PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 2
Liberty University
Question 1
Question
A heat engine operates between a hot reservoir at a temperature of 500 K
and a cold reservoir at a temperature of 300 K. If the engine absorbs 4000 J
of heat from the hot reservoir in each cycle, determine the maximum possible
efficiency of the engine. Explain whether this efficiency violates the second law
of thermodynamics.
Solution
Step 1: Calculate the maximum possible efficiency of the heat engine using the
Carnot efficiency formula:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir (in Kelvin) and This the
temperature of the hot reservoir (in Kelvin). In this case, Th= 500 K and
Tc= 300 K.
Step 2: Substitute the values into the formula:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Therefore, the maximum possible efficiency of the engine is 40
Step 3: The second law of thermodynamics states that no heat engine can
be 100
Step 4: Since the efficiency of the engine in this case is 40
Question 2
Question
A heat engine operates using a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 2200 J of heat from the hot reservoir per cycle and
exhausts 1400 J of heat to the cold reservoir per cycle. Calculate the efficiency
of the engine. Is this engine operating in accordance with the second law of
thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Calculate the heat input. Given that the engine absorbs 2200 J of
heat from the hot reservoir per cycle, the heat input is 2200 J.
Step 3: Calculate the heat output. Given that the engine exhausts 1400 J
of heat to the cold reservoir per cycle, the heat output is 1400 J.
Step 4: Substitute the values into the formula for efficiency:
Efficiency = 1 −1400
2200
Step 5: Calculate the efficiency:
Efficiency = 1 −1400
2200 = 1 −0.6364 = 0.3636
Therefore, the efficiency of the engine is 36.36
Step 6: Evaluate whether the engine is operating in accordance with the
second law of thermodynamics. The efficiency of the engine is less than 1,
which is expected according to the second law of thermodynamics. This law
states that no heat engine can have an efficiency of 100
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the high-temperature reservoir and
exhausts heat Qcto the low-temperature reservoir while performing work W
during each cycle. If the efficiency of the engine is η, show that the efficiency of
a reversible engine operating between the same two reservoirs is given by:
ηrev = 1 −Tc
Th
2
Solution
Step 1: First, we will express the efficiency of the heat engine in terms of the
heat flows: By the first law of thermodynamics:
Qh=W+Qc
The efficiency ηis defined as:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
Step 2: Next, we will express the heat flows in terms of the temperatures:
According to Carnot’s theorem, the efficiency of any reversible engine operating
between the same two reservoirs is given by:
ηrev = 1 −Tc
Th
Step 3: We will now show that η=ηrev. Substitute Qh=Qc+Winto the
expression for efficiency:
η= 1 −Qc
Qc+W
Step 4: Rearrange the expression using Qc=TcScand W= (Th−Tc)Sc
(where Scis the entropy change of the low-temperature reservoir):
η= 1 −TcSc
TcSc+ (Th−Tc)Sc
= 1 −Tc
Th
Therefore, the efficiency of the reversible engine operating between the same
two reservoirs is given by:
ηrev = 1 −Tc
Th
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
where Th> Tc. The engine absorbs 500 J of heat from the hot reservoir and
has an efficiency of 40
1. The work done by the engine.
2. The heat rejected to the cold reservoir.
3. The ratio Th/Tc.
3
Solution
1. Let Qh= 500 J be the heat absorbed from the hot reservoir, and let W
be the work done by the engine. The efficiency of the Carnot engine is
given by
η=W
Qh
=W
500 = 0.40
Solving for Wgives us
W=η×Qh= 0.40 ×500 = 200 J
Therefore, the work done by the engine is 200 J.
2. The heat rejected to the cold reservoir can be found using the first law of
thermodynamics:
Qc=Qh−W= 500 −200 = 300 J
Therefore, the heat rejected to the cold reservoir is 300 J.
3. The efficiency of a Carnot engine is given by
η= 1 −Tc
Th
Substituting the given efficiency of 40
0.40 = 1 −Tc
Th
Solving for the ratio Th/Tcgives us
Th
Tc
=1
0.60 =5
3
Therefore, the ratio Th/Tcis 5
3.
Question 5
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and expels heat Qc
to the cold reservoir. If the efficiency of the engine is η, prove that the efficiency
of the engine satisfies the inequality η < 1−Tc
Th.
4
Solution
Let’s start by defining the efficiency of the heat engine and then deriving the
desired inequality.
Step 1: Define the efficiency of the heat engine The efficiency ηof a
heat engine is defined as the ratio of the work output Wout to the heat input
Qin:
η=Wout
Qin
Step 2: Express work output in terms of heat input From the first
law of thermodynamics, we have:
Qin =Wout +Qout
Since the heat engine is operating in a cycle:
Wout =Qin −Qout
Step 3: Apply the second law of thermodynamics The second law
of thermodynamics states that no engine can be more efficient than a Carnot
engine operating between the same two temperature reservoirs. The efficiency
of a Carnot engine is:
ηCarnot = 1 −Tc
Th
Step 4: Prove the desired inequality Since the heat engine is less effi-
cient than a Carnot engine:
η < ηCarnot
Wout
Qin
<1−Tc
Th
Qin −Qout
Qin
<1−Tc
Th
1−Qout
Qin
<1−Tc
Th
Qout
Qin
>Tc
Th
Therefore, the efficiency of the engine satisfies the inequality η < 1−Tc
Th.
Question 6
Question
A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 4000 J of heat from the hot reservoir in each
cycle. a) Determine the efficiency of the engine. b) Calculate the amount of
heat rejected to the cold reservoir in each cycle. c) If the engine operates for
200 cycles, find the total work done by the engine.
5
Solution
a) To find the efficiency of the engine, we use the formula for Carnot engine
efficiency:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir.
Step 1: Calculate the efficiency of the engine.
Efficiency = 1 −300
800 = 1 −3
8=5
8= 62.5%
b) The amount of heat rejected to the cold reservoir in each cycle can be
determined using the efficiency formula:
Efficiency = Qh−Qc
Qh
where Qhis the heat absorbed from the hot reservoir, and Qcis the heat rejected
to the cold reservoir.
Step 2: Calculate the amount of heat rejected to the cold reservoir.
0.625 = 4000 −Qc
4000
0.625 ×4000 = 4000 −Qc
Qc= 4000 −2500 = 1500 J
c) The total work done by the engine over 200 cycles is equal to the total
heat absorbed minus the total heat rejected to the cold reservoir.
Step 3: Calculate the total work done by the engine.
Total work = 200(4000 −1500)
Total work = 200(2500)
Total work = 500000 J
Question 7
Question
A heat engine operates between two heat reservoirs at temperatures Th= 400
K and Tc= 200 K. If the engine absorbs 600 J of heat from the hot reservoir
in each cycle, calculate the efficiency of the engine. Also, discuss whether this
engine violates the second law of thermodynamics.
6
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency of
a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values of Th= 400 K and Tc= 200 K:
Efficiency = 1 −200
400 = 1 −1
2=1
2= 50%
Step 3: The efficiency of the engine is 50
Step 4: The second law of thermodynamics states that no heat engine can
have 100
Step 5: Since the efficiency of this engine is 50
Therefore, the efficiency of the engine is 50
Question 8
Question
A Carnot heat engine operates between a hot reservoir at 500◦C and a cold
reservoir at 0◦C. If the engine absorbs 6000 J of heat from the hot reservoir in
each cycle, determine:
(a) The maximum possible efficiency of this engine.
(b) The amount of heat expelled to the cold reservoir in each cycle.
Solution
(a) The maximum efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given TH= 500◦C = 500 + 273 = 773 K and TC= 0◦C = 273 K, the
efficiency of the Carnot engine is:
Efficiency = 1 −273
773 = 1 −273
773 = 1 −3
10 =7
10 = 70%
Therefore, the maximum possible efficiency of this heat engine is 70
(b) The amount of heat expelled to the cold reservoir in each cycle can be
determined using the conservation of energy. Since the heat engine absorbs 6000
J of heat from the hot reservoir and the efficiency is 70
Work = Efficiency ·Heat in = 0.7×6000 = 4200 J
7
Since energy is conserved, the amount of heat expelled to the cold reservoir
is:
Heat out = Heat in −Work = 6000 −4200 = 1800 J
Therefore, the amount of heat expelled to the cold reservoir in each cycle is
1800 J.
Question 9
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and expels heat
Qcto the cold reservoir. The engine does work Wduring each cycle.
Prove that the efficiency of the heat engine is given by η= 1 −Tc
Th.
Solution
Step 1: According to the first law of thermodynamics, the work done by the
engine in each cycle is equal to the difference between the heat absorbed and
the heat expelled:
W=Qh−Qc
Step 2: The efficiency of a heat engine is defined as the ratio of the work
output to the heat input:
η=W
Qh
Step 3: Substituting Wfrom Step 1 into the efficiency equation:
η=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: Using the definition of efficiency, we can rearrange the equation to
express ηin terms of the temperatures of the reservoirs:
η= 1 −Qc
Qh
Step 5: From the second law of thermodynamics, we have that the heat
flow from a hot reservoir to a cold reservoir during a cycle is related to the
temperatures by: Qc
Tc
=Qh
Th
Step 6: Rearranging the equation from Step 5 to solve for Qc
Qh:
Qc
Qh
=Tc
Th
8
Step 7: Substituting the result from Step 6 into the expression for efficiency
in Step 4:
η= 1 −Tc
Th
Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 10
Question
A Carnot heat engine operates between a high-temperature reservoir at 800 K
and a low-temperature reservoir at 300 K. The engine consumes 1500 J of heat
from the high-temperature reservoir for each cycle. Calculate the efficiency of
the engine and determine the amount of heat rejected to the low-temperature
reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency of a heat engine:
Efficiency = 1 −Tlow
Thigh
Step 2: Substitute Tlow = 300 K and Thigh = 800 K into the formula:
Efficiency = 1 −300
800 = 1 −3
8=5
8= 0.625
Therefore, the efficiency of the Carnot engine is 62.5
Step 3: Calculate the amount of heat rejected to the low-temperature reser-
voir during each cycle using the efficiency of the engine:
Efficiency = Work output
Heat input
Work output = Efficiency ×Heat input
Heat rejected = Heat input −Work output
Step 4: Substitute the efficiency and heat input values into the equations:
Work output = 0.625 ×1500 J = 937.5 J
Heat rejected = 1500 J −937.5 J = 562.5 J
Therefore, during each cycle, 562.5 J of heat is rejected to the low-temperature
reservoir.
9
Question 11
Question
A heat engine operates between a hot reservoir at a temperature of 700 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and the work output of the engine is 2000 J.
Calculate:
1. The efficiency of the heat engine.
2. The amount of heat rejected to the cold reservoir in each cycle.
Solution
1. To find the efficiency of the heat engine, we can use the formula:
Efficiency = 1 −Heat rejected
Heat absorbed
Step 1: Find the heat rejected The heat rejected to the cold reservoir
is equal to the heat absorbed minus the work output:
Heat rejected = Heat absorbed −Work output
Heat rejected = 4000 J −2000 J = 2000 J
Step 2: Calculate the efficiency Now, we can plug the values into the
efficiency formula:
Efficiency = 1 −2000 J
4000 J =1
2= 0.5
Therefore, the efficiency of the heat engine is 50%.
2. To find the amount of heat rejected to the cold reservoir in each cycle,
we have already calculated it in Step 1:
Heat rejected = 2000 J
So, the amount of heat rejected to the cold reservoir in each cycle is 2000 J.
Question 12
Question
A Carnot engine operates between a reservoir at 700 K and a reservoir at 300 K.
The engine absorbs 5000 J of heat from the high-temperature reservoir during
each cycle. Calculate the efficiency of this engine and determine the amount of
heat rejected to the low-temperature reservoir during each cycle.
10
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine can be calculated using the formula:
Efficiency = 1 −Tlow
Thigh
where Tlow and Thigh are the temperatures of the low and high-temperature
reservoirs, respectively.
Given: Thigh = 700 K, Tlow = 300 K Plugging these values into the formula:
Efficiency = 1 −300
700 = 1 −3
7=4
7≈0.5714
Therefore, the efficiency of the engine is 57.14%.
Step 2: Determine the amount of heat rejected to the low-temperature reser-
voir. The amount of heat Qout rejected to the low-temperature reservoir can be
calculated using the conservation of energy principle:
Qin −Qout =Wout
where Qin = 5000 J is the heat absorbed from the high-temperature reservoir,
and Wout =Qin −Qout is the work output of the engine.
Since the engine is reversible and operates as a Carnot engine, the work
output can be calculated as:
Wout = Efficiency ×Qin =4
7×5000 J = 2857.14 J
Therefore, the amount of heat rejected to the low-temperature reservoir is:
Qout =Qin −Wout = 5000 J −2857.14 J = 2142.86 J
Hence, the amount of heat rejected to the low-temperature reservoir during
each cycle is 2142.86 J.
Question 13
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc(with Th> Tc). The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir during each cycle. Find the efficiency
of the engine in terms of Th,Tc,Qh, and Qc.
11
Solution
To find the efficiency of the Carnot heat engine, we will use the formula for
efficiency of a heat engine:
Efficiency = 1 −Qc
Qh
Step 1: Determine the work done by the engine in terms of Qhand Qc.
The net work done by the engine is given by the difference between the heat
absorbed and the heat rejected:
W=Qh−Qc
Step 2: Find the efficiency of the Carnot heat engine.
The efficiency of a Carnot engine is given by:
Efficiency = 1 −Tc
Th
Thus, the efficiency of the Carnot heat engine in terms of Th,Tc,Qh, and
Qcis:
Efficiency = 1 −Qc
Qh
= 1 −Tc
Th
Question 14
Question
A Carnot heat engine operates between a hot reservoir at 600 K and a cold
reservoir at 300 K. If the engine absorbs 2400 J of heat from the hot reservoir
in each cycle, calculate:
1. The efficiency of the heat engine.
2. The heat rejected to the cold reservoir in each cycle.
3. Discuss how the second law of thermodynamics applies to this Carnot
engine.
Solution
1. To find the efficiency of the Carnot heat engine, we can use the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
12
Step 1: Calculate the efficiency of the heat engine.
Efficiency = 1 −300 K
600 K = 1 −1
2= 0.5 = 50%
2. The heat rejected to the cold reservoir can be determined using the
efficiency of the engine. Since the engine absorbs 2400 J of heat from the hot
reservoir, the heat rejected to the cold reservoir is given by:
Heat rejected = Heat absorbed ×Efficiency
Step 2: Calculate the heat rejected to the cold reservoir.
Heat rejected = 2400 J ×0.5 = 1200 J
3. The second law of thermodynamics states that heat will spontaneously
flow from a hotter body to a colder body, and not the other way around. In the
Carnot engine, the efficiency is limited by the temperature difference between
the hot and cold reservoirs. No heat engine can have an efficiency greater than
the Carnot efficiency. In this case, the engine operates at 50
Thus, the Carnot engine serves as an idealized model that illustrates the
limitations imposed by the second law of thermodynamics on the efficiency of
heat engines.
Question 15
Question
A heat engine operating between two reservoirs absorbs Q1Joules of heat from
a reservoir at temperature T1and exhausts Q2Joules of heat to a reservoir at
temperature T2. If Q1= 8000 J, T1= 500 K, T2= 300 K, and the engine does
4000 J of work per cycle, calculate the efficiency of the engine. Is this efficiency
possible according to the second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency = Work done per cycle
Heat absorbed per cycle
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 4000 J
8000 J = 0.5 = 50%
Step 3: According to the second law of thermodynamics, the maximum
possible efficiency of a heat engine operating between two reservoirs is given by
the Carnot efficiency formula:
EfficiencyCarnot = 1 −T2
T1
13
where T1is the temperature of the hot reservoir and T2is the temperature of
the cold reservoir.
Step 4: Substitute the values T1= 500 K and T2= 300 K into the Carnot
efficiency formula:
EfficiencyCarnot = 1 −300
500 = 1 −0.6=0.4 = 40%
Step 5: The calculated efficiency of the engine (50
Question 16
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc. The engine absorbs heat energy Qhfrom the high-temperature reservoir
and exhausts heat energy Qcto the low-temperature reservoir. If the efficiency
of this Carnot engine is 50
Solution
Step 1: Recall the Carnot efficiency formula, which is given by
η= 1 −Tc
Th
where ηis the efficiency of the Carnot engine operating between two reservoirs
at different temperatures Tcand Th.
Step 2: Given that the efficiency is 50
0.5=1−Tc
Th
Step 3: Rearranging the equation above, we have
Tc
Th
= 1 −0.5=0.5
Step 4: Now, let’s relate the heat absorbed and heat exhausted by the engine
to the temperatures of the reservoirs. Recall that for a Carnot engine,
Qc
Tc
=Qh
Th
Step 5: Given that Qcis the heat energy exhausted to the low-temperature
reservoir and Qhis the heat energy absorbed from the high-temperature reser-
voir, we have Qc
Tc
=Qh
Th
14
Step 6: From step 3, we found that Tc
Th= 0.5. Substituting this into the
equation above, we get Qc
0.5Th
=Qh
Th
Step 7: Simplifying the equation above, we find
Qc= 0.5Qh
Step 8: Therefore, the ratio Tc
Thin terms of Qcand Qhis 0.5 .
Question 17
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine converts 4000 J of heat from the hot reservoir into work
output of 1800 J. Calculate the efficiency of the heat engine.
Solution
Step 1: Recall the efficiency of a heat engine is given by the formula:
Efficiency = Work Output
Heat Input
Step 2: We are given the work output as 1800 J, and the heat input as 4000
J. Therefore, we can substitute these values into the formula:
Efficiency = 1800 J
4000 J
Step 3: Simplifying the expression gives:
Efficiency = 9
20 = 0.45
Step 4: Finally, the efficiency of the heat engine is 45
Question 18
Question
A Carnot heat engine operates between a hot reservoir at temperature Thand
a cold reservoir at temperature Tc. The engine absorbs 4000 J of heat from the
hot reservoir and has a work output of 2000 J. Determine the efficiency of the
engine. If the cold reservoir temperature is increased, will the efficiency of the
engine increase, decrease, or remain the same? Justify your answer.
15
Solution
Step 1: Recall that the efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute Tc= cold reservoir temperature, Th= hot reservoir
temperature into the efficiency formula:
Efficiency = 1 −Tc
Th
Step 3: Given that the engine absorbs 4000 J of heat from the hot reservoir
and has a work output of 2000 J, determine the heat rejected to the cold reservoir
using the first law of thermodynamics:
Qh=W+Qc
4000 = 2000 + Qc
Qc= 2000J
Step 4: Substitute the known values into the efficiency formula:
Efficiency = 1 −2000
4000
Efficiency = 1 −0.5=0.5 = 50%
Step 5: To determine the effect of increasing the cold reservoir temperature
on the efficiency of the engine, consider the efficiency formula. Since Tc
This
in the denominator, increasing Tcwould decrease the efficiency of the engine.
Therefore, increasing the cold reservoir temperature will decrease the efficiency
of the engine.
Therefore, the efficiency of the engine will decrease if the cold reservoir
temperature is increased.
Question 19
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine absorbs 500 J of heat from the reservoir at Th
in each cycle and exhausts 300 J to the reservoir at Tc, determine:
1. The efficiency of the heat engine
2. The net work output per cycle
16
Solution
1. To find the efficiency of the Carnot heat engine, we can use the formula for
Carnot efficiency:
Efficiency = 1 −Tc
Th
Step 1: Plug in the given values.
Efficiency = 1 −300 K
500 K = 1 −3
5=2
5= 0.4 = 40%
Therefore, the efficiency of the heat engine is 40
2. The net work output per cycle can be calculated using the efficiency
formula and the heat input from the hot reservoir:
Net work output = Efficiency ×Heat input from hot reservoir
Step 2: Calculate the net work output.
Net work output = 0.4×500 J = 200 J
Therefore, the net work output per cycle is 200 J.
Question 20
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 200 K. The engine absorbs 6000 J of heat from the hot reservoir in
each cycle. Calculate the efficiency of the engine and discuss how it relates to
the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine. Given: - Temperature of hot
reservoir, Th= 500 K - Temperature of cold reservoir, Tc= 200 K - Heat
absorbed from the hot reservoir, Qh= 6000 J
The efficiency, η, of a Carnot heat engine is given by the formula:
η= 1 −Tc
Th
Substitute the given values into the formula:
η= 1 −200
500 = 1 −0.4=0.6
So, the efficiency of the engine is 60%.
17
Step 2: Discuss the second law of thermodynamics. The second law of
thermodynamics states that no heat engine can have an efficiency of 100%,
meaning that it is impossible to convert all the input heat into work. In this
case, the efficiency of the Carnot engine is 60%, meaning that 40% of the input
heat is wasted or rejected to the cold reservoir. This demonstrates the second law
of thermodynamics, which shows that there will always be some heat dissipated
to a lower temperature reservoir in any heat engine cycle.
Question 21
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc. The efficiency of the engine is 40
1. The ratio Th
Tc.
2. The heat rejected to the cold reservoir in each cycle.
Solution
1. Let Qhbe the heat absorbed from the hot reservoir and Qcbe the heat
rejected to the cold reservoir. The efficiency of a Carnot engine is given by:
Efficiency = W
Qh
= 1 −Tc
Th
Given that the efficiency is 40
0.4=1−Tc
Th
Tc
Th
= 0.6
This is the ratio Tc
Th.
2. We are given that Qh= 600 J. From the first law of thermodynamics for
the engine, we have:
W=Qh−Qc
We know that W= Efficiency ×Qh= 0.4×600 = 240 J. Therefore,
Qc=Qh−W= 600 −240 = 360 J
So, the heat rejected to the cold reservoir in each cycle is 360 J.
18
Question 22
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 500 J of heat from the hot reservoir in each cycle
and exhausts 300 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine.
Solution
Step 1: Calculate the work done by the engine in each cycle using the first law
of thermodynamics: Given: - Heat absorbed from the hot reservoir, Qh= 500
J - Heat exhausted to the cold reservoir, Qc= 300 J
The work done by the engine in each cycle, W, can be calculated using the
equation:
W=Qh−Qc
Calculating the work done:
W= 500 J −300 J = 200 J
Step 2: Calculate the efficiency of the engine using the formula for efficiency:
The efficiency of the engine, η, is defined as the ratio of the work done by the
engine to the heat absorbed from the hot reservoir:
η=W
Qh
Substitute the values of Wand Qhinto the efficiency formula and calculate:
η=200 J
500 J = 0.4 = 40%
Therefore, the efficiency of the engine is 40
Question 23
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 500 J of heat from the hot reservoir in
each cycle and expels 300 J to the cold reservoir in each cycle. Calculate the
efficiency of the engine.
19
Solution
Step 1: We can calculate the efficiency of the Carnot engine using the formula
for the efficiency of a heat engine:
Efficiency = 1 −Tc
Th
Step 2: First, let’s calculate the efficiency using the given temperatures Th
and Tc:
Efficiency = 1 −Tc
Th
= 1 −300
500 = 1 −0.6=0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Question 24
Question
A Carnot engine operates between reservoirs at temperatures THand TC, with
TH> TC. The engine absorbs heat at a rate of QHfrom the hot reservoir and
exhausts heat at a rate of QCto the cold reservoir. If the engine does work at
a rate of W, prove that the efficiency of the Carnot engine is given by
η= 1 −TC
TH
.
Solution
Step 1: Recall the first law of thermodynamics for a cyclic process:
QH−QC=W.
Step 2: The efficiency of a heat engine is defined as the ratio of the net work
output to the heat input from the hot reservoir:
η=W
QH
.
Step 3: Substitute W=QH−QCinto the expression for efficiency to get
η=QH−QC
QH
.
Step 4: Rearrange the terms to simplify the expression for efficiency:
η= 1 −QC
QH
.
Step 5: Now, recall the definition of the efficiency of a Carnot engine, which
is given by
ηCarnot = 1 −TC
TH
.
20
Step 6: Notice that in a reversible process such as the Carnot cycle, the heat
exchanges QHand QCcan be related to the temperatures of the reservoirs:
QC
TC
=QH
TH
.
Step 7: Substitute QC
QH=TC
THinto the expression for the efficiency of the
engine to get
η= 1 −TC
TH
.
Step 8: Thus, we have proven that the efficiency of a Carnot engine is given
by η= 1 −TC
TH.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine absorbs Qhof heat at Thand exhausts Qcof heat at Tc. The
engine does 4000 J of work per cycle. Calculate the efficiency of the engine and
discuss whether this violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine.
Efficiency = Useful work done
Heat absorbed
Efficiency = 4000 J
Qh
Step 2: Apply the first law of thermodynamics:
Qh= Work done + Qc
Qh= 4000 J + Qc
Step 3: Express the efficiency in terms of Qcand Th:
Efficiency = 4000 J
4000 J + Qc
Efficiency = 1
1 + Qc
4000 J
Step 4: Use Carnot efficiency to relate Qcand Tc:
Carnot Efficiency = 1 −Tc
Th
1
1 + Qc
4000 J
= 1 −Tc
Th
21
Step 5: Since Th> Tc, the efficiency must be less than Carnot efficiency:
1−Qc
4000 J <1−Tc
Th
Tc
Th
<Qc
4000 J
Step 6: This shows that the efficiency is less than the Carnot efficiency,
satisfying the second law of thermodynamics.
Question 26
Question
A Carnot engine operates between a hot reservoir at a temperature of 600 K
and a cold reservoir at a temperature of 300 K. The engine absorbs 5000 J of
heat from the hot reservoir during each cycle. Calculate the efficiency of the
engine and determine the maximum amount of work it can perform per cycle.
Solution
Step 1: Determine the efficiency of the Carnot engine. Step 2: Calculate the
maximum work output of the engine per cycle.
Step 1: The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
Given that the hot reservoir temperature (Thot) is 600 K and the cold reser-
voir temperature (Tcold) is 300 K, we can substitute these values into the formula
to calculate the efficiency:
Efficiency = 1 −300
600 = 1 −0.5=0.5
So, the efficiency of the Carnot engine is 50
Step 2: The maximum work output of the engine per cycle can be calculated
using the formula for efficiency:
Efficiency = Output work
Input heat
Given that the engine absorbs 5000 J of heat from the hot reservoir during each
cycle and the efficiency is 0.5, we can rearrange the formula to solve for the
output work:
Output work = Efficiency ×Input heat = 0.5×5000 J = 2500 J
Therefore, the maximum work output of the engine per cycle is 2500 J.
22
Question 27
Question
A heat engine operates between two reservoirs at temperatures T1and T2where
T1> T2. The engine absorbs Q1amount of heat from T1and Q2amount of
heat from T2. If the engine does 2000 J of work per cycle and Q1= 8000 J,
calculate the efficiency of the engine.
Solution
Step 1: Calculate the heat rejected by the engine to the cold reservoir Q2using
the first law of thermodynamics:
Work done = Q1−Q2
Substitute the given values:
2000 = 8000 −Q2
Q2= 8000 −2000 = 6000 J
Step 2: Calculate the efficiency of the engine using the formula:
Efficiency = Work done
Heat input
Substitute the given values:
Efficiency = 2000
8000 =1
4= 0.25 = 25%
Therefore, the efficiency of the engine is 25% .
Question 28
Question
A Carnot engine operates between two heat reservoirs at temperatures Th= 600
K and Tc= 200 K. If the engine absorbs 2000 J of heat from the hot reservoir
per cycle, calculate the efficiency of the engine and determine the heat expelled
to the cold reservoir per cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
23
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −200
600
Step 3: Calculate the efficiency:
Efficiency = 1 −1
3=2
3≈0.67
Step 4: Calculate the heat expelled to the cold reservoir per cycle using the
formula:
Heat expelled = Heat absorbed ×Efficiency
Step 5: Substitute the given heat absorbed and efficiency into the formula:
Heat expelled = 2000 ×2
3= 1333.33 J
Therefore, the efficiency of the engine is 0.67 (or 67
Question 29
Question
A heat engine operates between two heat reservoirs at temperatures Thot and
Tcold. The engine absorbs Qhot of heat from the hot reservoir and rejects Qcold
of heat to the cold reservoir. The engine performs 2400 J of work during each
cycle. Given that Thot = 500 K and Tcold = 300 K, calculate the efficiency of this
engine and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Tcold
Thot
Step 2: Substitute Thot = 500 K and Tcold = 300 K into the formula:
Efficiency = 1 −300
500
Step 3: Calculate the efficiency:
Efficiency = 1 −3
5=2
5= 0.4
Therefore, the efficiency of the engine is 40
Step 4: Determine whether the engine violates the second law of thermody-
namics. According to the second law of thermodynamics, the efficiency of any
24
heat engine operating between two reservoirs cannot exceed the efficiency of a
Carnot engine operating between the same two reservoirs. The efficiency of a
Carnot engine is given by:
EfficiencyCarnot = 1 −Tcold
Thot
Step 5: Substitute Thot = 500 K and Tcold = 300 K into the formula for the
Carnot engine efficiency:
EfficiencyCarnot = 1 −300
500 = 1 −3
5=2
5= 0.4
Since the efficiency of the actual engine is equal to the efficiency of the
Carnot engine, it does not violate the second law of thermodynamics.
Question 30
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 300 K, and delivers 5000 J of work per cycle. Calculate the heat
supplied to the engine from the hot reservoir, the heat rejected to the cold
reservoir, and the efficiency of the engine.
Solution
Step 1: We first calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
where Tc= 300 K and Th= 500 K.
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
So, the efficiency of the Carnot engine is 40%.
Step 2: The work done by the engine per cycle is given as 5000 J. This work
is obtained from the heat absorbed from the hot reservoir (Qh) and the heat
rejected to the cold reservoir (Qc), so we have:
W=Qh−Qc
Step 3: Using the efficiency formula, we can express Qcin terms of Qh:
W=Qh−Tc
Th
·Qh
5000 = Qh−300
500 ·Qh
25
5000 = Qh−3
5·Qh
5000 = 2
5·Qh
Qh=5
2·5000 = 12,500 J
So, the heat supplied to the engine from the hot reservoir is 12,500 J.
Step 4: Finally, we can calculate the heat rejected to the cold reservoir using
the formula:
Qc=Qh−W
Qc= 12,500 −5000 = 7500 J
Therefore, the heat rejected to the cold reservoir is 7500 J.
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thot = 600 K
and Tcold = 300 K. The engine absorbs 1500 J of heat from the hot reservoir
and expels 800 J of heat to the cold reservoir during each cycle. Calculate the
efficiency of the engine.
Solution
Given: Hot reservoir temperature, Thot = 600 K
Cold reservoir temperature, Tcold = 300 K
Heat absorbed, Qin = 1500 J
Heat expelled, Qout = 800 J
The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
Step 1: Convert temperatures to Celsius Subtract 273 from each tempera-
ture value to convert from Kelvin to Celsius: Thot = 600 −273 = 327
°
C
Tcold = 300 −273 = 27
°
C
Step 2: Calculate the efficiency Substitute the given values into the effi-
ciency formula:
Efficiency = 1 −Tcold
Thot
= 1 −27
327 = 1 −9
109 =100
109 ≈0.9174
Therefore, the efficiency of the heat engine is approximately 0.9174 or 91.74
26
Question 32
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
350 K. The engine takes in 2000 J of heat from the hot reservoir and exhausts
1200 J of heat to the cold reservoir in each cycle.
a) Calculate the efficiency of this engine.
b) Determine if this engine violates the second law of thermodynamics.
Solution
a) The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Heat output
Heat input
Step 1: Find the efficiency Given that the heat input is 2000 J and the
heat output is 1200 J, we can substitute these values into the formula:
Efficiency = 1 −1200
2000 = 1 −0.6=0.4
Therefore, the efficiency of this engine is 40
b) According to the second law of thermodynamics, no heat engine can be
100
In this case, the engine’s efficiency is 40
Question 33
Question
A Carnot heat engine operates on a hot reservoir at a temperature of 600 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 2000 J of heat
from the hot reservoir in each cycle. Calculate the efficiency of the engine.
Solution
Step 1: Calculate the efficiency of the Carnot heat engine. The efficiency of a
Carnot heat engine is given by the formula:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Convert temperatures to Kelvin. Given that the hot reservoir tem-
perature Th= 600 K and the cold reservoir temperature Tc= 300 K.
27
Step 3: Calculate the efficiency. Substitute the temperatures into the for-
mula to find the efficiency:
Efficiency = 1 −300
600
Efficiency = 1 −1
2
Efficiency = 1
2
Step 4: Calculate the absorbed heat. Given that the engine absorbs 2000 J
of heat from the hot reservoir in each cycle.
Step 5: Calculate the work output. The work output of a heat engine can
be calculated using the formula:
W=Qh×Efficiency
where Qhis the heat absorbed from the hot reservoir.
Step 6: Substitute values and calculate the work output.
W= 2000 J ×1
2
W= 1000 J
Therefore, the efficiency of the Carnot heat engine is 1
2and the work output
in each cycle is 1000 J.
Question 34
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Calculate the heat absorbed from
the hot reservoir and the heat rejected to the cold reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Step 2: Substituting TC= 300 K, TH= 600 K, and Efficiency = 0.40 into
the efficiency formula:
0.40 = 1 −300
600
0.40 = 1 −0.50
0.40 = 0.50
28
Step 3: The efficiency of the heat engine has been calculated to be 0.50,
which does not match the given efficiency of 0.40. This is because the efficiency
is calculated based on Carnot’s efficiency, and the engine in this case is not ideal.
Step 4: To find the actual efficiency, we use the formula:
Efficiencyactual =Wout
Qin
= 0.40
Step 5: Since efficiency is also given by Efficiencyactual = 1 −QC
QH, we rear-
range to solve for QC:
1−0.40 = QC
QH
0.60 = QC
QH
Step 6: The heat rejected to the cold reservoir is equal to 0.60 times the
heat absorbed from the hot reservoir. In other words, QC= 0.60 ×QH.
Step 7: Let QHbe the heat absorbed from the hot reservoir. Then:
QC= 0.60 ×QH
Step 8: The total heat input must equal the sum of the heat converted to
work and the heat rejected, so QH=Wout +QC.
Step 9: Substituting QC= 0.60 ×QHinto QH=Wout +QC, we get:
QH=Wout + 0.60 ×QH
Step 10: Solving for Wout in terms of QHgives Wout = 0.40 ×QH. Thus,
40% of the heat absorbed is converted to work, and 60% is rejected.
Question 35
Question
A heat engine working between two reservoirs has a maximum efficiency of 40a)
The amount of heat released to the cold reservoir. b) The work done by the
engine. c) The temperature of the cold reservoir if the temperature of the hot
reservoir is 600 K.
Solution
Step 1: Recall that the maximum efficiency of a heat engine operating between
two reservoirs is given by
ηmax = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
29
Step 2: Given that the maximum efficiency is 40
0.4=1−Tcold
600 =⇒Tcold = 360 K
Step 3: Knowing the temperatures of both reservoirs, we can calculate the
amount of heat released to the cold reservoir using the conservation of energy
principle:
Qcold =Qhot −Wout
where Qhot = 2000 J is the heat absorbed from the hot reservoir, and Wout is
the work done by the engine.
Step 4: The work done by the engine can be calculated as:
Wout =ηmax ×Qhot
Step 5: Substituting the values, we find:
Wout = 0.4×2000 = 800 J
Step 6: Therefore, the amount of heat released to the cold reservoir is:
Qcold = 2000 −800 = 1200 J
Step 7: Finally, we can determine the work done by the engine from the cold
reservoir to the hot reservoir, using the first law of thermodynamics:
Qout =Wout +Qcold
Step 8: Since work done is coming out of the engine, we would have:
Qout =−Wout −Qcold
Step 9: Substituting the values, we find:
Qout =−800 −1200 = −2000 J
Step 10: Therefore, the work done by the engine is −2000 J. This means
that the work is done on the engine.
30
Question 2
Question
A heat engine operates using a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 2200 J of heat from the hot reservoir per cycle and
exhausts 1400 J of heat to the cold reservoir per cycle. Calculate the efficiency
of the engine. Is this engine operating in accordance with the second law of
thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Calculate the heat input. Given that the engine absorbs 2200 J of
heat from the hot reservoir per cycle, the heat input is 2200 J.
Step 3: Calculate the heat output. Given that the engine exhausts 1400 J
of heat to the cold reservoir per cycle, the heat output is 1400 J.
Step 4: Substitute the values into the formula for efficiency:
Efficiency = 1 −1400
2200
Step 5: Calculate the efficiency:
Efficiency = 1 −1400
2200 = 1 −0.6364 = 0.3636
Therefore, the efficiency of the engine is 36.36
Step 6: Evaluate whether the engine is operating in accordance with the
second law of thermodynamics. The efficiency of the engine is less than 1,
which is expected according to the second law of thermodynamics. This law
states that no heat engine can have an efficiency of 100
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the high-temperature reservoir and
exhausts heat Qcto the low-temperature reservoir while performing work W
during each cycle. If the efficiency of the engine is η, show that the efficiency of
a reversible engine operating between the same two reservoirs is given by:
ηrev = 1 −Tc
Th
2
Solution
Step 1: First, we will express the efficiency of the heat engine in terms of the
heat flows: By the first law of thermodynamics:
Qh=W+Qc
The efficiency ηis defined as:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
Step 2: Next, we will express the heat flows in terms of the temperatures:
According to Carnot’s theorem, the efficiency of any reversible engine operating
between the same two reservoirs is given by:
ηrev = 1 −Tc
Th
Step 3: We will now show that η=ηrev. Substitute Qh=Qc+Winto the
expression for efficiency:
η= 1 −Qc
Qc+W
Step 4: Rearrange the expression using Qc=TcScand W= (Th−Tc)Sc
(where Scis the entropy change of the low-temperature reservoir):
η= 1 −TcSc
TcSc+ (Th−Tc)Sc
= 1 −Tc
Th
Therefore, the efficiency of the reversible engine operating between the same
two reservoirs is given by:
ηrev = 1 −Tc
Th
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
where Th> Tc. The engine absorbs 500 J of heat from the hot reservoir and
has an efficiency of 40
1. The work done by the engine.
2. The heat rejected to the cold reservoir.
3. The ratio Th/Tc.
3
Solution
1. Let Qh= 500 J be the heat absorbed from the hot reservoir, and let W
be the work done by the engine. The efficiency of the Carnot engine is
given by
η=W
Qh
=W
500 = 0.40
Solving for Wgives us
W=η×Qh= 0.40 ×500 = 200 J
Therefore, the work done by the engine is 200 J.
2. The heat rejected to the cold reservoir can be found using the first law of
thermodynamics:
Qc=Qh−W= 500 −200 = 300 J
Therefore, the heat rejected to the cold reservoir is 300 J.
3. The efficiency of a Carnot engine is given by
η= 1 −Tc
Th
Substituting the given efficiency of 40
0.40 = 1 −Tc
Th
Solving for the ratio Th/Tcgives us
Th
Tc
=1
0.60 =5
3
Therefore, the ratio Th/Tcis 5
3.
Question 5
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and expels heat Qc
to the cold reservoir. If the efficiency of the engine is η, prove that the efficiency
of the engine satisfies the inequality η < 1−Tc
Th.
4
Solution
Let’s start by defining the efficiency of the heat engine and then deriving the
desired inequality.
Step 1: Define the efficiency of the heat engine The efficiency ηof a
heat engine is defined as the ratio of the work output Wout to the heat input
Qin:
η=Wout
Qin
Step 2: Express work output in terms of heat input From the first
law of thermodynamics, we have:
Qin =Wout +Qout
Since the heat engine is operating in a cycle:
Wout =Qin −Qout
Step 3: Apply the second law of thermodynamics The second law
of thermodynamics states that no engine can be more efficient than a Carnot
engine operating between the same two temperature reservoirs. The efficiency
of a Carnot engine is:
ηCarnot = 1 −Tc
Th
Step 4: Prove the desired inequality Since the heat engine is less effi-
cient than a Carnot engine:
η < ηCarnot
Wout
Qin
<1−Tc
Th
Qin −Qout
Qin
<1−Tc
Th
1−Qout
Qin
<1−Tc
Th
Qout
Qin
>Tc
Th
Therefore, the efficiency of the engine satisfies the inequality η < 1−Tc
Th.
Question 6
Question
A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 4000 J of heat from the hot reservoir in each
cycle. a) Determine the efficiency of the engine. b) Calculate the amount of
heat rejected to the cold reservoir in each cycle. c) If the engine operates for
200 cycles, find the total work done by the engine.
5
Solution
a) To find the efficiency of the engine, we use the formula for Carnot engine
efficiency:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir.
Step 1: Calculate the efficiency of the engine.
Efficiency = 1 −300
800 = 1 −3
8=5
8= 62.5%
b) The amount of heat rejected to the cold reservoir in each cycle can be
determined using the efficiency formula:
Efficiency = Qh−Qc
Qh
where Qhis the heat absorbed from the hot reservoir, and Qcis the heat rejected
to the cold reservoir.
Step 2: Calculate the amount of heat rejected to the cold reservoir.
0.625 = 4000 −Qc
4000
0.625 ×4000 = 4000 −Qc
Qc= 4000 −2500 = 1500 J
c) The total work done by the engine over 200 cycles is equal to the total
heat absorbed minus the total heat rejected to the cold reservoir.
Step 3: Calculate the total work done by the engine.
Total work = 200(4000 −1500)
Total work = 200(2500)
Total work = 500000 J
Question 7
Question
A heat engine operates between two heat reservoirs at temperatures Th= 400
K and Tc= 200 K. If the engine absorbs 600 J of heat from the hot reservoir
in each cycle, calculate the efficiency of the engine. Also, discuss whether this
engine violates the second law of thermodynamics.
6
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency of
a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values of Th= 400 K and Tc= 200 K:
Efficiency = 1 −200
400 = 1 −1
2=1
2= 50%
Step 3: The efficiency of the engine is 50
Step 4: The second law of thermodynamics states that no heat engine can
have 100
Step 5: Since the efficiency of this engine is 50
Therefore, the efficiency of the engine is 50
Question 8
Question
A Carnot heat engine operates between a hot reservoir at 500◦C and a cold
reservoir at 0◦C. If the engine absorbs 6000 J of heat from the hot reservoir in
each cycle, determine:
(a) The maximum possible efficiency of this engine.
(b) The amount of heat expelled to the cold reservoir in each cycle.
Solution
(a) The maximum efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given TH= 500◦C = 500 + 273 = 773 K and TC= 0◦C = 273 K, the
efficiency of the Carnot engine is:
Efficiency = 1 −273
773 = 1 −273
773 = 1 −3
10 =7
10 = 70%
Therefore, the maximum possible efficiency of this heat engine is 70
(b) The amount of heat expelled to the cold reservoir in each cycle can be
determined using the conservation of energy. Since the heat engine absorbs 6000
J of heat from the hot reservoir and the efficiency is 70
Work = Efficiency ·Heat in = 0.7×6000 = 4200 J
7
Since energy is conserved, the amount of heat expelled to the cold reservoir
is:
Heat out = Heat in −Work = 6000 −4200 = 1800 J
Therefore, the amount of heat expelled to the cold reservoir in each cycle is
1800 J.
Question 9
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and expels heat
Qcto the cold reservoir. The engine does work Wduring each cycle.
Prove that the efficiency of the heat engine is given by η= 1 −Tc
Th.
Solution
Step 1: According to the first law of thermodynamics, the work done by the
engine in each cycle is equal to the difference between the heat absorbed and
the heat expelled:
W=Qh−Qc
Step 2: The efficiency of a heat engine is defined as the ratio of the work
output to the heat input:
η=W
Qh
Step 3: Substituting Wfrom Step 1 into the efficiency equation:
η=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: Using the definition of efficiency, we can rearrange the equation to
express ηin terms of the temperatures of the reservoirs:
η= 1 −Qc
Qh
Step 5: From the second law of thermodynamics, we have that the heat
flow from a hot reservoir to a cold reservoir during a cycle is related to the
temperatures by: Qc
Tc
=Qh
Th
Step 6: Rearranging the equation from Step 5 to solve for Qc
Qh:
Qc
Qh
=Tc
Th
8
Step 7: Substituting the result from Step 6 into the expression for efficiency
in Step 4:
η= 1 −Tc
Th
Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 10
Question
A Carnot heat engine operates between a high-temperature reservoir at 800 K
and a low-temperature reservoir at 300 K. The engine consumes 1500 J of heat
from the high-temperature reservoir for each cycle. Calculate the efficiency of
the engine and determine the amount of heat rejected to the low-temperature
reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency of a heat engine:
Efficiency = 1 −Tlow
Thigh
Step 2: Substitute Tlow = 300 K and Thigh = 800 K into the formula:
Efficiency = 1 −300
800 = 1 −3
8=5
8= 0.625
Therefore, the efficiency of the Carnot engine is 62.5
Step 3: Calculate the amount of heat rejected to the low-temperature reser-
voir during each cycle using the efficiency of the engine:
Efficiency = Work output
Heat input
Work output = Efficiency ×Heat input
Heat rejected = Heat input −Work output
Step 4: Substitute the efficiency and heat input values into the equations:
Work output = 0.625 ×1500 J = 937.5 J
Heat rejected = 1500 J −937.5 J = 562.5 J
Therefore, during each cycle, 562.5 J of heat is rejected to the low-temperature
reservoir.
9
Question 11
Question
A heat engine operates between a hot reservoir at a temperature of 700 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and the work output of the engine is 2000 J.
Calculate:
1. The efficiency of the heat engine.
2. The amount of heat rejected to the cold reservoir in each cycle.
Solution
1. To find the efficiency of the heat engine, we can use the formula:
Efficiency = 1 −Heat rejected
Heat absorbed
Step 1: Find the heat rejected The heat rejected to the cold reservoir
is equal to the heat absorbed minus the work output:
Heat rejected = Heat absorbed −Work output
Heat rejected = 4000 J −2000 J = 2000 J
Step 2: Calculate the efficiency Now, we can plug the values into the
efficiency formula:
Efficiency = 1 −2000 J
4000 J =1
2= 0.5
Therefore, the efficiency of the heat engine is 50%.
2. To find the amount of heat rejected to the cold reservoir in each cycle,
we have already calculated it in Step 1:
Heat rejected = 2000 J
So, the amount of heat rejected to the cold reservoir in each cycle is 2000 J.
Question 12
Question
A Carnot engine operates between a reservoir at 700 K and a reservoir at 300 K.
The engine absorbs 5000 J of heat from the high-temperature reservoir during
each cycle. Calculate the efficiency of this engine and determine the amount of
heat rejected to the low-temperature reservoir during each cycle.
10
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine can be calculated using the formula:
Efficiency = 1 −Tlow
Thigh
where Tlow and Thigh are the temperatures of the low and high-temperature
reservoirs, respectively.
Given: Thigh = 700 K, Tlow = 300 K Plugging these values into the formula:
Efficiency = 1 −300
700 = 1 −3
7=4
7≈0.5714
Therefore, the efficiency of the engine is 57.14%.
Step 2: Determine the amount of heat rejected to the low-temperature reser-
voir. The amount of heat Qout rejected to the low-temperature reservoir can be
calculated using the conservation of energy principle:
Qin −Qout =Wout
where Qin = 5000 J is the heat absorbed from the high-temperature reservoir,
and Wout =Qin −Qout is the work output of the engine.
Since the engine is reversible and operates as a Carnot engine, the work
output can be calculated as:
Wout = Efficiency ×Qin =4
7×5000 J = 2857.14 J
Therefore, the amount of heat rejected to the low-temperature reservoir is:
Qout =Qin −Wout = 5000 J −2857.14 J = 2142.86 J
Hence, the amount of heat rejected to the low-temperature reservoir during
each cycle is 2142.86 J.
Question 13
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc(with Th> Tc). The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir during each cycle. Find the efficiency
of the engine in terms of Th,Tc,Qh, and Qc.
11
Solution
To find the efficiency of the Carnot heat engine, we will use the formula for
efficiency of a heat engine:
Efficiency = 1 −Qc
Qh
Step 1: Determine the work done by the engine in terms of Qhand Qc.
The net work done by the engine is given by the difference between the heat
absorbed and the heat rejected:
W=Qh−Qc
Step 2: Find the efficiency of the Carnot heat engine.
The efficiency of a Carnot engine is given by:
Efficiency = 1 −Tc
Th
Thus, the efficiency of the Carnot heat engine in terms of Th,Tc,Qh, and
Qcis:
Efficiency = 1 −Qc
Qh
= 1 −Tc
Th
Question 14
Question
A Carnot heat engine operates between a hot reservoir at 600 K and a cold
reservoir at 300 K. If the engine absorbs 2400 J of heat from the hot reservoir
in each cycle, calculate:
1. The efficiency of the heat engine.
2. The heat rejected to the cold reservoir in each cycle.
3. Discuss how the second law of thermodynamics applies to this Carnot
engine.
Solution
1. To find the efficiency of the Carnot heat engine, we can use the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
12
Step 1: Calculate the efficiency of the heat engine.
Efficiency = 1 −300 K
600 K = 1 −1
2= 0.5 = 50%
2. The heat rejected to the cold reservoir can be determined using the
efficiency of the engine. Since the engine absorbs 2400 J of heat from the hot
reservoir, the heat rejected to the cold reservoir is given by:
Heat rejected = Heat absorbed ×Efficiency
Step 2: Calculate the heat rejected to the cold reservoir.
Heat rejected = 2400 J ×0.5 = 1200 J
3. The second law of thermodynamics states that heat will spontaneously
flow from a hotter body to a colder body, and not the other way around. In the
Carnot engine, the efficiency is limited by the temperature difference between
the hot and cold reservoirs. No heat engine can have an efficiency greater than
the Carnot efficiency. In this case, the engine operates at 50
Thus, the Carnot engine serves as an idealized model that illustrates the
limitations imposed by the second law of thermodynamics on the efficiency of
heat engines.
Question 15
Question
A heat engine operating between two reservoirs absorbs Q1Joules of heat from
a reservoir at temperature T1and exhausts Q2Joules of heat to a reservoir at
temperature T2. If Q1= 8000 J, T1= 500 K, T2= 300 K, and the engine does
4000 J of work per cycle, calculate the efficiency of the engine. Is this efficiency
possible according to the second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency = Work done per cycle
Heat absorbed per cycle
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 4000 J
8000 J = 0.5 = 50%
Step 3: According to the second law of thermodynamics, the maximum
possible efficiency of a heat engine operating between two reservoirs is given by
the Carnot efficiency formula:
EfficiencyCarnot = 1 −T2
T1
13
where T1is the temperature of the hot reservoir and T2is the temperature of
the cold reservoir.
Step 4: Substitute the values T1= 500 K and T2= 300 K into the Carnot
efficiency formula:
EfficiencyCarnot = 1 −300
500 = 1 −0.6=0.4 = 40%
Step 5: The calculated efficiency of the engine (50
Question 16
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc. The engine absorbs heat energy Qhfrom the high-temperature reservoir
and exhausts heat energy Qcto the low-temperature reservoir. If the efficiency
of this Carnot engine is 50
Solution
Step 1: Recall the Carnot efficiency formula, which is given by
η= 1 −Tc
Th
where ηis the efficiency of the Carnot engine operating between two reservoirs
at different temperatures Tcand Th.
Step 2: Given that the efficiency is 50
0.5=1−Tc
Th
Step 3: Rearranging the equation above, we have
Tc
Th
= 1 −0.5=0.5
Step 4: Now, let’s relate the heat absorbed and heat exhausted by the engine
to the temperatures of the reservoirs. Recall that for a Carnot engine,
Qc
Tc
=Qh
Th
Step 5: Given that Qcis the heat energy exhausted to the low-temperature
reservoir and Qhis the heat energy absorbed from the high-temperature reser-
voir, we have Qc
Tc
=Qh
Th
14
Step 6: From step 3, we found that Tc
Th= 0.5. Substituting this into the
equation above, we get Qc
0.5Th
=Qh
Th
Step 7: Simplifying the equation above, we find
Qc= 0.5Qh
Step 8: Therefore, the ratio Tc
Thin terms of Qcand Qhis 0.5 .
Question 17
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine converts 4000 J of heat from the hot reservoir into work
output of 1800 J. Calculate the efficiency of the heat engine.
Solution
Step 1: Recall the efficiency of a heat engine is given by the formula:
Efficiency = Work Output
Heat Input
Step 2: We are given the work output as 1800 J, and the heat input as 4000
J. Therefore, we can substitute these values into the formula:
Efficiency = 1800 J
4000 J
Step 3: Simplifying the expression gives:
Efficiency = 9
20 = 0.45
Step 4: Finally, the efficiency of the heat engine is 45
Question 18
Question
A Carnot heat engine operates between a hot reservoir at temperature Thand
a cold reservoir at temperature Tc. The engine absorbs 4000 J of heat from the
hot reservoir and has a work output of 2000 J. Determine the efficiency of the
engine. If the cold reservoir temperature is increased, will the efficiency of the
engine increase, decrease, or remain the same? Justify your answer.
15
Solution
Step 1: Recall that the efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute Tc= cold reservoir temperature, Th= hot reservoir
temperature into the efficiency formula:
Efficiency = 1 −Tc
Th
Step 3: Given that the engine absorbs 4000 J of heat from the hot reservoir
and has a work output of 2000 J, determine the heat rejected to the cold reservoir
using the first law of thermodynamics:
Qh=W+Qc
4000 = 2000 + Qc
Qc= 2000J
Step 4: Substitute the known values into the efficiency formula:
Efficiency = 1 −2000
4000
Efficiency = 1 −0.5=0.5 = 50%
Step 5: To determine the effect of increasing the cold reservoir temperature
on the efficiency of the engine, consider the efficiency formula. Since Tc
This
in the denominator, increasing Tcwould decrease the efficiency of the engine.
Therefore, increasing the cold reservoir temperature will decrease the efficiency
of the engine.
Therefore, the efficiency of the engine will decrease if the cold reservoir
temperature is increased.
Question 19
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine absorbs 500 J of heat from the reservoir at Th
in each cycle and exhausts 300 J to the reservoir at Tc, determine:
1. The efficiency of the heat engine
2. The net work output per cycle
16
Solution
1. To find the efficiency of the Carnot heat engine, we can use the formula for
Carnot efficiency:
Efficiency = 1 −Tc
Th
Step 1: Plug in the given values.
Efficiency = 1 −300 K
500 K = 1 −3
5=2
5= 0.4 = 40%
Therefore, the efficiency of the heat engine is 40
2. The net work output per cycle can be calculated using the efficiency
formula and the heat input from the hot reservoir:
Net work output = Efficiency ×Heat input from hot reservoir
Step 2: Calculate the net work output.
Net work output = 0.4×500 J = 200 J
Therefore, the net work output per cycle is 200 J.
Question 20
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 200 K. The engine absorbs 6000 J of heat from the hot reservoir in
each cycle. Calculate the efficiency of the engine and discuss how it relates to
the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine. Given: - Temperature of hot
reservoir, Th= 500 K - Temperature of cold reservoir, Tc= 200 K - Heat
absorbed from the hot reservoir, Qh= 6000 J
The efficiency, η, of a Carnot heat engine is given by the formula:
η= 1 −Tc
Th
Substitute the given values into the formula:
η= 1 −200
500 = 1 −0.4=0.6
So, the efficiency of the engine is 60%.
17
Step 2: Discuss the second law of thermodynamics. The second law of
thermodynamics states that no heat engine can have an efficiency of 100%,
meaning that it is impossible to convert all the input heat into work. In this
case, the efficiency of the Carnot engine is 60%, meaning that 40% of the input
heat is wasted or rejected to the cold reservoir. This demonstrates the second law
of thermodynamics, which shows that there will always be some heat dissipated
to a lower temperature reservoir in any heat engine cycle.
Question 21
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc. The efficiency of the engine is 40
1. The ratio Th
Tc.
2. The heat rejected to the cold reservoir in each cycle.
Solution
1. Let Qhbe the heat absorbed from the hot reservoir and Qcbe the heat
rejected to the cold reservoir. The efficiency of a Carnot engine is given by:
Efficiency = W
Qh
= 1 −Tc
Th
Given that the efficiency is 40
0.4=1−Tc
Th
Tc
Th
= 0.6
This is the ratio Tc
Th.
2. We are given that Qh= 600 J. From the first law of thermodynamics for
the engine, we have:
W=Qh−Qc
We know that W= Efficiency ×Qh= 0.4×600 = 240 J. Therefore,
Qc=Qh−W= 600 −240 = 360 J
So, the heat rejected to the cold reservoir in each cycle is 360 J.
18
Question 22
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 500 J of heat from the hot reservoir in each cycle
and exhausts 300 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine.
Solution
Step 1: Calculate the work done by the engine in each cycle using the first law
of thermodynamics: Given: - Heat absorbed from the hot reservoir, Qh= 500
J - Heat exhausted to the cold reservoir, Qc= 300 J
The work done by the engine in each cycle, W, can be calculated using the
equation:
W=Qh−Qc
Calculating the work done:
W= 500 J −300 J = 200 J
Step 2: Calculate the efficiency of the engine using the formula for efficiency:
The efficiency of the engine, η, is defined as the ratio of the work done by the
engine to the heat absorbed from the hot reservoir:
η=W
Qh
Substitute the values of Wand Qhinto the efficiency formula and calculate:
η=200 J
500 J = 0.4 = 40%
Therefore, the efficiency of the engine is 40
Question 23
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 500 J of heat from the hot reservoir in
each cycle and expels 300 J to the cold reservoir in each cycle. Calculate the
efficiency of the engine.
19
Solution
Step 1: We can calculate the efficiency of the Carnot engine using the formula
for the efficiency of a heat engine:
Efficiency = 1 −Tc
Th
Step 2: First, let’s calculate the efficiency using the given temperatures Th
and Tc:
Efficiency = 1 −Tc
Th
= 1 −300
500 = 1 −0.6=0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Question 24
Question
A Carnot engine operates between reservoirs at temperatures THand TC, with
TH> TC. The engine absorbs heat at a rate of QHfrom the hot reservoir and
exhausts heat at a rate of QCto the cold reservoir. If the engine does work at
a rate of W, prove that the efficiency of the Carnot engine is given by
η= 1 −TC
TH
.
Solution
Step 1: Recall the first law of thermodynamics for a cyclic process:
QH−QC=W.
Step 2: The efficiency of a heat engine is defined as the ratio of the net work
output to the heat input from the hot reservoir:
η=W
QH
.
Step 3: Substitute W=QH−QCinto the expression for efficiency to get
η=QH−QC
QH
.
Step 4: Rearrange the terms to simplify the expression for efficiency:
η= 1 −QC
QH
.
Step 5: Now, recall the definition of the efficiency of a Carnot engine, which
is given by
ηCarnot = 1 −TC
TH
.
20
Step 6: Notice that in a reversible process such as the Carnot cycle, the heat
exchanges QHand QCcan be related to the temperatures of the reservoirs:
QC
TC
=QH
TH
.
Step 7: Substitute QC
QH=TC
THinto the expression for the efficiency of the
engine to get
η= 1 −TC
TH
.
Step 8: Thus, we have proven that the efficiency of a Carnot engine is given
by η= 1 −TC
TH.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine absorbs Qhof heat at Thand exhausts Qcof heat at Tc. The
engine does 4000 J of work per cycle. Calculate the efficiency of the engine and
discuss whether this violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine.
Efficiency = Useful work done
Heat absorbed
Efficiency = 4000 J
Qh
Step 2: Apply the first law of thermodynamics:
Qh= Work done + Qc
Qh= 4000 J + Qc
Step 3: Express the efficiency in terms of Qcand Th:
Efficiency = 4000 J
4000 J + Qc
Efficiency = 1
1 + Qc
4000 J
Step 4: Use Carnot efficiency to relate Qcand Tc:
Carnot Efficiency = 1 −Tc
Th
1
1 + Qc
4000 J
= 1 −Tc
Th
21
Step 5: Since Th> Tc, the efficiency must be less than Carnot efficiency:
1−Qc
4000 J <1−Tc
Th
Tc
Th
<Qc
4000 J
Step 6: This shows that the efficiency is less than the Carnot efficiency,
satisfying the second law of thermodynamics.
Question 26
Question
A Carnot engine operates between a hot reservoir at a temperature of 600 K
and a cold reservoir at a temperature of 300 K. The engine absorbs 5000 J of
heat from the hot reservoir during each cycle. Calculate the efficiency of the
engine and determine the maximum amount of work it can perform per cycle.
Solution
Step 1: Determine the efficiency of the Carnot engine. Step 2: Calculate the
maximum work output of the engine per cycle.
Step 1: The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
Given that the hot reservoir temperature (Thot) is 600 K and the cold reser-
voir temperature (Tcold) is 300 K, we can substitute these values into the formula
to calculate the efficiency:
Efficiency = 1 −300
600 = 1 −0.5=0.5
So, the efficiency of the Carnot engine is 50
Step 2: The maximum work output of the engine per cycle can be calculated
using the formula for efficiency:
Efficiency = Output work
Input heat
Given that the engine absorbs 5000 J of heat from the hot reservoir during each
cycle and the efficiency is 0.5, we can rearrange the formula to solve for the
output work:
Output work = Efficiency ×Input heat = 0.5×5000 J = 2500 J
Therefore, the maximum work output of the engine per cycle is 2500 J.
22
Question 27
Question
A heat engine operates between two reservoirs at temperatures T1and T2where
T1> T2. The engine absorbs Q1amount of heat from T1and Q2amount of
heat from T2. If the engine does 2000 J of work per cycle and Q1= 8000 J,
calculate the efficiency of the engine.
Solution
Step 1: Calculate the heat rejected by the engine to the cold reservoir Q2using
the first law of thermodynamics:
Work done = Q1−Q2
Substitute the given values:
2000 = 8000 −Q2
Q2= 8000 −2000 = 6000 J
Step 2: Calculate the efficiency of the engine using the formula:
Efficiency = Work done
Heat input
Substitute the given values:
Efficiency = 2000
8000 =1
4= 0.25 = 25%
Therefore, the efficiency of the engine is 25% .
Question 28
Question
A Carnot engine operates between two heat reservoirs at temperatures Th= 600
K and Tc= 200 K. If the engine absorbs 2000 J of heat from the hot reservoir
per cycle, calculate the efficiency of the engine and determine the heat expelled
to the cold reservoir per cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
23
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −200
600
Step 3: Calculate the efficiency:
Efficiency = 1 −1
3=2
3≈0.67
Step 4: Calculate the heat expelled to the cold reservoir per cycle using the
formula:
Heat expelled = Heat absorbed ×Efficiency
Step 5: Substitute the given heat absorbed and efficiency into the formula:
Heat expelled = 2000 ×2
3= 1333.33 J
Therefore, the efficiency of the engine is 0.67 (or 67
Question 29
Question
A heat engine operates between two heat reservoirs at temperatures Thot and
Tcold. The engine absorbs Qhot of heat from the hot reservoir and rejects Qcold
of heat to the cold reservoir. The engine performs 2400 J of work during each
cycle. Given that Thot = 500 K and Tcold = 300 K, calculate the efficiency of this
engine and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Tcold
Thot
Step 2: Substitute Thot = 500 K and Tcold = 300 K into the formula:
Efficiency = 1 −300
500
Step 3: Calculate the efficiency:
Efficiency = 1 −3
5=2
5= 0.4
Therefore, the efficiency of the engine is 40
Step 4: Determine whether the engine violates the second law of thermody-
namics. According to the second law of thermodynamics, the efficiency of any
24
heat engine operating between two reservoirs cannot exceed the efficiency of a
Carnot engine operating between the same two reservoirs. The efficiency of a
Carnot engine is given by:
EfficiencyCarnot = 1 −Tcold
Thot
Step 5: Substitute Thot = 500 K and Tcold = 300 K into the formula for the
Carnot engine efficiency:
EfficiencyCarnot = 1 −300
500 = 1 −3
5=2
5= 0.4
Since the efficiency of the actual engine is equal to the efficiency of the
Carnot engine, it does not violate the second law of thermodynamics.
Question 30
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 300 K, and delivers 5000 J of work per cycle. Calculate the heat
supplied to the engine from the hot reservoir, the heat rejected to the cold
reservoir, and the efficiency of the engine.
Solution
Step 1: We first calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
where Tc= 300 K and Th= 500 K.
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
So, the efficiency of the Carnot engine is 40%.
Step 2: The work done by the engine per cycle is given as 5000 J. This work
is obtained from the heat absorbed from the hot reservoir (Qh) and the heat
rejected to the cold reservoir (Qc), so we have:
W=Qh−Qc
Step 3: Using the efficiency formula, we can express Qcin terms of Qh:
W=Qh−Tc
Th
·Qh
5000 = Qh−300
500 ·Qh
25
5000 = Qh−3
5·Qh
5000 = 2
5·Qh
Qh=5
2·5000 = 12,500 J
So, the heat supplied to the engine from the hot reservoir is 12,500 J.
Step 4: Finally, we can calculate the heat rejected to the cold reservoir using
the formula:
Qc=Qh−W
Qc= 12,500 −5000 = 7500 J
Therefore, the heat rejected to the cold reservoir is 7500 J.
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thot = 600 K
and Tcold = 300 K. The engine absorbs 1500 J of heat from the hot reservoir
and expels 800 J of heat to the cold reservoir during each cycle. Calculate the
efficiency of the engine.
Solution
Given: Hot reservoir temperature, Thot = 600 K
Cold reservoir temperature, Tcold = 300 K
Heat absorbed, Qin = 1500 J
Heat expelled, Qout = 800 J
The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
Step 1: Convert temperatures to Celsius Subtract 273 from each tempera-
ture value to convert from Kelvin to Celsius: Thot = 600 −273 = 327
°
C
Tcold = 300 −273 = 27
°
C
Step 2: Calculate the efficiency Substitute the given values into the effi-
ciency formula:
Efficiency = 1 −Tcold
Thot
= 1 −27
327 = 1 −9
109 =100
109 ≈0.9174
Therefore, the efficiency of the heat engine is approximately 0.9174 or 91.74
26
Question 32
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
350 K. The engine takes in 2000 J of heat from the hot reservoir and exhausts
1200 J of heat to the cold reservoir in each cycle.
a) Calculate the efficiency of this engine.
b) Determine if this engine violates the second law of thermodynamics.
Solution
a) The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Heat output
Heat input
Step 1: Find the efficiency Given that the heat input is 2000 J and the
heat output is 1200 J, we can substitute these values into the formula:
Efficiency = 1 −1200
2000 = 1 −0.6=0.4
Therefore, the efficiency of this engine is 40
b) According to the second law of thermodynamics, no heat engine can be
100
In this case, the engine’s efficiency is 40
Question 33
Question
A Carnot heat engine operates on a hot reservoir at a temperature of 600 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 2000 J of heat
from the hot reservoir in each cycle. Calculate the efficiency of the engine.
Solution
Step 1: Calculate the efficiency of the Carnot heat engine. The efficiency of a
Carnot heat engine is given by the formula:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Convert temperatures to Kelvin. Given that the hot reservoir tem-
perature Th= 600 K and the cold reservoir temperature Tc= 300 K.
27
Step 3: Calculate the efficiency. Substitute the temperatures into the for-
mula to find the efficiency:
Efficiency = 1 −300
600
Efficiency = 1 −1
2
Efficiency = 1
2
Step 4: Calculate the absorbed heat. Given that the engine absorbs 2000 J
of heat from the hot reservoir in each cycle.
Step 5: Calculate the work output. The work output of a heat engine can
be calculated using the formula:
W=Qh×Efficiency
where Qhis the heat absorbed from the hot reservoir.
Step 6: Substitute values and calculate the work output.
W= 2000 J ×1
2
W= 1000 J
Therefore, the efficiency of the Carnot heat engine is 1
2and the work output
in each cycle is 1000 J.
Question 34
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Calculate the heat absorbed from
the hot reservoir and the heat rejected to the cold reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Step 2: Substituting TC= 300 K, TH= 600 K, and Efficiency = 0.40 into
the efficiency formula:
0.40 = 1 −300
600
0.40 = 1 −0.50
0.40 = 0.50
28
Step 3: The efficiency of the heat engine has been calculated to be 0.50,
which does not match the given efficiency of 0.40. This is because the efficiency
is calculated based on Carnot’s efficiency, and the engine in this case is not ideal.
Step 4: To find the actual efficiency, we use the formula:
Efficiencyactual =Wout
Qin
= 0.40
Step 5: Since efficiency is also given by Efficiencyactual = 1 −QC
QH, we rear-
range to solve for QC:
1−0.40 = QC
QH
0.60 = QC
QH
Step 6: The heat rejected to the cold reservoir is equal to 0.60 times the
heat absorbed from the hot reservoir. In other words, QC= 0.60 ×QH.
Step 7: Let QHbe the heat absorbed from the hot reservoir. Then:
QC= 0.60 ×QH
Step 8: The total heat input must equal the sum of the heat converted to
work and the heat rejected, so QH=Wout +QC.
Step 9: Substituting QC= 0.60 ×QHinto QH=Wout +QC, we get:
QH=Wout + 0.60 ×QH
Step 10: Solving for Wout in terms of QHgives Wout = 0.40 ×QH. Thus,
40% of the heat absorbed is converted to work, and 60% is rejected.
Question 35
Question
A heat engine working between two reservoirs has a maximum efficiency of 40a)
The amount of heat released to the cold reservoir. b) The work done by the
engine. c) The temperature of the cold reservoir if the temperature of the hot
reservoir is 600 K.
Solution
Step 1: Recall that the maximum efficiency of a heat engine operating between
two reservoirs is given by
ηmax = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
29
Step 2: Given that the maximum efficiency is 40
0.4=1−Tcold
600 =⇒Tcold = 360 K
Step 3: Knowing the temperatures of both reservoirs, we can calculate the
amount of heat released to the cold reservoir using the conservation of energy
principle:
Qcold =Qhot −Wout
where Qhot = 2000 J is the heat absorbed from the hot reservoir, and Wout is
the work done by the engine.
Step 4: The work done by the engine can be calculated as:
Wout =ηmax ×Qhot
Step 5: Substituting the values, we find:
Wout = 0.4×2000 = 800 J
Step 6: Therefore, the amount of heat released to the cold reservoir is:
Qcold = 2000 −800 = 1200 J
Step 7: Finally, we can determine the work done by the engine from the cold
reservoir to the hot reservoir, using the first law of thermodynamics:
Qout =Wout +Qcold
Step 8: Since work done is coming out of the engine, we would have:
Qout =−Wout −Qcold
Step 9: Substituting the values, we find:
Qout =−800 −1200 = −2000 J
Step 10: Therefore, the work done by the engine is −2000 J. This means
that the work is done on the engine.
30
