heat transferred to the rod. Finally, we can
substitute the values obtained into the formula for heat energy:
Q=mc∆T.
Question 34
Question
A copper rod of length 2 m and cross-sectional area 0.01 m2is initially at a
temperature of 100◦C. If one end of the rod is heated to a temperature of 200◦C
and the other end is kept at 100◦C, find the rate at which heat is conducted
along the rod. The thermal conductivity of copper is 400 W/(m K).
Solution
Step 1: Calculate the temperature difference between the two ends of the rod.
Given that one end is at 200◦Cand the other end is at 100◦C, the temperature
difference (∆T) is:
∆T=Thot −Tcold = 200◦C−100◦C= 100◦C
31
Step 2: Calculate the rate of heat conduction using the formula:
Q=−kA∆T
L
where: Q= rate of heat conduction, k= thermal conductivity of copper, A=
cross-sectional area of the rod, ∆T= temperature difference, L= length of the
rod.
Given that: k= 400 W/(m K), A= 0.01m2, ∆T= 100 K, L= 2 m,
we can plug in these values to find the rate of heat conduction:
Q=−400 ×0.01 ×100
2=−400 ×0.01 ×50 = −20 W
Therefore, the rate at which heat is conducted along the rod is 20 W.
Question 35
Question
A gas is compressed adiabatically from an initial volume of 2.00 m3to a final
volume of 0.50 m3. The initial pressure and temperature are 2.00 atm and 300
K, respectively. Compute the final pressure and temperature of the gas.
Solution
Step 1: Find the final pressure using the adiabatic compression formula:
P1
P2
γ−1
γ=V2
V1
where P1and P2are the initial and final pressures, V1and V2are the initial and
final volumes, and γis the adiabatic index (approximated as 1.4 for diatomic
gases).
Given V1= 2.00 m3,V2= 0.50 m3,P1= 2.00 atm, and γ= 1.4, we can plug
the values into the formula to solve for P2:
2.00 atm
P2
0.4
1.4=0.50 m3
2.00 m3
2.00
P20.286
= 0.25
2.00
P2
= 0.253.497
2.00
P2
≈0.00391
32
Step 3: Setting up an equation for the heat transfer to the water: As no
heat is lost to the surroundings, the heat lost by the copper block is equal to
the heat gained by the water. This allows us to write the equation:
2×390 ×(T−100) = 1000 ×4186 ×(T−20)
Step 4: Solve the equation to find the equilibrium temperature T: Solving
the equation from step 3:
780(T−100) = 4186000(T−20)
Solving for Tgives T= 24.6
°
C.
Therefore, the final equilibrium temperature when the copper block and
water reach thermal equilibrium is 24.6
°
C.
Question 2
Question
A piece of copper of mass 200 g at 200
°
C is placed in a container with 800
g of water at 20
°
C. If no heat is lost to the surroundings, what is the final
temperature of the system? Assume the specific heat capacities of copper and
water are cCu = 0.385 J/g
°
C and cH2O = 4.18 J/g
°
C, respectively, and the latent
heat of fusion for water is 334 J/g.
Solution
Step 1: Calculate the heat lost by the copper and the heat gained by the water.
The heat lost by the copper (QCu) can be calculated using the formula:
QCu =mCu ·cCu ·∆TCu
where: mCu = mass of copper = 200 g, cCu = specific heat capacity of copper
= 0.385 J/g
°
C, ∆TCu = change in temperature of copper = final temperature -
initial temperature.
Similarly, the heat gained by the water (QH2O) can be calculated using the
formula:
QH2O =mH2O ·cH2O ·∆TH2O
where: mH2O = mass of water = 800 g, cH2O = specific heat capacity of water
= 4.18 J/g
°
C, ∆TH2O = change in temperature of water = final temperature -
initial temperature.
Since the system is isolated and no heat is lost to the surroundings, QCu =
−QH2O. Let the final temperature be Tf.
Step 2: Set up the equation using the heat lost and gained expressions. Since
QCu =−QH2O, we have:
mCu ·cCu ·(Tf−200) = −mH2O ·cH2O ·(Tf−20)
2
Step 3: Solve for the final temperature. Plug in the given values:
200 ·0.385 ·(Tf−200) = −800 ·4.18 ·(Tf−20)
Solving this equation will give the final temperature Tf.
Question 3
Question
A piece of iron has a mass of 500 g and is initially at a temperature of 100
°
C.
It is dropped into a bucket containing 2 kg of water at 20
°
C. Assuming no heat
is lost to the surroundings, calculate the final equilibrium temperature of the
system. The specific heat capacity of iron is 450 J/kg◦C and that of water is
4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the iron.
The heat lost by the iron can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the iron (in kg), - cis the specific heat capacity of
iron (in J/kg◦C), and - ∆Tis the change in temperature of the iron (initial
temperature - final temperature).
Converting the mass of the iron to kilograms:
m=500 g
1000 = 0.5 kg
Plugging in the values:
Qiron = 0.5×450 ×(100 −Tf)
Qiron = 225 ×(100 −Tf)
Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
Q=mc∆T
where: - mis the mass of the water (in kg), - cis the specific heat capacity of
water (in J/kg◦C), and - ∆Tis the change in temperature of the water (final
temperature - initial temperature).
Converting the mass of the water to kilograms:
m= 2 kg
3
Plugging in the values:
Qwater = 2 ×4186 ×(Tf−20)
Qwater = 8372 ×(Tf−20)
Step 3: Setting up the energy balance equation.
Since there is no heat lost to the surroundings, the heat lost by the iron is equal
to the heat gained by the water:
Qiron =Qwater
225 ×(100 −Tf) = 8372 ×(Tf−20)
Step 4: Solve for Tf.
Expanding and simplifying the equation:
22500 −225Tf= 8372Tf−167440
225Tf+ 8372Tf= 167440 + 22500
8597Tf= 190940
Tf=190940
8597
Tf≈22.23◦C
Therefore, the final equilibrium temperature of the system is approximately
22.23◦C.
Question 4
Question
A copper rod of length 2 meters and diameter 1 centimeter is heated from 20
°
C
to 80
°
C. If the thermal conductivity of copper is 390 W/mK, calculate the rate
at which heat flows through the rod.
Solution
Step 1: Calculate the cross-sectional area of the copper rod. The cross-sectional
area of the rod can be calculated using the formula for the area of a circle:
A=πr2, where ris the radius of the rod. Given that the diameter of the rod
is 1 centimeter, the radius ris 0.5 cm (or 0.005 m). The cross-sectional area is
then:
A=π(0.005)2= 0.00007854 m2
Step 2: Calculate the temperature difference. The temperature difference
∆Tis given by:
∆T=T2−T1= 80C−20C= 60C
4
Step 3: Calculate the rate of heat flow. The rate of heat flow (Q) through
the rod is given by Fourier’s Law:
Q=k·A·∆T
L
where: k= thermal conductivity of copper (390 W/mK), A= cross-sectional
area of the rod (0.00007854 m2), ∆T= temperature difference (60C), L= length
of the rod (2 m).
Substitute the values into the formula:
Q= 390 ×0.00007854 ×60
2= 0.921 W
Therefore, the rate at which heat flows through the rod is 0.921 W.
Question 5
Question
A copper rod of length 2 meters and cross-sectional area 2 ×10−4m2is initially
at a temperature of 200◦C. If 5000 J of heat is added to the rod, what is the
final temperature of the rod? (Note: The specific heat capacity of copper is
390 J/kg ·K and the density of copper is 8900 kg/m3.)
Solution
Step 1: Find the mass of the copper rod. The mass of the copper rod can be
calculated using the formula:
mass = density ×volume = ρ×A×L
where: ρ= density of copper = 8900 kg/m3,A= cross-sectional area = 2 ×
10−4m2,L= length of the rod = 2 meters.
Substitute the given values into the formula:
mass = 8900 ×2×10−4×2=3.56 kg
Step 2: Calculate the change in temperature. The heat added to the rod
can be expressed as:
Q=mc∆T
where: Q= heat added = 5000 J, m= mass of the rod = 3.56 kg, c= specific
heat capacity of copper = 390 J/kg ·K, ∆T= change in temperature.
Solve for ∆T:
∆T=Q
mc =5000
3.56 ×390 ≈3.24 K
Step 3: Find the final temperature of the rod. The final temperature Tfcan
be calculated as:
Tf=Ti+ ∆T
5
where: Ti= initial temperature of the rod = 200◦C⇒200 + 273 = 473 K.
Substitute the values into the formula:
Tf= 473 + 3.24 = 476.24 K
Convert the final temperature from Kelvin back to degrees Celsius:
Tf= 476.24 −273 ≈203.24 ◦C
Therefore, the final temperature of the copper rod is approximately 203.24◦C.
Question 6
Question
A metal block of mass 2 kg is heated to a temperature of 80
°
C and then placed in
a large container of water at 20
°
C. The initial temperature of the water is 25
°
C
and its mass is 5 kg. Assuming no heat loss to the surroundings, calculate the
final temperature of the system (metal block + water) when thermal equilibrium
is reached. The specific heat capacity of the metal is 500 J/kg
°
C and that of
water is 4186 J/kg
°
C.
Solution
Let the final temperature of the system be T
°
C. The heat lost by the metal
block must equal the heat gained by the water.
Step 1: Calculate the heat lost by the metal block. The heat lost by the
metal block can be calculated using the formula:
Qlost =mc∆T
where: - mis the mass of the metal block, - cis the specific heat capacity of
the metal, and - ∆Tis the change in temperature of the metal block.
Substitute the given values:
Qlost = 2 kg ×500 J/kg
°
C×(80 −T)
°
C
Step 2: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the same formula:
Qgained =mc∆T
where: - mis the mass of the water, - cis the specific heat capacity of water,
and - ∆Tis the change in temperature of the water.
Substitute the given values:
Qgained = 5 kg ×4186 J/kg
°
C×(T−25)
°
C
6
Step 3: Set up the equation for heat equilibrium. Since there is no heat
loss to the surroundings, the heat lost by the metal block must equal the heat
gained by the water:
2×500 ×(80 −T)=5×4186 ×(T−25)
Step 4: Solve the equation for T. Simplify and solve the equation to find
the final temperature Tof the system. This involves expanding and rearranging
terms before solving for T.
Question 7
Question
A sample of aluminum initially at a temperature of 100
°
C is placed in contact
with a reservoir at 0
°
C. The sample exchanges heat with the reservoir until it
reaches thermal equilibrium. Given that the mass of the aluminum sample is
500g and the specific heat capacity of aluminum is 0.9 J/g
°
C, calculate the final
temperature of the aluminum sample.
Solution
Step 1: Calculate the heat transferred from the aluminum sample to the reser-
voir using the formula:
Q=mc∆T
where: Q= heat transferred, m= mass of the aluminum sample = 500g, c=
specific heat capacity of aluminum = 0.9 J/g
°
C, ∆T= change in temperature.
The initial temperature of the aluminum sample is 100
°
C and the final tem-
perature is θ
°
C. Therefore, the change in temperature, ∆T=θ−100. Substitute
the given values into the formula:
Q= 500 ×0.9×(θ−100)
Step 2: Calculate the heat absorbed by the reservoir, which is equal in
magnitude to the heat lost by the aluminum sample.
Q=−mc∆T
Substitute the given values into the formula:
Q=−500 ×0.9×θ
Step 3: Set the two equations equal to each other and solve for θ.
500 ×0.9×(θ−100) = −500 ×0.9×θ
Step 4: Solve the equation for θ.
450(θ−100) = −450θ
7
450θ−45000 = −450θ
450θ+ 450θ= 45000
900θ= 45000
θ=45000
900
θ= 50
Therefore, the final temperature of the aluminum sample is 50
°
C.
Question 8
Question
A copper bar of length 2 m is initially at a uniform temperature of 400 K. One
end of the bar is then heated until its temperature is 600 K. If the specific heat
capacity of copper is 390 J/kg·K and its density is 8930 kg/m3, find the amount
of heat energy transferred to the bar during this process. Assume no heat is
lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper bar. Given that the density of copper
is 8930 kg/m3, the mass of the copper bar can be calculated using the formula:
mass = density ×volume
The volume of the copper bar is V= 2 m ×A, where Ais the cross-sectional
area of the bar. Since the bar is uniform, we can assume the cross-sectional area
is constant.
Step 2: Calculate the amount of heat energy transferred. The amount of
heat energy transferred can be calculated using the formula:
Q=mc∆T
where mis the mass of the copper bar, cis the specific heat capacity of copper,
and ∆Tis the change in temperature of the copper bar.
Step 3: Substitute the known values into the formula. We have already
calculated the mass of the copper bar in Step 1. The specific heat capacity of
copper is 390 J/kg·K. The initial temperature is 400 K and the final temperature
is 600 K, so the change in temperature is ∆T= 600 K −400 K = 200 K.
Step 4: Calculate the amount of heat energy transferred. Substitute the
values into the formula to find the amount of heat energy transferred:
Q=mc∆T
Q= mass ×c×∆T
8
Q= (8930 kg/m3×2 m ×A)×390 J/kg ·K×200 K
Q= 17,860 AJ
Therefore, the amount of heat energy transferred to the copper bar during
this process is 17,860 AJ.
Question 9
Question
A 2 kg block of copper at 100◦C is placed into a container filled with 1 kg of
water at 20◦C. Assuming no heat is lost to the surroundings, what will be the
final temperature of the system? (Specific heat capacity of copper = 390 J/kg◦C,
specific heat capacity of water = 4186 J/kg◦C)
Solution
Let the final temperature of the system be T◦C.
Step 1: Write the heat equation for the system. The heat gained by the
copper block (cooling down from 100◦C to T◦C) is equal to the heat lost by
the water (heating up from 20◦C to T◦C). This can be expressed as:
mcopper ·ccopper ·(T−100) = −mwater ·cwater ·(20 −T)
where: mcopper = 2 kg (mass of copper block), ccopper = 390 J/kg◦C (specific
heat capacity of copper), mwater = 1 kg (mass of water), cwater = 4186 J/kg◦C
(specific heat capacity of water).
Step 2: Solve the heat equation. Substitute the given values into the heat
equation and solve for T:
2×390 ×(T−100) = −1×4186 ×(20 −T)
780 ×(T−100) = −4186 ×(20 −T)
780T−78000 = −83720 + 4186T
3614T= 5720
T=5720
3614 ≈1.58
°
C
Step 3: Answer Therefore, the final temperature of the system will be
approximately 1.58◦C.
9
Question 10
Question
A metal bar of mass 0.5 kg at an initial temperature of 100
°
C is placed in a
large ice bath at 0
°
C. The specific heat of the metal is 450 J/kg
°
C. Assuming
no heat is lost to the surroundings, calculate the final temperature of the metal
bar when it reaches thermal equilibrium with the ice bath.
Solution
Step 1: Calculate the heat lost by the metal bar as it cools down from 100
°
C
to the final temperature. Given: - Initial temperature, Ti= 100C- Final
temperature, Tf=? - Specific heat, c= 450J/kgC - Mass of the metal bar,
m= 0.5kg - Heat lost, Q=mc∆T
From the formula,
Q=mc∆T
we have,
Q= 0.5×450 ×(100 −Tf)
Q= 225(100 −Tf)
Step 2: Calculate the heat gained by the metal bar as it absorbs heat from
the ice bath to raise its temperature to the final temperature. Given: - Initial
temperature of the ice bath, Tice = 0C- Final temperature, Tf- Heat gained,
Q=mc∆T
From the formula,
Q=mc∆T
we have,
Q= 0.5×450 ×(Tf−0)
Q= 225Tf
Step 3: Equate the heat lost and heat gained equations to find the final
temperature.
225(100 −Tf) = 225Tf
22500 −225Tf= 225Tf
225Tf+ 225Tf= 22500
450Tf= 22500
Tf= 50C
Therefore, the final temperature of the metal bar when it reaches thermal
equilibrium with the ice bath is 50
°
C.
10
Question 11
Question
A copper pot contains 0.5 kg of water at 20
°
C. How much heat is required to
raise the temperature of the water to 100
°
C? (Specific heat capacity of copper
= 390 J/kg◦C, specific heat capacity of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat required to raise the temperature of the water to
100
°
C.
The heat required to raise the temperature of a substance can be calculated
using the formula:
Q=mc∆T
where: Q= heat energy (Joules), m= mass of the substance (kg), c= spe-
cific heat capacity of the substance (J/kg◦C), and ∆T= change in temperature
(◦C).
Given: m= 0.5 kg, cwater = 4186 J/kg◦C, ∆T= 100 −20 = 80◦C.
Substitute these values into the formula to find the heat required for the
water:
Qwater = 0.5 kg ×4186 J/kg◦C×80◦C
Qwater = 167440 J
Therefore, the heat required to raise the temperature of the water to 100
°
C
is 167440 Joules.
Step 2: Calculate the heat capacity of the copper pot.
The heat capacity of the copper pot can be calculated using the formula:
Q=mc∆T
where: Q= heat energy (Joules), m= mass of the substance (kg), c= spe-
cific heat capacity of the substance (J/kg◦C), and ∆T= change in temperature
(◦C).
Given: mcopper ≈mwater,ccopper = 390 J/kg◦C, ∆Tcopper = 100−20 = 80◦C.
Substitute these values into the formula to find the heat capacity of the
copper pot:
Qcopper = 0.5 kg ×390 J/kg◦C×80◦C
Qcopper = 15600 J
Therefore, the heat capacity of the copper pot is 15600 Joules.
Step 3: The total heat required is the sum of the heat required for the water
and the copper pot.
11
Qtotal =Qwater +Qcopper
Qtotal = 167440 J + 15600 J
Qtotal = 183040 J
Therefore, the total heat required to raise the temperature of the water to
100
°
C is 183040 Joules.
Question 12
Question
A 0.5 kg block of copper is heated from 20
°
C to 50
°
C. Calculate the amount of
heat transferred to the block. (Specific heat capacity of copper is 390 J/kg ·K)
Solution
Step 1: Calculate the change in temperature of the block. Given: Initial tem-
perature (Ti) = 20
°
C Final temperature (Tf) = 50
°
C
The change in temperature (∆T) can be calculated as:
∆T=Tf−Ti= 50C−20C= 30C
Step 2: Use the formula for heat transfer to find the amount of heat trans-
ferred. The formula for heat transfer is:
Q=mc∆T
where: Q= amount of heat transferred m= mass of the block (0.5 kg) c=
specific heat capacity of copper (390 J/kg*K) ∆T= change in temperature
(30
°
C)
Substitute the given values into the formula:
Q= 0.5 kg ×390 J/kg ·K×30 K
Q= 5850 J
Therefore, the amount of heat transferred to the block is 5850 J.
Question 13
Question
A copper ball at a temperature of 200
°
C is placed into a container of water at
20
°
C. The mass of the copper ball is 500 g and the mass of the water is 2 kg.
Assuming no heat is lost to the surroundings and that the specific heat capacities
of copper and water are 0.385 J/g
°
C and 4.18 J/g
°
C respectively, calculate the
final temperature of the system after thermal equilibrium is reached.
12
Solution
Step 1: Calculate the heat lost by the copper ball as it cools down to the final
temperature. The formula to calculate heat is:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity, and - ∆Tis the change in temperature.
The heat lost by the copper ball can be calculated as:
Qcopper =mcopper ×ccopper ×(200 −Tf)
Qcopper = 500 ×0.385 ×(200 −Tf)
Step 2: Calculate the heat gained by the water as it warms up to the final
temperature. Using the same formula as above:
Qwater =mwater ×cwater ×(Tf−20)
Qwater = 2000 ×4.18 ×(Tf−20)
Step 3: Since energy is conserved in this closed system, the heat lost by the
copper ball is equal to the heat gained by the water. Therefore, we can set the
two equations equal to each other and solve for the final temperature (Tf).
500 ×0.385 ×(200 −Tf) = 2000 ×4.18 ×(Tf−20)
Step 4: Solve for Tf.
192.5×(200 −Tf) = 8360 ×(Tf−20)
38500 −192.5Tf= 8360Tf−167200
38500 + 167200 = 192.5Tf+ 8360Tf
205700 = 8552.5Tf
Tf=205700
8552.5
Tf≈24.06C
Therefore, the final temperature of the system after thermal equilibrium is
reached is approximately 24.06C.
Question 14
Question
A 0.5 kg block of copper is initially at a temperature of 50◦C. How much heat
needs to be added to the block to raise its temperature to 100◦C? The specific
heat capacity of copper is 390 J/kg ·K.
13
Solution
Step 1: Determine the change in temperature. The change in temperature (∆T)
can be calculated using the formula:
∆T=Tf−Ti
where Tfis the final temperature and Tiis the initial temperature. Given that
the initial temperature Ti= 50◦Cand the final temperature Tf= 100◦C, we
have:
∆T= 100◦C−50◦C= 50◦C
Step 2: Calculate the amount of heat using the specific heat capacity formula.
The amount of heat (Q) needed to change the temperature of a substance can
be calculated using the formula:
Q=mc∆T
where: m= 0.5 kg (mass of the copper block), c= 390 J/kg ·K (specific heat
capacity of copper), and ∆T= 50◦C(change in temperature).
Substitute the values into the formula:
Q= (0.5 kg)(390 J/kg ·K)(50 ◦C)
Step 3: Calculate the amount of heat.
Q= 0.5×390 ×50 = 9750 J
Therefore, 9750 J of heat needs to be added to the copper block to raise its
temperature to 100◦C.
Question 15
Question
A 2 kg block of copper at an initial temperature of 300 K is placed in a container
with 1 kg of water at 350 K. The thermal conductivity of copper is 400 W/mK
and that of water is 0.6 W/mK. Assuming no heat is lost to the surroundings,
calculate the final equilibrium temperature of the system.
Solution
Step 1: Calculate the heat gained or lost by each substance using the equation
Q=mc∆T, where mis mass, cis specific heat capacity, and ∆Tis the change
in temperature.
For copper: Qcopper =mcopperccopper∆Tcopper
Given: mcopper = 2 kg, ccopper = 390 J/kgK, ∆Tcopper =T−300 K.
Step 2: Calculate Qcopper.
14
Qcopper = 2 ×390 ×(T−300)
Step 3: For water: Qwater =mwatercwater∆Twater
Given: mwater = 1 kg, cwater = 4186 J/kgK, ∆Twater =T−350 K.
Step 4: Calculate Qwater.
Qwater = 1 ×4186 ×(T−350)
Step 5: Since there is no heat loss, Qcopper =−Qwater.
Step 6: Solve the equation for T.
2×390 ×(T−300) = −1×4186 ×(T−350)
Step 7: Solve for Tto find the final equilibrium temperature.
Question 16
Question
An iron bar of length 1.5 m and cross-sectional area 4 cm2is heated at one end
to 100◦C. If the coefficient of linear expansion of iron is 1.2×10−5◦C−1and
its Young’s modulus is 2 ×1011 N/m2, determine the increase in length of the
bar due to heating.
Solution
Step 1: Calculate the initial length change of the iron bar due to heating. Given
that the coefficient of linear expansion of iron is α= 1.2×10−5◦C−1, the change
in length ∆Lof the iron bar when heated from 0◦C to 100◦C can be calculated
using the formula:
∆L=L·α·∆T
where Lis the initial length, αis the coefficient of linear expansion, and ∆Tis
the change in temperature. Substitute the values L= 1.5 m, α= 1.2×10−5
◦C−1, and ∆T= 100 ◦C into the formula:
∆L= 1.5 m ·1.2×10−5◦C−1·100 ◦C
∆L= 1.8×10−3m=1.8 mm
Step 2: Calculate the increase in length of the iron bar due to heating. The
increase in length of the iron bar due to heating is related to the stress and
Young’s modulus by the formula:
Stress = F orce
Area =Y·∆L
L
where Yis the Young’s modulus, ∆Lis the change in length, and Lis the initial
length. Rearrange the formula to solve for the increase in length ∆Lincrease:
∆Lincrease =Stress ·L
Y
15
Substitute the values Y= 2 ×1011 N/m2,L= 1.5 m, and Stress = F orce
Area into
the formula:
∆Lincrease =
F
Area ·1.5
2×1011
Since F= 0 initially and Fonly arises due to the increase in length,
∆Lincrease =
F
Area ·1.5
2×1011 =
F
4×10−4·1.5
2×1011 =1.5·F
8×107
∆Lincrease =1.5·F
8×107
Therefore, the increase in length of the iron bar due to heating is 1.5·F
8×107
meters.
Question 17
Question
A 0.2 kg block of ice at -10
°
C is placed on a stove where it absorbs 5000 J of
heat. The specific heat capacity of ice is 2100 J/kg ·K, the specific heat capacity
of water is 4200 J/kg ·K, and the latent heat of fusion of ice is 3.34 ×105J/kg.
Determine the final temperature of the water.
Solution
Step 1: Calculate the heat required for the ice to reach 0
°
C. The specific heat
equation is given by:
Q=mc∆T
where Qis the heat absorbed, mis the mass of the substance, cis the specific
heat capacity, and ∆Tis the change in temperature.
Substitute the given values for ice:
5000 = 0.2×2100 ×(0 −(−10))
5000 = 0.2×2100 ×10
5000 = 4200 ×2
∆Q= 8400 J
Step 2: Calculate the heat required for the ice to melt at 0
°
C. The heat
required for a change of state is given by:
Q=mL
where Qis the heat absorbed, mis the mass of the substance, and Lis the
latent heat of fusion.
16
Substitute the given values for ice:
8400 = 0.2×3.34 ×105
8400 = 66800 J
Step 3: Calculate the heat required for the water to increase to the final
temperature. The specific heat equation is given by:
Q=mc∆T
Substitute the given values for water:
5000 = 0.2×4200 ×(Tfinal −0)
5000 = 840 ×Tfinal
Tfinal =5000
840
Tfinal ≈5.95
°
C
Therefore, the final temperature of the water is approximately 5.95
°
C.
Question 18
Question
A copper rod of length 2.0 m and cross-sectional area 4.0 cm2is initially at
a temperature of 100◦C. It is placed in a large bath of water at 10◦C until
it reaches thermal equilibrium. If the rod has a thermal conductivity of 400
W/(m*K) and a specific heat capacity of 385 J/(kg*K), determine the heat
transfer in the rod during this process.
Solution
Step 1: Calculate the volume of the copper rod. The volume Vof the copper
rod can be calculated as:
V= Area ×Length
V= 4.0×10−4m2×2.0 m
V= 8 ×10−4m3
Step 2: Calculate the mass of the copper rod. The mass mof the copper
rod can be calculated using the density of copper, which is 8900 kg/m3:
m= Density ×Volume
m= 8900 kg/m3×8×10−4m3
m= 7.12 kg
17
Step 3: Calculate the initial energy in the copper rod. The initial energy Qi
in the copper rod can be calculated using the specific heat capacity of copper:
Qi=m×c×∆T
Qi= 7.12 kg ×385 J/(kg*K) ×(100 −10) K
Qi= 2.34 ×104J
Step 4: Calculate the rate of heat flow out of the rod. The rate of heat flow
Pout of the rod can be calculated using the thermal conductivity of copper:
P=k×Area ×∆T
L
P=400 W/(m*K) ×4.0×10−4m2×(100 −10) K
2.0 m
P= 1400 W
Step 5: Determine 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 initial energy and the rate of heat flow:
Qi=P×t
2.34 ×104J = 1400 W ×t
t≈16.7 s
Step 6: Calculate the total heat transfer in the rod during this process. The
total heat transfer Qin the rod can be calculated as:
Q=P×t
Q= 1400 W ×16.7 s
Q= 2.34 ×104J
Therefore, the heat transfer in the rod during this process is 23,400 J.
Question 19
Question
A copper block of mass 500 g is initially at a temperature of 80◦C. It is placed
in a calorimeter containing 200 g of water at 20◦C. If the final temperature of
the system is 30◦C, calculate the specific heat capacity of copper. Assume no
heat is lost to the surroundings.
18
Solution
Step 1: Calculate the heat lost by the copper block.
Heat lost by copper = mass ×specific heat capacity ×∆temperature
Qcopper =mcopper ×ccopper ×∆T
Given:
mcopper = 500 g = 0.5 kg
cwater = 4200 J/kg ·◦C
Initial temperature of copper, Tinitial = 80◦C Final temperature of system,
Tfinal = 30◦C
∆T=Tfinal −Tinitial = 30 −80 = −50 C
Qcopper = 0.5×ccopper ×(−50)
Step 2: Calculate the heat gained by the water and calorimeter.
Qwater =mwater ×cwater ×∆T
Qcalorimeter =Qwater
Given:
mwater = 200 g = 0.2 kg
∆T=Tfinal −Tinitial = 30 −20 = 10 C
Qwater = 0.2×4200 ×10
Step 3: To find the specific heat capacity of copper, set the total heat lost
by copper equal to the total heat gained by water and calorimeter.
Qcopper =Qwater +Qcalorimeter
0.5×ccopper ×(−50) = 0.2×4200 ×10
ccopper =0.2×4200 ×10
0.5×(−50)
ccopper = 1680 J/kg·C
Therefore, the specific heat capacity of copper is 1680 J/kg·◦C.
Question 20
Question
A gas expands at constant pressure while absorbing 1500 J of heat. If the
internal energy of the gas increases by 2500 J, determine the work done by the
gas during the process.
19
Solution
Step 1: Recall that the first law of thermodynamics states that the change in
internal energy (∆U) of a system is equal to the heat added to the system minus
the work done by the system. Mathematically, this can be expressed as:
∆U=Q−W
where ∆Uis the change in internal energy, Qis the heat added to the system,
and Wis the work done by the system.
Step 2: From the information given in the question, we know that Q= 1500
J (heat absorbed by the gas) and ∆U= 2500 J (increase in internal energy).
Step 3: Substituting the known values into the first law of thermodynamics
equation, we have:
2500 J = 1500 J −W
Step 4: Rearranging the equation to solve for W, we get:
W= 1500 J −2500 J = −1000 J
Step 5: Therefore, the work done by the gas during the process is −1000 J.
This negative sign indicates that work is done on the gas (the gas is compressed)
rather than by the gas (the gas expands).
Question 21
Question
A copper sphere of radius 5 cm is heated to a temperature of 120◦C. If the
specific heat capacity of copper is 0.385 J/g◦C and the density of copper is
8.96 g/cm3, calculate the amount of heat required to raise the temperature of
the sphere to 120◦C.
Solution
Step 1: Calculate the mass of the copper sphere. Given that the density of
copper is 8.96 g/cm3, we can use the formula for the volume of a sphere to find
the mass of the sphere. The volume of a sphere is given by V=4
3πr3, where r
is the radius of the sphere. Substitute r= 5 cm into the formula to get:
V=4
3×π×(5)3cm3=500π
3cm3
Since density = mass
volume , we can rearrange the formula to find the mass:
mass = density ×volume = 8.96 g/cm3×500π
3cm3
20
Step 2: Calculate the amount of heat required. The specific heat formula is
Q=mc∆T, where: - Qis the heat energy, - mis the mass of the substance, -
cis the specific heat capacity, and - ∆Tis the change in temperature.
Given that the specific heat capacity (c) is 0.385 J/g◦C and the temperature
change (∆T) is 120◦C, we can substitute these values along with the mass we
calculated in Step 1 into the formula to find the amount of heat required.
Question 22
Question
A copper block initially at a temperature of 100
°
C is dropped into a calorimeter
containing 500 g of water at 20
°
C. If the final temperature of the system is
25
°
C, calculate the specific heat capacity of the copper block. The specific heat
capacity of water is 4.18 J/g
°
C and no heat is lost to the surroundings.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by
the water. Given: Mass of water, mw= 500 g Initial temperature of water,
Tinitial, water = 20◦C Final temperature of the system, Tfinal = 25◦C Specific
heat capacity of water, cwater = 4.18 J/g
°
C
Let the specific heat capacity of the copper block be denoted as ccopper.
The heat lost by the copper block is equal to the heat gained by the water:
mccopper(Tfinal −Tinitial, copper) = mcwater(Tfinal −Tinitial, water)
Step 2: Substitute the known values into the equation and solve for ccopper.
ccopper(25 −100) = 500 ×4.18 ×(25 −20)
ccopper(−75) = 500 ×4.18 ×5
ccopper =500 ×4.18 ×5
75
ccopper =1045 ×5
3
ccopper =5225
3
ccopper ≈1741.67 J/kg
°
C
Therefore, the specific heat capacity of the copper block is approximately
1741.67 J/kg
°
C.
21
Question 23
Question
A block of copper with a mass of 500 g is heated from 20
°
C to 100
°
C. Calculate
the amount of heat required to raise its temperature. The specific heat capacity
of copper is 0.385 J/g
°
C.
Solution
Step 1: Calculate the change in temperature. Step 2: Use the formula Q=
mc∆Tto find the amount of heat required.
Step 1: The change in temperature can be calculated as:
∆T=Tfinal −Tinitial = 100C−20C= 80C
Step 2: Using the formula Q=mc∆T, where: - mis the mass of the block
(500 g), - cis the specific heat capacity of copper (0.385 J/g
°
C), - ∆Tis the
change in temperature (80
°
C),
we can substitute the values and solve for Q:
Q= (500 g)(0.385 J/g
°
C)(80
°
C)
Q= 15400 J
Therefore, the amount of heat required to raise the temperature of the copper
block from 20
°
C to 100
°
C is 15400 J.
Question 24
Question
A 500 g aluminum pot at an initial temperature of 20
°
C contains 0.50 kg of
water at 10
°
C. How much heat is required to raise the temperature of the pot
and its contents to 100
°
C? Assume the specific heat capacity of aluminum is
900 J/kg
°
C, and the specific heat capacity of water is 4186 J/kg
°
C.
Solution
Step 1: Calculate the heat required to raise the temperature of the aluminum
pot: The formula to calculate heat is: Q=mc∆T, where Qis the heat, mis
the mass, cis the specific heat capacity, and ∆Tis the change in temperature.
Given mAl = 0.5 kg, cAl = 900 J/kg
°
C, Ti,Al = 20
°
C, and Tf= 100
°
C. We
can calculate ∆TAl as follows:
∆TAl =Tf−Ti,Al = 100 −20 = 80
°
C
22
Substitute the values into the formula:
QAl =mAlcAl∆TAl
QAl = 0.5×900 ×80
QAl = 36,000 J
Therefore, the heat required to raise the temperature of the aluminum pot
to 100
°
C is 36,000 J.
Step 2: Calculate the heat required to raise the temperature of the water:
Given mwater = 0.5 kg, cwater = 4186 J/kg
°
C, Ti,water = 10
°
C, and Tf= 100
°
C.
We can calculate ∆Twater as follows:
∆Twater =Tf−Ti,water = 100 −10 = 90
°
C
Substitute the values into the formula:
Qwater =mwatercwater∆Twater
Qwater = 0.5×4186 ×90
Qwater = 188,370 J
Therefore, the heat required to raise the temperature of the water to 100
°
C
is 188,370 J.
Step 3: Calculate the total heat required for both the pot and water:
Qtotal =QAl +Qwater
Qtotal = 36,000 + 188,370
Qtotal = 224,370 J
Therefore, the total heat required to raise the temperature of the pot and
its contents to 100
°
C is 224,370 J.
Question 25
Question
An aluminum bar of mass 0.5 kg and specific heat capacity 900 J/kg·K is initially
at a temperature of 200◦C. The bar is placed in a container of water at 20◦C.
The final equilibrium temperature of the system is 25◦C. What is the mass of
the water in the container? Assume no heat is lost to the surroundings.
23
Solution
Step 1: Calculate the heat lost by the aluminum bar as it cools down from
200◦C to 25◦C. The formula for heat transfer is given by Q=mc∆T, where
-Qis the heat transfer, - mis the mass of the object, - cis the specific heat
capacity of the object, and - ∆Tis the change in temperature of the object.
Given that m= 0.5 kg, c= 900 J/kg·K, and ∆T= 200 −25 = 175 K, we
can calculate the heat lost by the aluminum bar:
Qaluminum = (0.5)(900)(175)
Step 2: Calculate the heat gained by the water in the container as it warms
up from 20◦C to 25◦C. Let Mbe the mass of water in the container. The heat
gained by the water can be calculated using the same formula as above:
Qwater = (M)(4200)(5)
Step 3: The heat lost by the aluminum bar is equal to the heat gained by
the water.
Qaluminum =Qwater
(0.5)(900)(175) = (M)(4200)(5)
Step 4: Solve for the mass of water, M.
0.5×900 ×175 = M×4200 ×5
78750 = 21000M
M=78750
21000 = 3.75 kg
Therefore, the mass of water in the container is 3.75 kg.
Question 26
Question
A 1 kg block of copper at an initial temperature of 100◦C is placed in 2 kg of
water at 20◦C. The specific heat capacity of copper is 0.385 J/g◦C and that of
water is 4.18 J/g◦C. Assuming no heat is lost to the surroundings, calculate the
final temperature when the system reaches thermal equilibrium.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
water.
Heat lost by copper = mcopper ·ccopper ·∆T
Heat gained by water = mwater ·cwater ·∆T
24
where m= mass, c= specific heat capacity, ∆T= change in temperature.
Substitute the given values:
Heat lost by copper = 1 kg ×0.385 J/g◦C×(Tfinal −100◦C)
Heat gained by water = 2 kg ×4.18 J/g◦C×(Tfinal −20◦C)
Step 2: Set the heat lost equal to the heat gained to find the final tempera-
ture.
1×0.385 ×(Tfinal −100) = 2 ×4.18 ×(Tfinal −20)
Step 3: Solve for Tfinal.
⇒0.385Tfinal −38.5=8.36Tfinal −167.2
⇒8.36Tfinal −0.385Tfinal = 167.2−38.5
⇒7.975Tfinal = 128.7
⇒Tfinal =128.7
7.975
⇒Tfinal ≈16.1◦C
Therefore, the final temperature when the system reaches thermal equilib-
rium is approximately 16.1◦C.
Question 27
Question
A copper container of mass 0.5 kg contains 0.2 kg of water at 30
°
C. How much
heat must be supplied to the system in order to raise the temperature of the wa-
ter in the container to 90
°
C? The specific heat capacity of copper is 390 J/kg◦C
and the specific heat capacity of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
30
°
C to 90
°
C. The formula to calculate the heat required to change the temper-
ature of a substance is given by:
Q=mc∆T
where: - Qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity of the substance, and - ∆Tis the change in temperature.
For the water:
Qwater = 0.2 kg ×4186 J/kg◦C×(90 −30)◦C
Qwater = 0.2×4186 ×60
25
Qwater = 50232 J
Step 2: Calculate the heat required to raise the temperature of the copper
container from 30
°
C to 90
°
C. Using the same formula, for the copper container:
Qcopper = 0.5 kg ×390 J/kg◦C×(90 −30)◦C
Qcopper = 0.5×390 ×60
Qcopper = 11700 J
Step 3: Calculate the total heat required for both the water and the copper
container.
Qtotal =Qwater +Qcopper
Qtotal = 50232 + 11700
Qtotal = 61932 J
Therefore, the total heat that must be supplied to the system is 61932 J.
Question 28
Question
A 200 g piece of aluminum at 80
°
C is placed in a calorimeter containing 800
g of water at 20
°
C. The specific heat capacity of aluminum is 0.897 J/g
°
C
and the specific heat capacity of water is 4.18 J/g
°
C. If no heat is lost to the
surroundings, what is the final equilibrium temperature of the system?
Solution
Step 1: Calculate the heat lost by the aluminum as it cools down to the final
equilibrium temperature. Using the formula for heat transfer: Q=mc∆T,
where Qis the heat, mis the mass, cis the specific heat capacity, and ∆Tis
the change in temperature. The heat lost by the aluminum can be calculated
as:
QAluminum = (200 g)(0.897 J/g
°
C)(80 −T)
°
C
where Tis the final equilibrium temperature of the system.
Step 2: Calculate the heat gained by the water as it warms up to the final
equilibrium temperature. Using the same formula as above, the heat gained by
the water can be calculated as:
QWater = (800 g)(4.18 J/g
°
C)(T−20)
°
C
Step 3: Since no heat is lost to the surroundings, the heat lost by the alu-
minum must be equal to the heat gained by the water. Therefore, we set the
two equations equal to each other:
(200 g)(0.897 J/g
°
C)(80 −T) = (800 g)(4.18 J/g
°
C)(T−20)
26
Step 4: Solve the equation for the final equilibrium temperature T.
179.4(80 −T) = 3344(T−20)
14352 −179.4T= 3344T−66880
69816 = 513.4T
T=69816
513.4
T≈136
°
C
Therefore, the final equilibrium temperature of the system is approximately
136
°
C.
Question 29
Question
A copper rod of length 2.5 m and diameter 2 cm is initially at a temperature of
25
°
C. If 6.0 x 105Jofheatenergyissuppliedtotherod, calculatethef inaltemperatureoftherod, giventhatthespecificheatcapacityof copperis390
J/kg ·K and the density of copper is 8.96 g/cm3. Assume the rod is perfectly
insulated.
Solution
Step 1: Calculate the mass of the copper rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder:
V=πr2h,
where ris the radius and his the height (or length) of the rod. Given that
the diameter of the rod is 2 cm, the radius r= 1 cm = 0.01 m and the height
h= 2.5 m. Hence,
V=π×(0.01 m)2×2.5 m.
Calculating this gives us the volume of the rod.
Step 2: Calculate the mass of the copper rod. The density of copper is given
as 8.96 g/cm3. Converting this to kg/m3, we get 8960 kg/m3. The mass mof
the rod can be calculated as
m= density ×volume.
Step 3: Calculate the specific heat energy absorbed by the copper rod. The
specific heat energy Qabsorbed by the rod can be calculated using the formula
Q=mc∆T,
where mis the mass of the rod, cis the specific heat capacity of copper, and
∆Tis the change in temperature.
27
Step 4: Calculate the final temperature of the rod. Since the rod is perfectly
insulated, all the heat supplied to it will result in a temperature increase. Thus,
the final temperature can be calculated using the formula
Q=mc(Tf−Ti),
where Tfis the final temperature and Tiis the initial temperature. Rearranging
this formula gives us the final temperature Tf.
Now we can substitute the given values into the equations to find the final
temperature of the rod.
Question 30
Question
A copper rod of length 2 m and diameter 1 cm is initially at a temperature of
100◦C. It is then placed in an ice bath at 0◦C. If the rod loses heat energy at
a rate of 200 W, determine the time it takes for the temperature of the rod to
drop to 10◦C. Assume the specific heat capacity of copper is 390 J/kg ·K and
the density of copper is 8900 kg/m3.
Solution
Step 1: Calculate the mass of the copper rod. The volume of the rod can be
calculated using the formula for the volume of a cylinder: V=πr2h, where
r= 0.005 m and h= 2 m.
V=π×(0.005 m)2×2 m = 0.000157 m3
The mass can be calculated using the formula m= density ×volume.
m= 8900 kg/m3×0.000157 m3= 1.3973 kg
Step 2: Calculate the heat energy required to change the temperature of the
copper rod from 100◦C to 10◦C. The formula for heat energy is Q=mc∆T,
where c= 390J/kg·K is the specific heat capacity of copper and ∆T= 100−10 =
90 K.
Q= 1.3973 kg ×390 J/kg ·K×90 K = 49257.3 J
Step 3: Calculate the time taken for the rod to cool from 100◦C to 10◦C.
Given that heat energy is lost at a rate of 200 W, the time taken can be calcu-
lated using the formula t=Q
P, where P= 200 W.
t=49257.3 J
200 W = 246.29 s
Therefore, it will take approximately 246.29 seconds for the temperature of
the rod to drop from 100◦C to 10◦C.
28
Question 31
Question
A copper rod of length 2 m and diameter 1 cm is initially at a uniform tempera-
ture of 100◦C. It is then placed in an ice-water mixture at a temperature of 0◦C.
Assuming the rod loses heat only through its lateral surface, calculate the time
it takes for the temperature at the center of the rod to drop to 10◦C. Given:
Thermal conductivity of copper = 400 W/mK, density of copper = 8900 kg/m3,
specific heat capacity of copper = 390 J/kgK.
Solution
Step 1: The initial temperature distribution in the rod will be given by the
one-dimensional heat conduction equation:
∂T
∂t =α∂2T
∂x2
where Tis the temperature, tis time, xis the distance along the rod, and αis
the thermal diffusivity given by α=k
ρc . Since the rod loses heat only through
its lateral surface, we consider heat conduction only along the length of the rod
(x-direction). The boundary conditions are:
At x=0: T(0, t) = 100◦C
At x=L/2 : T(L/2, t) = 10◦C
At t=0: T(x, 0) = 100◦C
Step 2: Let’s find the thermal diffusivity of copper using the given data:
α=k
ρc =400 W/mK
8900 kg/m3×390 J/kgK
α≈1.034 ×10−5m2/s
Step 3: The general solution to the one-dimensional heat conduction equa-
tion with the given boundary conditions is:
T(x, t) =
∞
X
n=1 Ancos nπx
Le−(nπ
L)2αt
where Lis the length of the rod.
Step 4: For T(L/2, t) = 10◦C, we have:
10 =
∞
X
n=1 Ancos nπ
4e−(nπ
2)2αt
Step 5: To find the time it takes for the temperature at the center of the rod
to drop to 10◦C, we would need to solve for tin the above equation to find the
sum of the series. This involves quite complex math and may require numerical
methods for the summation.
29
Question 32
Question
A copper rod of length 2 m is initially at a temperature of 100
°
C. One end of
the rod is then placed in an ice-water mixture at 0
°
C, while the other end is
placed in a furnace at 500
°
C. If the thermal conductivity of copper is 400 W/(m
K), determine the time taken for the whole rod to reach a uniform temperature.
Assume the rod is well-insulated along its length except at the ends.
Solution
Step 1: Calculate the rate of heat transfer per unit length for the rod.
The rate of heat transfer per unit length (Q) through the rod is given by
Fourier’s law as:
Q=kA∆T
L
where: - kis the thermal conductivity of copper (400 W/(m K)), - Ais the
cross-sectional area of the rod, - ∆Tis the temperature difference (500
°
C - 0
°
C
= 500 K), - Lis the length of the rod (2 m).
Step 2: Calculate the rate of heat transfer for the entire rod.
The rate of heat transfer for the entire rod is given by:
Qtotal =Q·Length of the rod
Step 3: Apply the principle of conservation of energy to find the time taken
for the rod to reach a uniform temperature.
The rate of heat transfer at the furnace end is equal to the rate of heat
transfer at the ice-water end. This implies that the total heat received by the
ice end is equal to the total heat loss at the furnace end.
The equation for this conservation of energy is:
Qtotal =mc∆T
where: - mis the mass of the rod, - cis the specific heat capacity of copper,
- ∆Tis the final temperature of the rod.
Solve for the time taken for the whole rod to reach a uniform temperature.
Question 33
Question
A steel rod of length 2.5 m and diameter 5 cm is initially at a temperature of
200
°
C. If the rod is heated until its temperature reaches 500
°
C, determine the
amount of heat transferred to the rod. Assume the thermal conductivity of steel
is 50 W/(m
·
K) and its specific heat capacity is 450 J/(kg
·
K).
30
Solution
Step 1: Calculate the mass of the steel rod. The volume of the steel rod can be
calculated using the formula for the volume of a cylinder:
V=πr2h,
where ris the radius of the rod and his the length of the rod. Given that
the diameter of the rod is 5 cm, the radius r=5
2cm = 0.025 m and the length
h= 2.5 m. Now, we can calculate the volume:
V=π(0.025)2×2.5.
Step 2: Calculate the mass of the steel rod. The density of steel is approxi-
mately 7850 kg/m
³
. Therefore, the mass of the steel rod can be calculated using
the formula:
mass = density ×volume.
Step 3: Calculate the heat energy absorbed by the steel rod. The heat energy
absorbed by the rod can be calculated using the formula:
Q=mc∆T,
where mis the mass of the rod, cis the specific heat capacity of steel, and ∆T
is the change in temperature.
Step 4: Calculate the change in temperature. The change in temperature,
∆T, is equal to the final temperature minus the initial temperature:
∆T= 500 −200.
Step 5: Find the amount of heat transferred to the rod. Finally, we can
substitute the values obtained into the formula for heat energy:
Q=mc∆T.
Question 34
Question
A copper rod of length 2 m and cross-sectional area 0.01 m2is initially at a
temperature of 100◦C. If one end of the rod is heated to a temperature of 200◦C
and the other end is kept at 100◦C, find the rate at which heat is conducted
along the rod. The thermal conductivity of copper is 400 W/(m K).
Solution
Step 1: Calculate the temperature difference between the two ends of the rod.
Given that one end is at 200◦Cand the other end is at 100◦C, the temperature
difference (∆T) is:
∆T=Thot −Tcold = 200◦C−100◦C= 100◦C
31
Step 2: Calculate the rate of heat conduction using the formula:
Q=−kA∆T
L
where: Q= rate of heat conduction, k= thermal conductivity of copper, A=
cross-sectional area of the rod, ∆T= temperature difference, L= length of the
rod.
Given that: k= 400 W/(m K), A= 0.01m2, ∆T= 100 K, L= 2 m,
we can plug in these values to find the rate of heat conduction:
Q=−400 ×0.01 ×100
2=−400 ×0.01 ×50 = −20 W
Therefore, the rate at which heat is conducted along the rod is 20 W.
Question 35
Question
A gas is compressed adiabatically from an initial volume of 2.00 m3to a final
volume of 0.50 m3. The initial pressure and temperature are 2.00 atm and 300
K, respectively. Compute the final pressure and temperature of the gas.
Solution
Step 1: Find the final pressure using the adiabatic compression formula:
P1
P2
γ−1
γ=V2
V1
where P1and P2are the initial and final pressures, V1and V2are the initial and
final volumes, and γis the adiabatic index (approximated as 1.4 for diatomic
gases).
Given V1= 2.00 m3,V2= 0.50 m3,P1= 2.00 atm, and γ= 1.4, we can plug
the values into the formula to solve for P2:
2.00 atm
P2
0.4
1.4=0.50 m3
2.00 m3
2.00
P20.286
= 0.25
2.00
P2
= 0.253.497
2.00
P2
≈0.00391
32
P2≈2.00
0.00391 ≈511 atm
Step 2: Use the ideal gas law to find the final temperature:
P1V1=nRT1
P2V2=nRT2
Since the number of moles nand the gas constant Rremain constant, we
can set the initial and final equations equal to each other:
P1V1=P2V2
2.00 atm ×2.00 m3= 511 atm ×T2×0.50 m3
8.00 atm ·m3= 255.5 atm ·m3×T2
T2=8.00
255.5≈0.0313 K−1≈31.3 K
Therefore, the final pressure of the gas is approximately 511 atm and the
final temperature is approximately 31.3 K.
33
