PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 7
Liberty University
Question 1
Question
A Carnot heat engine operates between two reservoirs at temperatures T1and
T2with T1> T2. If the engine absorbs 600 J of heat from the reservoir at
temperature T1in each cycle, determine:
(a) The efficiency of the engine in terms of T1and T2.
(b) The heat rejected by the engine to the reservoir at temperature T2in each
cycle.
(c) If the engine completes 100 cycles, calculate the total work done by the
engine.
Solution
Let’s go through each part of the question step by step.
Part (a)
Step 1: The efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −T2
T1
Step 2: Substituting T1and T2into the formula, we get:
Efficiency = 1 −T2
T1
= 1 −T2
T1
= 1 −T2
T1
So, the efficiency of the engine in terms of T1and T2is 1 −T2
T1.
Part (b)
Step 3: The heat rejected by the engine to the reservoir at temperature T2
in each cycle can be calculated using the formula for efficiency:
Efficiency = Work output
Heat input
Step 4: Rearranging the formula, we get:
Heat rejected = Heat input −Work output
Step 5: Substituting the given values, we have:
Heat rejected = 600 J −1−T2
T1×600 J
Therefore, the heat rejected by the engine to the reservoir at temperature
T2in each cycle is 600 J −600 J 1−T2
T1.
Part (c)
Step 6: The total work done by the engine after completing 100 cycles can
be calculated using the relationship between work and heat:
Work output = Heat input −Heat rejected
Step 7: For 100 cycles, the total work done is:
Total work done = 100 ×Heat input −1−T2
T1×Heat input
Step 8: Substituting the given values, we get:
Total work done = 100 ×600 J −600 J 1−T2
T1
Therefore, the total work done by the engine after completing 100 cycles is
100 ×h600 J −600 J 1−T2
T1i.
Question 2
Question
A Carnot heat engine operates between a high temperature reservoir at 500 K
and a low temperature reservoir at 300 K. If the engine absorbs 1,000 J of heat
from the high temperature reservoir in each cycle, calculate the following:
(a) The efficiency of the engine.
(b) The amount of heat rejected to the low temperature reservoir in each
cycle.
2
Solution
Step 1: Calculate the efficiency of the engine using the Carnot efficiency for-
mula:
Efficiency = 1 −Tlow
Thigh
Given Thigh = 500 K and Tlow = 300 K,
Efficiency = 1 −300
500 = 1 −0.6=0.4 = 40%
Step 2: Calculate the amount of heat rejected to the low temperature
reservoir in each cycle. Since the engine absorbs 1,000 J of heat from the high
temperature reservoir in each cycle, the amount of heat rejected to the low
temperature reservoir is equal to the energy not converted to work:
Qrejected =Qin −Wout
Where Qin is the heat absorbed from the high temperature reservoir and Wout
is the work output. Since the engine is reversible (Carnot engine), the work
output can be calculated using the Carnot efficiency:
Wout = Efficiency ×Qin = 0.4×1000 J = 400 J
Therefore,
Qrejected = 1000 J −400 J = 600 J
Therefore,
(a) The efficiency of the engine is 40
(b) The amount of heat rejected to the low temperature reservoir in each cycle
is 600 J.
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, such
that the engine absorbs heat energy Qhfrom the hot reservoir and releases
heat energy Qcto the cold reservoir. If the engine does 6000 J of work while
absorbing 9000 J of heat energy from the hot reservoir, what is the efficiency of
the engine? Is this result consistent with the second law of thermodynamics?
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = W
Qh
3
where Wis the work done by the engine and Qhis the heat energy absorbed
from the hot reservoir.
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 6000 J
9000 J
Step 3: Calculate the efficiency:
Efficiency = 2
3≈0.67
Step 4: The efficiency of the engine is 0.67 or 67
Step 5: According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Therefore, the efficiency of the engine is 67
Question 4
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
expels heat Qcto the low-temperature reservoir. If the efficiency of the engine
is ηand the net work output is W, prove that the efficiency of the engine can
be expressed in terms of Qhand Qcas:
η= 1 −TcQc
ThQh
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η=W
Qh
Step 2: Since W=Qh−Qc, we can rewrite the efficiency formula as:
η=Qh−Qc
Qh
Step 3: Simplify the expression by dividing both the numerator and denom-
inator by Qh:
η= 1 −Qc
Qh
Step 4: Recall from the first law of thermodynamics that for a cyclic process,
the net heat input must be equal to the net work output:
4
Qh−Qc=W
Step 5: Rearrange the equation above for Qh:
Qh=W+Qc
Step 6: Substitute this expression for Qhinto the efficiency formula:
η= 1 −Qc
W+Qc
Step 7: Substitute W=Qh−Qcinto the equation above:
η= 1 −Qc
Qh
Step 8: Finally, use the given relation between Qhand Qcin terms of tem-
peratures Thand Tc:
Qh=ThQc
η= 1 −Qc
ThQc
= 1 −TcQc
ThQh
Therefore, the efficiency of the engine can be expressed in terms of Qhand
Qcas η= 1 −TcQc
ThQh.
Question 5
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine is able to convert 400 J of heat to work during
each cycle while absorbing 600 J of heat from the hot reservoir, what is the
efficiency of the engine? Additionally, discuss how this efficiency relates to the
second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
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.
5
Given that the engine absorbs 600 J of heat from the hot reservoir (Qh= 600
J) and converts 400 J of heat to work per cycle (W= 400 J), we know that the
remaining 200 J is rejected to the cold reservoir.
Step 2: Calculate the temperatures of the hot and cold reservoirs. We can
use the fact that the work output is related to the difference in heat between
the hot and cold reservoirs:
W=Qh−Qc
where Qcis the heat rejected to the cold reservoir.
Since Qh= 600 J, Qc= 200 J, and W= 400 J, we can rewrite the formula
as:
400 = 600 −200
Step 3: Calculate the efficiency of the engine. Using the formula for efficiency
of the Carnot engine, we can substitute the given temperatures Tcand Thto
find the efficiency:
Efficiency = 1 −200
600 = 1 −1
3=2
3
Step 4: Discuss the relationship between the efficiency and the second law
of thermodynamics. The efficiency of the Carnot engine, often referred to as
the Carnot efficiency, is the maximum possible efficiency that any heat engine
operating between the same two temperatures can achieve. This efficiency is a
consequence of the second law of thermodynamics, which states that no engine
can be 100
In this case, the calculated efficiency of 2
3indicates that the engine is able
to convert 2/3 of the input heat into work, while the remaining 1/3 is rejected
as waste heat to the cold reservoir. This demonstrates the limitation imposed
by the second law of thermodynamics on the efficiency of heat engines.
Question 6
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir. Calculate the efficiency of the Carnot
engine in terms of Th,Tc,Qh, and Qc.
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
6
Step 2: The heat absorbed by the engine from the hot reservoir is equal
to the work done by the engine plus the heat rejected to the cold reservoir:
Qh=W+Qc.
Step 3: Since the Carnot engine is reversible, the work done by the engine
is equal to the difference in heat absorbed and heat rejected: W=Qh−Qc.
Step 4: Substitute W=Qh−Qcinto the efficiency formula:
Efficiency = 1 −Tc
Th
= 1 −Tc
Th
= 1 −Qc
Qh−Qc
= 1 −Qc
Qh
−Qc
Qh
Step 5: Simplify the expression to obtain the efficiency in terms of Th,Tc,
Qh, and Qc:
Efficiency = 1 −Qc
Qh
−Qc
Qh
= 1 −2Qc
Qh
Therefore, the efficiency of the Carnot engine in terms of Th,Tc,Qh, and
Qcis 1 −2Qc
Qh.
Question 7
Question
A heat engine operates between a hot reservoir at temperature T1= 500 K and
a cold reservoir at temperature T2= 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and discharges 2500 J to the cold reservoir.
Calculate the efficiency of the heat engine and discuss its relation to the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −T2
T1
Let’s substitute T1= 500 K and T2= 300 K into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
So, the efficiency of the heat engine is 40
Step 2: Discuss the relation of the efficiency to the second law of thermody-
namics. According to the second law of thermodynamics, no engine can have
an efficiency of 100
7
Question 8
Question
A Carnot engine operates between two heat reservoirs at temperatures Th= 500
K and Tc= 200 K. The engine absorbs 4000 J of heat from the hot reservoir in
each cycle. Calculate:
1. The heat rejected by the engine to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
3. The efficiency of the engine.
4. Discuss whether this Carnot engine violates the second law of thermody-
namics.
Solution
1. To find the heat rejected by the engine to the cold reservoir in each cycle, we
can use the conservation of energy:
Qh=W+Qc
where Qhis the heat absorbed from the hot reservoir, Wis the work done by the
engine, and Qcis the heat rejected to the cold reservoir. Given that Qh= 4000
J, we can calculate Qc:
4000 J = W+Qc
2. The work done by the engine in each cycle can be calculated using the
efficiency of the engine:
e=W
Qh
where eis the efficiency of the engine. Using the Carnot efficiency formula:
e= 1 −Tc
Th
we can solve for W.
3. The efficiency of the Carnot engine is calculated by substituting the given
temperatures into the Carnot efficiency formula:
e= 1 −200
500
4. The second law of thermodynamics states that no engine operating be-
tween two heat reservoirs can be more efficient than a Carnot engine operating
between the same reservoirs. Therefore, this Carnot engine does not violate the
second law of thermodynamics since it is the most efficient engine possible.
Therefore, to summarize: 1. The heat rejected by the engine to the cold
reservoir in each cycle is 1600 J. 2. The work done by the engine in each cycle
is 2400 J. 3. The efficiency of the engine is 0.6 or 604. The Carnot engine does
not violate the second law of thermodynamics.
8
Question 9
Question
A heat engine operates between two reservoirs at temperatures Th= 500 K and
Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir per cycle
and exhausts 1200 J to the cold reservoir per cycle. Calculate the efficiency of
the engine and discuss whether it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input = 1 −Heat rejected
Heat input
Step 2: Calculate the useful work output. Since the engine’s cycle absorbs
2000 J of heat and exhausts 1200 J to the cold reservoir, the useful work output
can be calculated as the net heat flow into the engine:
Useful work output = Heat input −Heat rejected = 2000 J −1200 J
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −1200 J
2000 J
Step 4: Calculate the efficiency:
Efficiency = 1 −0.6 = 0.4 or 40%
Step 5: Discuss whether the engine violates the second law of thermody-
namics. The efficiency of the engine is 40
Question 10
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine delivers 1000 J of work during each 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 engine using the formula
Efficiency = 1 −Qcoulomb
Qhot
9
where Qcoulomb is the heat expelled to the cold reservoir and Qhot is the heat
absorbed from the hot reservoir. We can relate the work done to the heat
exchanged using the first law of thermodynamics: Qhot −Qcoulomb =W.
Step 2: Since the engine delivers 1000 J of work during each cycle, we have
Qhot −Qcoulomb = 1000 J.
Step 3: Express Qhot and Qcoulomb in terms of the reservoir temperatures,
Thot and Tcold, and the efficiency, η.
Qhot =Qcoulomb
1−η
Qcoulomb =η·Qhot
Step 4: Substitute Qhot =Qcoulomb + 1000 into Qhot =Qcoulomb
1−η.
Qcoulomb + 1000 = Qcoulomb
1−η
Step 5: Solve for Qcoulomb in terms of η.
Qcoulomb =1000
1 + 1
η
Step 6: Substitute Qcoulomb =1000
1+ 1
η
into the efficiency formula to find the
efficiency of the engine.
Step 7: With the efficiency calculated, discuss whether the engine violates
the second law of thermodynamics by considering if the efficiency is less than 1.
Question 11
Question
A heat engine operates between two reservoirs at temperatures 700 K and 400
K. The engine absorbs 2400 J of heat from the high-temperature reservoir and
produces 1600 J of work. Calculate the efficiency of the engine. Is this engine
violating the second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency
Efficiency = Useful work output
Heat input
Step 2: Given that the engine absorbs 2400 J of heat from the high-temperature
reservoir and produces 1600 J of work, we have:
Useful work output = 1600 J
10
Heat input = 2400 J
Step 3: Substitute the values into the efficiency formula and calculate
Efficiency = 1600
2400 =2
3= 0.67 = 67%
Therefore, the efficiency of the engine is 67
Step 4: According to the second law of thermodynamics, no heat engine can
be 100
Thus, the engine is not violating the second law of thermodynamics.
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwhere
Th> Tc. If the efficiency of the engine is 40
Solution
Let’s denote the efficiency of the heat engine as η. The efficiency of a heat
engine is given by the formula:
η= 1 −Tc
Th
Given that the efficiency is 40
Step 1: Substitute the given efficiency value into the efficiency formula:
0.40 = 1 −Tc
Th
Step 2: Rearrange the equation to solve for Tc
Th:
Tc
Th
= 1 −0.40
Step 3: Simplify the expression:
Tc
Th
= 0.60
Step 4: Hence, the ratio Tc
Thin terms of This 0.60 .
Question 13
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
If the efficiency of the engine is 30
11
Solution
Let’s denote the efficiency of the engine as η.
Step 1: Recall the formula for the efficiency of a heat engine:
η= 1 −Tcold
Thot
Step 2: Given that η= 30% = 0.3, we can rewrite the efficiency formula
as:
0.3=1−Tcold
Thot
Step 3: Solve for the ratio Thot/Tcold by rearranging the equation:
Tcold
Thot
= 1 −0.3 = 0.7
Step 4: Finally, the ratio Thot/Tcold is given by:
Thot
Tcold
=1
0.7=10
7
Therefore, the ratio Thot/Tcold is 10
7.
Question 14
Question
A Carnot heat engine operates between two heat reservoirs at temperatures TH
and TC, with TH> TC. If the efficiency of the engine is 40
Solution
Step 1: Recall the formula for the efficiency of a Carnot heat engine:
Efficiency = 1 −TC
TH
Given that the efficiency is 40
0.40 = 1 −TC
TH
Step 2: Rearrange the equation to get TH
TCin terms of TC:
TC
TH
= 1 −0.40 = 0.60
TH
TC
=1
0.60 =5
3
Therefore, the ratio TH
TCin terms of TCis 5
3.
12
Question 15
Question
A heat engine operates between two reservoirs at temperatures TH= 500 K
and TC= 300 K. The engine absorbs 1500 J of heat from the hot reservoir and
rejects 900 J of heat to the cold reservoir during each cycle.
a) Calculate the efficiency of the engine.
b) Determine the maximum work that the engine can perform during each
cycle.
Solution
a) Given: Hot reservoir temperature, TH= 500 K Cold reservoir temperature,
TC= 300 K Heat absorbed, QH= 1500 J Heat rejected, QC= 900 J
Efficiency of a heat engine is given by the formula:
Efficiency, η= 1 −QC
QH
Step 1: Substitute the given values into the formula.
η= 1 −900
1500
Step 2: Calculate the efficiency.
η= 1 −0.6=0.4 = 40%
Therefore, the efficiency of the engine is 40
b) The maximum work that can be performed by the engine is given by the
formula:
Wmax =QH−QC
Step 1: Substitute the given values into the formula.
Wmax = 1500 −900
Step 2: Calculate the maximum work performed by the engine.
Wmax = 600 J
Therefore, the maximum work that the engine can perform during each cycle
is 600 J.
Question 16
Question
A heat engine operates between two reservoirs at 600K and 400K. The engine
absorbs 4000 J of heat from the high-temperature reservoir in each cycle. Cal-
culate the maximum efficiency of the engine.
13
Solution
To find the maximum efficiency of the heat engine, we can use the Carnot
efficiency formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the low-temperature reservoir and THis the
temperature of the high-temperature reservoir.
Step 1: Calculate the temperature of the low-temperature reservoir TC=
400 K.
Step 2: Calculate the temperature of the high-temperature reservoir TH=
600 K.
Step 3: Substitute the values of TCand THinto the Carnot efficiency
formula:
Efficiency = 1 −400 K
600 K
Step 4: Calculate the efficiency:
Efficiency = 1 −2
3=1
3≈33.3%
Therefore, the maximum efficiency of the heat engine is approximately 33.3
Question 17
Question
A heat engine operates between two reservoirs at temperatures T1= 500 K and
T2= 300 K. If the engine absorbs 5,000 J of heat from the reservoir at T1in
each cycle, calculate the maximum theoretical efficiency of the engine. Explain
why this efficiency cannot be achieved in practice.
Solution
Step 1: Determine the maximum efficiency of the heat engine using Carnot’s
theorem. Carnot’s theorem states that the maximum efficiency of a heat engine
operating between two reservoirs at temperatures T1and T2is given by:
η= 1 −T2
T1
Step 2: Substitute the values into the formula. Substitute T1= 500 K and
T2= 300 K into the formula:
η= 1 −300
500 = 1 −0.6=0.4
Therefore, the maximum theoretical efficiency of the heat engine is 40%.
14
Step 3: Explain why this efficiency cannot be achieved in practice. In prac-
tice, the efficiency of real heat engines is always less than the maximum theo-
retical efficiency given by Carnot’s theorem due to various irreversible processes
such as friction, heat loss to the surroundings, and internal energy losses in the
engine components. These losses decrease the efficiency of the engine, making
it impossible to achieve the ideal efficiency predicted by Carnot’s theorem.
Question 18
Question
A heat engine operates between a hot reservoir at a temperature of 600 K and a
cold reservoir at a temperature of 200 K. The engine takes in 6000 J of heat from
the hot reservoir and expels 2000 J to the cold reservoir per cycle. Calculate
the efficiency of the heat engine and discuss whether this violates the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Qc
Qh
where Qcis the heat expelled to the cold reservoir and Qhis the heat taken in
from the hot reservoir.
Step 2: Substitute the given values into the formula. Given: Hot reservoir
temperature (Th) = 600 K Cold reservoir temperature (Tc) = 200 K Heat taken
in from hot reservoir (Qh) = 6000 J Heat expelled to cold reservoir (Qc) = 2000
J
So, Qc= 2000 J and Qh= 6000 J.
Step 3: Calculate the efficiency.
Efficiency = 1 −2000
6000 = 1 −1
3=2
3
Hence, the efficiency of the heat engine is 2
3or 66.67
Step 4: Discuss the violation of the second law of thermodynamics. The
efficiency obtained (66.67
EfficiencyCarnot = 1 −Tc
Th
= 1 −200
600 =2
3
This means that the heat engine is operating at the theoretical maximum effi-
ciency possible for engines operating between these two temperatures. There-
fore, the second law of thermodynamics is not violated.
15
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine absorbs heat Qhfrom the hot reservoir and rejects heat Qcto
the cold reservoir in each cycle. If the efficiency of the engine is η, show that
the efficiency can be expressed as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input:
η=W
Qh
Step 2: By the first law of thermodynamics, the net work output can be
expressed as:
W=Qh−Qc
Step 3: Substituting W=Qh−Qcback into the efficiency formula, we get:
η=Qh−Qc
Qh
Step 4: Rearranging the equation, we have:
η= 1 −Qc
Qh
Step 5: By the second law of thermodynamics, for a reversible engine, the
ratio of heat absorbed to the temperature at which it is absorbed is the same
as the ratio of heat rejected to the temperature at which it is rejected, i.e.
Qh
Th=Qc
Tc.
Step 6: Using this relation, we can rewrite Qc
Qhin terms of temperatures:
Qc
Qh
=Tc
Th
Step 7: Substituting Qc
Qh=Tc
Thback into our efficiency formula, we get:
η= 1 −Tc
Th
Therefore, the efficiency of the engine can be expressed as η= 1 −Tc
Th.
16
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir.
Prove that the efficiency of the heat engine is bounded by the Carnot effi-
ciency, which is given by 1 −Tc
Th.
Solution
Let Wbe the work done by the engine, Qhbe the heat absorbed from the hot
reservoir, and Qcbe the heat exhausted to the cold reservoir.
Step 1: Write the first law of thermodynamics. According to the
first law of thermodynamics, the net work done by the engine is equal to the
difference between the heat absorbed and the heat exhausted:
W=Qh−Qc
Step 2: Write the efficiency of the engine. The efficiency of the engine
is defined as the ratio of the work done by the engine to the heat absorbed from
the hot reservoir:
Efficiency, η=W
Qh
Step 3: Express work done in terms of heat. The work done by the
engine can be expressed in terms of the heat absorbed and the heat exhausted:
W=Qh−Qc
Step 4: Substitute work done into the efficiency equation. Substi-
tute the expression for work done into the efficiency equation:
η=Qh−Qc
Qh
Step 5: Simplify the efficiency equation. Simplify the efficiency equa-
tion:
η= 1 −Qc
Qh
Step 6: Apply the second law of thermodynamics. According to the
second law of thermodynamics, the efficiency of any heat engine is bounded by
the Carnot efficiency:
η≤1−Tc
Th
17
Step 7: Substitute Qcand Qhwith temperature differentials. Since
Qh=ThShand Qc=TcSc(where Shand Scare the entropies of the hot and
cold reservoirs, respectively), substitute these values into the efficiency equation:
η≤1−Tc
Th
Therefore, the efficiency of the heat engine is bounded by the Carnot effi-
ciency.
Question 21
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine takes in 4000 J of heat from the hot reservoir and exhausts
2500 J of heat to the cold reservoir during each cycle. Calculate the efficiency
of the engine. Is this engine violating the second law of thermodynamics?
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = 1 −Qc
Qh
where Qhis the heat absorbed from the hot reservoir and Qcis the heat ex-
hausted to the cold reservoir.
Step 2: Identify the values given in the question. Qh= 4000 J and Qc= 2500
J.
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −2500
4000 = 1 −5
8=3
8
Step 4: Thus, the efficiency of the engine is 3
8or 37.5
Step 5: The engine is not violating the second law of thermodynamics be-
cause the efficiency obtained is less than 100
Therefore, the efficiency of the engine is 3
8or 37.5
Question 22
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. If 1500 J of work is done by the
engine during each cycle, calculate the amount of heat absorbed from the hot
reservoir during each cycle. Also, determine the efficiency of the engine.
18
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir. Substitute Tc= 300 K and Th= 600 K into the equation:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Since efficiency is defined as the ratio of work done to heat absorbed
from the hot reservoir, we can find the amount of heat absorbed as follows: Let
Qhbe the heat absorbed from the hot reservoir. Then, efficiency = W
Qh, where
W= 1500 J.
Efficiency = W
Qh
⇒Qh=W
Efficiency
Substitute W= 1500 J and efficiency = 0.5 into the equation:
Qh=1500
0.5= 3000 J
Therefore, the amount of heat absorbed from the hot reservoir during each
cycle is 3000 J, and the efficiency of the Carnot engine is 50
Question 23
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir. The engine produces a net work output of W. Show
that the efficiency of the engine, η, is given by η= 1 −Tc
Thand discuss the
implications of the second law of thermodynamics on this expression.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency of a
heat engine is defined as the ratio of the work output to the heat input from
the hot reservoir. Mathematically, efficiency is given by:
η=W
Qh
Step 2: Use the first law of thermodynamics for the heat engine: According
to the first law of thermodynamics, the net work output of the engine is equal
19
to the difference between the heat absorbed from the hot reservoir and the heat
exhausted to the cold reservoir:
W=Qh−Qc
Step 3: Substitute the given expression for work into the efficiency equation:
Substitute W=Qh−Qcinto the expression for efficiency:
η=Qh−Qc
Qh
Step 4: Rearrange the expression: Divide both terms in the numerator by
Qh:
η=Qh
Qh
−Qc
Qh
Step 5: Consider the heat absorbed and heat exhausted in terms of temper-
atures: According to the second law of thermodynamics, the heat flow is related
to the temperatures of the reservoirs:
Qh
Th
−Qc
Tc
≤0
Step 6: Substitute the relationship between Qhand Qcusing the second law:
From Step 5, we have: Qh
Th
−Qc
Tc
≤0
Qh−Th
Tc
Qc≤0
Qh≤Th
Tc
Qc
Step 7: Substitute the inequality into the efficiency equation: Substitute
Qh≤Th
TcQcinto the efficiency equation:
η= 1 −Qc
Qh
≥1−Tc
Th
Step 8: Conclusion: Therefore, we have shown that the efficiency of the heat
engine is given by η= 1 −Tc
Th. This expression indicates that the efficiency of a
heat engine is limited by the temperature difference between the hot and cold
reservoirs, in accordance with the second law of thermodynamics.
Question 24
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhamount of heat from the high-
temperature reservoir and expels Qcamount of heat to the low-temperature
reservoir. Calculate the efficiency of the Carnot engine in terms of Qhand Th.
20
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: We can express the efficiency in terms of the heat absorbed from the
high-temperature reservoir Qhby using the relation Qh= (1 −Tc
Th)Qh.
Step 3: Simplify the expression:
Efficiency = 1 −Tc
Th
= 1 −Qc
Qh
Therefore, the efficiency of the Carnot engine in terms of Qhand This
1−Qc
Qh
.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc), where Qhis the heat absorbed from the hot reservoir and Wis the
work done by the engine. Prove that the efficiency of the engine is given by
η= 1 −Tc
Th
.
Solution
Step 1: Recall that the First Law of Thermodynamics states that the net work
done by a heat engine is equal to the heat absorbed from the hot reservoir minus
the heat rejected to the cold reservoir:
W=Qh−Qc.
Step 2: The efficiency of a heat engine is defined as the ratio of the work
done by the engine to the heat absorbed from the hot reservoir:
η=W
Qh
.
Step 3: Substitute W=Qh−Qcinto the efficiency formula:
η=Qh−Qc
Qh
.
Step 4: Rearrange terms and factor out Qhto simplify the expression:
η= 1 −Qc
Qh
.
21
Step 5: Recall the definition of efficiency in terms of temperatures:
η= 1 −Tc
Th
.
Step 6: Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 26
Question
A Carnot engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc. The engine absorbs 6000 J of heat from the hot
reservoir and discharges 4000 J to the cold reservoir in each cycle. Calculate
the efficiency of the engine in terms of Thand Tc.
Solution
Step 1: Recall that the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: First, let’s find the value of Tc
Th. Given that the engine absorbs 6000
J from the hot reservoir and discharges 4000 J to the cold reservoir in each cycle,
we have: Qh
Qc
=6000
4000 =Th
Tc
Step 3: Simplify the expression to find Tc
Th:
Tc
Th
=4000
6000 =2
3
Step 4: Substitute Tc
Thinto the efficiency formula:
Efficiency = 1 −2
3=1
3
Therefore, the efficiency of the engine in terms of Thand Tcis 1
3.
Question 27
Question
A Carnot engine operating between two reservoirs produces 500 J of work for
every 1000 J of heat absorbed. What is the efficiency of this engine? In addition,
if the colder reservoir is at 50◦C, what is the minimum temperature of the hotter
reservoir?
22
Solution
Step 1: Recall the formula for efficiency of a heat engine:
Efficiency = Useful work output
Heat input
Step 2: Given that the Carnot engine produces 500 J of work for every 1000
J of heat absorbed, we have:
Efficiency = 500 J
1000 J = 0.5 = 50%
Therefore, the efficiency of the engine is 50
Step 3: The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the colder reservoir and This the abso-
lute temperature of the hotter reservoir.
Step 4: We can rearrange the formula to solve for Th:
Tc
Th
= 1 −Efficiency
Th=Tc
1−Efficiency
Step 5: Convert the temperature in Celsius to Kelvin:
Tc= 50◦C + 273 = 323 K
Step 6: Substitute the values into the formula:
Th=323
1−0.5= 646 K
Therefore, the minimum temperature of the hotter reservoir is 646 K.
Question 28
Question
A heat engine operates between a hot reservoir at 700 K and a cold reservoir at
300 K. The engine absorbs 2000 J of heat from the hot reservoir in each cycle
and exhausts 1200 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine. Is this engine in violation of the second law of thermodynamics?
23
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Substitute the values for heat input and heat output into the equa-
tion above:
Efficiency = 1 −1200
2000
Step 3: Perform the calculation to find the efficiency of the engine:
Efficiency = 1 −1200
2000 = 1 −0.6=0.4
Step 4: Therefore, the efficiency of the engine is 40
Step 5: The efficiency of the engine is less than 100
Step 6: The second law of thermodynamics states that heat always flows
from a hotter body to a colder body, and no heat engine can be 100
Carnot efficiency = 1 −Tcold
Thot
Step 7: For this case, the Carnot efficiency would be:
Carnot efficiency = 1 −300
700 = 1 −0.4286 = 0.5714 = 57.14%
Step 8: Comparing the efficiency of the engine (40
Question 29
Question
A heat engine operates between two reservoirs at temperatures Thand Tc. The
engine absorbs Qhof heat from the high-temperature reservoir and does Wof
work. The efficiency of the engine is defined as η=W
Qh. Show that the maximum
possible efficiency of any heat engine operating between two reservoirs is given
by ηmax = 1−Tc
Th, where This the absolute temperature of the high-temperature
reservoir and Tcis the absolute temperature of the low-temperature reservoir.
Solution
Step 1: Let’s start with the First Law of Thermodynamics:
∆U=Q−W
where ∆Uis the change in internal energy, Qis the heat absorbed, and Wis
the work done by the engine.
24
Step 2: For a heat engine, the net work done by the engine is the difference
between the heat absorbed from the high-temperature reservoir and the heat
rejected to the low-temperature reservoir:
W=Qh−Qc
where Qhis the heat absorbed from the high-temperature reservoir and Qcis
the heat rejected to the low-temperature reservoir.
Step 3: The efficiency of the heat engine is defined as:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: By the Second Law of Thermodynamics, the heat engine cannot
operate at 100
Step 5: From the definition of temperature in the Carnot cycle:
Qc
Tc
=Qh
Th
Dividing through by Qh:Qc
Qh
=Tc
Th
Step 6: Substituting into the efficiency expression:
η≤1−Tc
Th
Hence, the maximum possible efficiency of any heat engine operating between
two reservoirs is ηmax = 1 −Tc
Th.
Question 30
Question
A heat 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 3500 J of heat
from the hot reservoir and delivers 2000 J of work output during each cycle.
Calculate: a) The efficiency of the engine. b) The amount of heat rejected to
the cold reservoir during each cycle. c) Discuss whether the engine adheres to
the second law of thermodynamics.
Solution
a) The efficiency of the engine is given by the formula:
Efficiency = Work Output
Heat Input
25
Step 1: Calculate the efficiency of the engine. The work output is 2000 J
and the heat input is 3500 J.
Efficiency = 2000 J
3500 J = 0.5714 or 57.14%
b) The amount of heat rejected to the cold reservoir during each cycle is
given by:
Heat Rejected = Heat Input −Work Output
Step 2: Calculate the amount of heat rejected to the cold reservoir. The
heat input is 3500 J and the work output is 2000 J.
Heat Rejected = 3500 J −2000 J = 1500 J
c) A heat engine operates by converting heat into work, but it cannot convert
all the heat it absorbs into work. According to the second law of thermodynam-
ics, the efficiency of a heat engine is never 100Therefore, the engine adheres to
the second law of thermodynamics as it cannot convert all the input heat into
work and some heat is always rejected to the cold reservoir.
Question 31
Question
A heat engine operates between a high temperature reservoir at 500 K and a low
temperature reservoir at 300 K. The engine absorbs 4000 J of heat from the high
temperature reservoir in each cycle and produces 2000 J of work. Calculate the
efficiency of the engine and discuss whether the operation of this engine violates
the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency (%) = Work output
Heat input ×100%
Step 2: Substitute the given values into the formula:
Efficiency (%) = 2000
4000 ×100% = 50%
Therefore, the efficiency of the heat engine is 50%.
Step 3: According to the second law of thermodynamics, no heat engine
operating in a cycle can absorb heat from a single reservoir and convert it
completely into work without producing any other effect.
Step 4: In this case, the engine violates the second law of thermodynamics
because it is operating at 100
Therefore, the operation of this engine violates the second law of thermody-
namics.
26
Question 32
Question
A heat engine operates between a source temperature of 600 K and a sink
temperature of 300 K. The engine absorbs 2000 J of heat from the source in
each cycle. Calculate the efficiency of the engine. Is this engine violating the
second law of thermodynamics?
Solution
Step 1: Calculate the work done by the engine during each cycle using the
formula for efficiency.
Efficiency = 1 −Heat rejected
Heat absorbed
Efficiency = 1 −Tsink
Tsource
Step 2: Substitute the source and sink temperatures into the formula.
Efficiency = 1 −300
600
Efficiency = 1 −0.5
Efficiency = 0.5
Step 3: Determine if the engine is violating the second law of thermodynam-
ics. According to the second law of thermodynamics, the efficiency of a heat
engine cannot be 100
Therefore, the efficiency of the engine is 50
Question 33
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
releases heat Qcto the low-temperature reservoir. If the efficiency of the engine
is η, show that the efficiency of the engine can be expressed in terms of the
temperatures of the reservoirs as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
η= 1 −|Qc|
|Qh|.
Step 2: We know that the work output of the engine is W=|Qh|−|Qc|.
27
Step 3: Using the first law of thermodynamics, we have W=|Qh|−|Qc|=
|Qh|(1 −|Qc|
|Qh|).
Step 4: Now, we can rewrite the efficiency in terms of the temperatures of
the reservoirs. The heat absorbed Qhis related to Thand Tcby the equation
Qh=ThS, where Sis the entropy change of the hot reservoir.
Step 5: Similarly, the heat rejected Qcis related to Thand Tcby the equation
Qc=TcS, where Sis the entropy change of the cold reservoir.
Step 6: Plugging these expressions back into the efficiency formula, we get
η= 1 −TcS
ThS= 1 −Tc
Th.
Step 7: Therefore, the efficiency of the heat engine can be expressed in terms
of the temperatures of the reservoirs as η= 1 −Tc
Th.
Question 34
Question
A heat engine operates between a high-temperature reservoir at 600 K and a
low-temperature reservoir at 300 K. The engine absorbs 600 J of heat from the
high-temperature reservoir in each cycle.
1. Calculate the maximum efficiency of this engine.
2. If the engine releases 300 J of heat to the low-temperature reservoir in
each cycle, calculate the work done by the engine in each cycle.
Solution
1. To find the maximum efficiency of the heat engine, we can use the formula
for the efficiency of a Carnot engine:
η= 1 −Tlow
Thigh
where Tlow and Thigh are the absolute temperatures of the low-temperature and
high-temperature reservoirs, respectively.
Given Thigh = 600 K and Tlow = 300 K, we can plug in these values to find
the efficiency:
η= 1 −300
600 = 1 −0.5 = 0.5 = 50%
Therefore, the maximum efficiency of this engine is 50%.
2. The work done by the engine in each cycle can be calculated using the
formula:
Work done = Heat absorbed −Heat released
Given that the engine absorbs 600 J from the high-temperature reservoir and
releases 300 J to the low-temperature reservoir, we can substitute these values
into the formula:
Work done = 600 J −300 J = 300 J
28
Therefore, the work done by the engine in each cycle is 300 J.
Question 35
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine takes in heat energy QHfrom the hot reservoir
and produces work output W. If the efficiency of the engine is ηand the engine
follows the Carnot cycle, prove that the efficiency of the engine can be expressed
as η= 1 −TC
TH.
Solution
Step 1: Recall the definition of efficiency for a heat engine:
η=Useful work output
Heat energy input
Step 2: For the Carnot engine, the efficiency is given by:
η= 1 −TC
TH
Step 3: The efficiency of a Carnot engine is based on the fact that the work
done by the engine is the difference between the heat energy input from the hot
reservoir and the heat energy rejected to the cold reservoir. Mathematically,
this can be expressed as:
W=QH−QC
Step 4: The efficiency can then be expressed as:
η=W
QH
Step 5: Substituting QC=QH−Winto the efficiency equation:
η=QH−(QH−W)
QH
Step 6: Simplifying the equation, we get:
η=W
QH
=W
W+QC
Step 7: Using the equation QC=TC(S2−S1) and the definition of entropy,
QC=TC∆S, where ∆Sis the change in entropy, we can write the equation in
terms of temperature:
QC=TC∆S=TC(S2−S1)
29
So, the efficiency of the engine in terms of T1and T2is 1 −T2
T1.
Part (b)
Step 3: The heat rejected by the engine to the reservoir at temperature T2
in each cycle can be calculated using the formula for efficiency:
Efficiency = Work output
Heat input
Step 4: Rearranging the formula, we get:
Heat rejected = Heat input −Work output
Step 5: Substituting the given values, we have:
Heat rejected = 600 J −1−T2
T1×600 J
Therefore, the heat rejected by the engine to the reservoir at temperature
T2in each cycle is 600 J −600 J 1−T2
T1.
Part (c)
Step 6: The total work done by the engine after completing 100 cycles can
be calculated using the relationship between work and heat:
Work output = Heat input −Heat rejected
Step 7: For 100 cycles, the total work done is:
Total work done = 100 ×Heat input −1−T2
T1×Heat input
Step 8: Substituting the given values, we get:
Total work done = 100 ×600 J −600 J 1−T2
T1
Therefore, the total work done by the engine after completing 100 cycles is
100 ×h600 J −600 J 1−T2
T1i.
Question 2
Question
A Carnot heat engine operates between a high temperature reservoir at 500 K
and a low temperature reservoir at 300 K. If the engine absorbs 1,000 J of heat
from the high temperature reservoir in each cycle, calculate the following:
(a) The efficiency of the engine.
(b) The amount of heat rejected to the low temperature reservoir in each
cycle.
2
Solution
Step 1: Calculate the efficiency of the engine using the Carnot efficiency for-
mula:
Efficiency = 1 −Tlow
Thigh
Given Thigh = 500 K and Tlow = 300 K,
Efficiency = 1 −300
500 = 1 −0.6=0.4 = 40%
Step 2: Calculate the amount of heat rejected to the low temperature
reservoir in each cycle. Since the engine absorbs 1,000 J of heat from the high
temperature reservoir in each cycle, the amount of heat rejected to the low
temperature reservoir is equal to the energy not converted to work:
Qrejected =Qin −Wout
Where Qin is the heat absorbed from the high temperature reservoir and Wout
is the work output. Since the engine is reversible (Carnot engine), the work
output can be calculated using the Carnot efficiency:
Wout = Efficiency ×Qin = 0.4×1000 J = 400 J
Therefore,
Qrejected = 1000 J −400 J = 600 J
Therefore,
(a) The efficiency of the engine is 40
(b) The amount of heat rejected to the low temperature reservoir in each cycle
is 600 J.
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, such
that the engine absorbs heat energy Qhfrom the hot reservoir and releases
heat energy Qcto the cold reservoir. If the engine does 6000 J of work while
absorbing 9000 J of heat energy from the hot reservoir, what is the efficiency of
the engine? Is this result consistent with the second law of thermodynamics?
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = W
Qh
3
where Wis the work done by the engine and Qhis the heat energy absorbed
from the hot reservoir.
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 6000 J
9000 J
Step 3: Calculate the efficiency:
Efficiency = 2
3≈0.67
Step 4: The efficiency of the engine is 0.67 or 67
Step 5: According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Therefore, the efficiency of the engine is 67
Question 4
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
expels heat Qcto the low-temperature reservoir. If the efficiency of the engine
is ηand the net work output is W, prove that the efficiency of the engine can
be expressed in terms of Qhand Qcas:
η= 1 −TcQc
ThQh
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η=W
Qh
Step 2: Since W=Qh−Qc, we can rewrite the efficiency formula as:
η=Qh−Qc
Qh
Step 3: Simplify the expression by dividing both the numerator and denom-
inator by Qh:
η= 1 −Qc
Qh
Step 4: Recall from the first law of thermodynamics that for a cyclic process,
the net heat input must be equal to the net work output:
4
Qh−Qc=W
Step 5: Rearrange the equation above for Qh:
Qh=W+Qc
Step 6: Substitute this expression for Qhinto the efficiency formula:
η= 1 −Qc
W+Qc
Step 7: Substitute W=Qh−Qcinto the equation above:
η= 1 −Qc
Qh
Step 8: Finally, use the given relation between Qhand Qcin terms of tem-
peratures Thand Tc:
Qh=ThQc
η= 1 −Qc
ThQc
= 1 −TcQc
ThQh
Therefore, the efficiency of the engine can be expressed in terms of Qhand
Qcas η= 1 −TcQc
ThQh.
Question 5
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine is able to convert 400 J of heat to work during
each cycle while absorbing 600 J of heat from the hot reservoir, what is the
efficiency of the engine? Additionally, discuss how this efficiency relates to the
second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
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.
5
Given that the engine absorbs 600 J of heat from the hot reservoir (Qh= 600
J) and converts 400 J of heat to work per cycle (W= 400 J), we know that the
remaining 200 J is rejected to the cold reservoir.
Step 2: Calculate the temperatures of the hot and cold reservoirs. We can
use the fact that the work output is related to the difference in heat between
the hot and cold reservoirs:
W=Qh−Qc
where Qcis the heat rejected to the cold reservoir.
Since Qh= 600 J, Qc= 200 J, and W= 400 J, we can rewrite the formula
as:
400 = 600 −200
Step 3: Calculate the efficiency of the engine. Using the formula for efficiency
of the Carnot engine, we can substitute the given temperatures Tcand Thto
find the efficiency:
Efficiency = 1 −200
600 = 1 −1
3=2
3
Step 4: Discuss the relationship between the efficiency and the second law
of thermodynamics. The efficiency of the Carnot engine, often referred to as
the Carnot efficiency, is the maximum possible efficiency that any heat engine
operating between the same two temperatures can achieve. This efficiency is a
consequence of the second law of thermodynamics, which states that no engine
can be 100
In this case, the calculated efficiency of 2
3indicates that the engine is able
to convert 2/3 of the input heat into work, while the remaining 1/3 is rejected
as waste heat to the cold reservoir. This demonstrates the limitation imposed
by the second law of thermodynamics on the efficiency of heat engines.
Question 6
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir. Calculate the efficiency of the Carnot
engine in terms of Th,Tc,Qh, and Qc.
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
6
Step 2: The heat absorbed by the engine from the hot reservoir is equal
to the work done by the engine plus the heat rejected to the cold reservoir:
Qh=W+Qc.
Step 3: Since the Carnot engine is reversible, the work done by the engine
is equal to the difference in heat absorbed and heat rejected: W=Qh−Qc.
Step 4: Substitute W=Qh−Qcinto the efficiency formula:
Efficiency = 1 −Tc
Th
= 1 −Tc
Th
= 1 −Qc
Qh−Qc
= 1 −Qc
Qh
−Qc
Qh
Step 5: Simplify the expression to obtain the efficiency in terms of Th,Tc,
Qh, and Qc:
Efficiency = 1 −Qc
Qh
−Qc
Qh
= 1 −2Qc
Qh
Therefore, the efficiency of the Carnot engine in terms of Th,Tc,Qh, and
Qcis 1 −2Qc
Qh.
Question 7
Question
A heat engine operates between a hot reservoir at temperature T1= 500 K and
a cold reservoir at temperature T2= 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and discharges 2500 J to the cold reservoir.
Calculate the efficiency of the heat engine and discuss its relation to the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −T2
T1
Let’s substitute T1= 500 K and T2= 300 K into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
So, the efficiency of the heat engine is 40
Step 2: Discuss the relation of the efficiency to the second law of thermody-
namics. According to the second law of thermodynamics, no engine can have
an efficiency of 100
7
Question 8
Question
A Carnot engine operates between two heat reservoirs at temperatures Th= 500
K and Tc= 200 K. The engine absorbs 4000 J of heat from the hot reservoir in
each cycle. Calculate:
1. The heat rejected by the engine to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
3. The efficiency of the engine.
4. Discuss whether this Carnot engine violates the second law of thermody-
namics.
Solution
1. To find the heat rejected by the engine to the cold reservoir in each cycle, we
can use the conservation of energy:
Qh=W+Qc
where Qhis the heat absorbed from the hot reservoir, Wis the work done by the
engine, and Qcis the heat rejected to the cold reservoir. Given that Qh= 4000
J, we can calculate Qc:
4000 J = W+Qc
2. The work done by the engine in each cycle can be calculated using the
efficiency of the engine:
e=W
Qh
where eis the efficiency of the engine. Using the Carnot efficiency formula:
e= 1 −Tc
Th
we can solve for W.
3. The efficiency of the Carnot engine is calculated by substituting the given
temperatures into the Carnot efficiency formula:
e= 1 −200
500
4. The second law of thermodynamics states that no engine operating be-
tween two heat reservoirs can be more efficient than a Carnot engine operating
between the same reservoirs. Therefore, this Carnot engine does not violate the
second law of thermodynamics since it is the most efficient engine possible.
Therefore, to summarize: 1. The heat rejected by the engine to the cold
reservoir in each cycle is 1600 J. 2. The work done by the engine in each cycle
is 2400 J. 3. The efficiency of the engine is 0.6 or 604. The Carnot engine does
not violate the second law of thermodynamics.
8
Question 9
Question
A heat engine operates between two reservoirs at temperatures Th= 500 K and
Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir per cycle
and exhausts 1200 J to the cold reservoir per cycle. Calculate the efficiency of
the engine and discuss whether it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input = 1 −Heat rejected
Heat input
Step 2: Calculate the useful work output. Since the engine’s cycle absorbs
2000 J of heat and exhausts 1200 J to the cold reservoir, the useful work output
can be calculated as the net heat flow into the engine:
Useful work output = Heat input −Heat rejected = 2000 J −1200 J
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −1200 J
2000 J
Step 4: Calculate the efficiency:
Efficiency = 1 −0.6 = 0.4 or 40%
Step 5: Discuss whether the engine violates the second law of thermody-
namics. The efficiency of the engine is 40
Question 10
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine delivers 1000 J of work during each 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 engine using the formula
Efficiency = 1 −Qcoulomb
Qhot
9
where Qcoulomb is the heat expelled to the cold reservoir and Qhot is the heat
absorbed from the hot reservoir. We can relate the work done to the heat
exchanged using the first law of thermodynamics: Qhot −Qcoulomb =W.
Step 2: Since the engine delivers 1000 J of work during each cycle, we have
Qhot −Qcoulomb = 1000 J.
Step 3: Express Qhot and Qcoulomb in terms of the reservoir temperatures,
Thot and Tcold, and the efficiency, η.
Qhot =Qcoulomb
1−η
Qcoulomb =η·Qhot
Step 4: Substitute Qhot =Qcoulomb + 1000 into Qhot =Qcoulomb
1−η.
Qcoulomb + 1000 = Qcoulomb
1−η
Step 5: Solve for Qcoulomb in terms of η.
Qcoulomb =1000
1 + 1
η
Step 6: Substitute Qcoulomb =1000
1+ 1
η
into the efficiency formula to find the
efficiency of the engine.
Step 7: With the efficiency calculated, discuss whether the engine violates
the second law of thermodynamics by considering if the efficiency is less than 1.
Question 11
Question
A heat engine operates between two reservoirs at temperatures 700 K and 400
K. The engine absorbs 2400 J of heat from the high-temperature reservoir and
produces 1600 J of work. Calculate the efficiency of the engine. Is this engine
violating the second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency
Efficiency = Useful work output
Heat input
Step 2: Given that the engine absorbs 2400 J of heat from the high-temperature
reservoir and produces 1600 J of work, we have:
Useful work output = 1600 J
10
Heat input = 2400 J
Step 3: Substitute the values into the efficiency formula and calculate
Efficiency = 1600
2400 =2
3= 0.67 = 67%
Therefore, the efficiency of the engine is 67
Step 4: According to the second law of thermodynamics, no heat engine can
be 100
Thus, the engine is not violating the second law of thermodynamics.
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwhere
Th> Tc. If the efficiency of the engine is 40
Solution
Let’s denote the efficiency of the heat engine as η. The efficiency of a heat
engine is given by the formula:
η= 1 −Tc
Th
Given that the efficiency is 40
Step 1: Substitute the given efficiency value into the efficiency formula:
0.40 = 1 −Tc
Th
Step 2: Rearrange the equation to solve for Tc
Th:
Tc
Th
= 1 −0.40
Step 3: Simplify the expression:
Tc
Th
= 0.60
Step 4: Hence, the ratio Tc
Thin terms of This 0.60 .
Question 13
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
If the efficiency of the engine is 30
11
Solution
Let’s denote the efficiency of the engine as η.
Step 1: Recall the formula for the efficiency of a heat engine:
η= 1 −Tcold
Thot
Step 2: Given that η= 30% = 0.3, we can rewrite the efficiency formula
as:
0.3=1−Tcold
Thot
Step 3: Solve for the ratio Thot/Tcold by rearranging the equation:
Tcold
Thot
= 1 −0.3 = 0.7
Step 4: Finally, the ratio Thot/Tcold is given by:
Thot
Tcold
=1
0.7=10
7
Therefore, the ratio Thot/Tcold is 10
7.
Question 14
Question
A Carnot heat engine operates between two heat reservoirs at temperatures TH
and TC, with TH> TC. If the efficiency of the engine is 40
Solution
Step 1: Recall the formula for the efficiency of a Carnot heat engine:
Efficiency = 1 −TC
TH
Given that the efficiency is 40
0.40 = 1 −TC
TH
Step 2: Rearrange the equation to get TH
TCin terms of TC:
TC
TH
= 1 −0.40 = 0.60
TH
TC
=1
0.60 =5
3
Therefore, the ratio TH
TCin terms of TCis 5
3.
12
Question 15
Question
A heat engine operates between two reservoirs at temperatures TH= 500 K
and TC= 300 K. The engine absorbs 1500 J of heat from the hot reservoir and
rejects 900 J of heat to the cold reservoir during each cycle.
a) Calculate the efficiency of the engine.
b) Determine the maximum work that the engine can perform during each
cycle.
Solution
a) Given: Hot reservoir temperature, TH= 500 K Cold reservoir temperature,
TC= 300 K Heat absorbed, QH= 1500 J Heat rejected, QC= 900 J
Efficiency of a heat engine is given by the formula:
Efficiency, η= 1 −QC
QH
Step 1: Substitute the given values into the formula.
η= 1 −900
1500
Step 2: Calculate the efficiency.
η= 1 −0.6=0.4 = 40%
Therefore, the efficiency of the engine is 40
b) The maximum work that can be performed by the engine is given by the
formula:
Wmax =QH−QC
Step 1: Substitute the given values into the formula.
Wmax = 1500 −900
Step 2: Calculate the maximum work performed by the engine.
Wmax = 600 J
Therefore, the maximum work that the engine can perform during each cycle
is 600 J.
Question 16
Question
A heat engine operates between two reservoirs at 600K and 400K. The engine
absorbs 4000 J of heat from the high-temperature reservoir in each cycle. Cal-
culate the maximum efficiency of the engine.
13
Solution
To find the maximum efficiency of the heat engine, we can use the Carnot
efficiency formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the low-temperature reservoir and THis the
temperature of the high-temperature reservoir.
Step 1: Calculate the temperature of the low-temperature reservoir TC=
400 K.
Step 2: Calculate the temperature of the high-temperature reservoir TH=
600 K.
Step 3: Substitute the values of TCand THinto the Carnot efficiency
formula:
Efficiency = 1 −400 K
600 K
Step 4: Calculate the efficiency:
Efficiency = 1 −2
3=1
3≈33.3%
Therefore, the maximum efficiency of the heat engine is approximately 33.3
Question 17
Question
A heat engine operates between two reservoirs at temperatures T1= 500 K and
T2= 300 K. If the engine absorbs 5,000 J of heat from the reservoir at T1in
each cycle, calculate the maximum theoretical efficiency of the engine. Explain
why this efficiency cannot be achieved in practice.
Solution
Step 1: Determine the maximum efficiency of the heat engine using Carnot’s
theorem. Carnot’s theorem states that the maximum efficiency of a heat engine
operating between two reservoirs at temperatures T1and T2is given by:
η= 1 −T2
T1
Step 2: Substitute the values into the formula. Substitute T1= 500 K and
T2= 300 K into the formula:
η= 1 −300
500 = 1 −0.6=0.4
Therefore, the maximum theoretical efficiency of the heat engine is 40%.
14
Step 3: Explain why this efficiency cannot be achieved in practice. In prac-
tice, the efficiency of real heat engines is always less than the maximum theo-
retical efficiency given by Carnot’s theorem due to various irreversible processes
such as friction, heat loss to the surroundings, and internal energy losses in the
engine components. These losses decrease the efficiency of the engine, making
it impossible to achieve the ideal efficiency predicted by Carnot’s theorem.
Question 18
Question
A heat engine operates between a hot reservoir at a temperature of 600 K and a
cold reservoir at a temperature of 200 K. The engine takes in 6000 J of heat from
the hot reservoir and expels 2000 J to the cold reservoir per cycle. Calculate
the efficiency of the heat engine and discuss whether this violates the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Qc
Qh
where Qcis the heat expelled to the cold reservoir and Qhis the heat taken in
from the hot reservoir.
Step 2: Substitute the given values into the formula. Given: Hot reservoir
temperature (Th) = 600 K Cold reservoir temperature (Tc) = 200 K Heat taken
in from hot reservoir (Qh) = 6000 J Heat expelled to cold reservoir (Qc) = 2000
J
So, Qc= 2000 J and Qh= 6000 J.
Step 3: Calculate the efficiency.
Efficiency = 1 −2000
6000 = 1 −1
3=2
3
Hence, the efficiency of the heat engine is 2
3or 66.67
Step 4: Discuss the violation of the second law of thermodynamics. The
efficiency obtained (66.67
EfficiencyCarnot = 1 −Tc
Th
= 1 −200
600 =2
3
This means that the heat engine is operating at the theoretical maximum effi-
ciency possible for engines operating between these two temperatures. There-
fore, the second law of thermodynamics is not violated.
15
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine absorbs heat Qhfrom the hot reservoir and rejects heat Qcto
the cold reservoir in each cycle. If the efficiency of the engine is η, show that
the efficiency can be expressed as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input:
η=W
Qh
Step 2: By the first law of thermodynamics, the net work output can be
expressed as:
W=Qh−Qc
Step 3: Substituting W=Qh−Qcback into the efficiency formula, we get:
η=Qh−Qc
Qh
Step 4: Rearranging the equation, we have:
η= 1 −Qc
Qh
Step 5: By the second law of thermodynamics, for a reversible engine, the
ratio of heat absorbed to the temperature at which it is absorbed is the same
as the ratio of heat rejected to the temperature at which it is rejected, i.e.
Qh
Th=Qc
Tc.
Step 6: Using this relation, we can rewrite Qc
Qhin terms of temperatures:
Qc
Qh
=Tc
Th
Step 7: Substituting Qc
Qh=Tc
Thback into our efficiency formula, we get:
η= 1 −Tc
Th
Therefore, the efficiency of the engine can be expressed as η= 1 −Tc
Th.
16
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir.
Prove that the efficiency of the heat engine is bounded by the Carnot effi-
ciency, which is given by 1 −Tc
Th.
Solution
Let Wbe the work done by the engine, Qhbe the heat absorbed from the hot
reservoir, and Qcbe the heat exhausted to the cold reservoir.
Step 1: Write the first law of thermodynamics. According to the
first law of thermodynamics, the net work done by the engine is equal to the
difference between the heat absorbed and the heat exhausted:
W=Qh−Qc
Step 2: Write the efficiency of the engine. The efficiency of the engine
is defined as the ratio of the work done by the engine to the heat absorbed from
the hot reservoir:
Efficiency, η=W
Qh
Step 3: Express work done in terms of heat. The work done by the
engine can be expressed in terms of the heat absorbed and the heat exhausted:
W=Qh−Qc
Step 4: Substitute work done into the efficiency equation. Substi-
tute the expression for work done into the efficiency equation:
η=Qh−Qc
Qh
Step 5: Simplify the efficiency equation. Simplify the efficiency equa-
tion:
η= 1 −Qc
Qh
Step 6: Apply the second law of thermodynamics. According to the
second law of thermodynamics, the efficiency of any heat engine is bounded by
the Carnot efficiency:
η≤1−Tc
Th
17
Step 7: Substitute Qcand Qhwith temperature differentials. Since
Qh=ThShand Qc=TcSc(where Shand Scare the entropies of the hot and
cold reservoirs, respectively), substitute these values into the efficiency equation:
η≤1−Tc
Th
Therefore, the efficiency of the heat engine is bounded by the Carnot effi-
ciency.
Question 21
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). The engine takes in 4000 J of heat from the hot reservoir and exhausts
2500 J of heat to the cold reservoir during each cycle. Calculate the efficiency
of the engine. Is this engine violating the second law of thermodynamics?
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = 1 −Qc
Qh
where Qhis the heat absorbed from the hot reservoir and Qcis the heat ex-
hausted to the cold reservoir.
Step 2: Identify the values given in the question. Qh= 4000 J and Qc= 2500
J.
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −2500
4000 = 1 −5
8=3
8
Step 4: Thus, the efficiency of the engine is 3
8or 37.5
Step 5: The engine is not violating the second law of thermodynamics be-
cause the efficiency obtained is less than 100
Therefore, the efficiency of the engine is 3
8or 37.5
Question 22
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. If 1500 J of work is done by the
engine during each cycle, calculate the amount of heat absorbed from the hot
reservoir during each cycle. Also, determine the efficiency of the engine.
18
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir. Substitute Tc= 300 K and Th= 600 K into the equation:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Since efficiency is defined as the ratio of work done to heat absorbed
from the hot reservoir, we can find the amount of heat absorbed as follows: Let
Qhbe the heat absorbed from the hot reservoir. Then, efficiency = W
Qh, where
W= 1500 J.
Efficiency = W
Qh
⇒Qh=W
Efficiency
Substitute W= 1500 J and efficiency = 0.5 into the equation:
Qh=1500
0.5= 3000 J
Therefore, the amount of heat absorbed from the hot reservoir during each
cycle is 3000 J, and the efficiency of the Carnot engine is 50
Question 23
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir. The engine produces a net work output of W. Show
that the efficiency of the engine, η, is given by η= 1 −Tc
Thand discuss the
implications of the second law of thermodynamics on this expression.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency of a
heat engine is defined as the ratio of the work output to the heat input from
the hot reservoir. Mathematically, efficiency is given by:
η=W
Qh
Step 2: Use the first law of thermodynamics for the heat engine: According
to the first law of thermodynamics, the net work output of the engine is equal
19
to the difference between the heat absorbed from the hot reservoir and the heat
exhausted to the cold reservoir:
W=Qh−Qc
Step 3: Substitute the given expression for work into the efficiency equation:
Substitute W=Qh−Qcinto the expression for efficiency:
η=Qh−Qc
Qh
Step 4: Rearrange the expression: Divide both terms in the numerator by
Qh:
η=Qh
Qh
−Qc
Qh
Step 5: Consider the heat absorbed and heat exhausted in terms of temper-
atures: According to the second law of thermodynamics, the heat flow is related
to the temperatures of the reservoirs:
Qh
Th
−Qc
Tc
≤0
Step 6: Substitute the relationship between Qhand Qcusing the second law:
From Step 5, we have: Qh
Th
−Qc
Tc
≤0
Qh−Th
Tc
Qc≤0
Qh≤Th
Tc
Qc
Step 7: Substitute the inequality into the efficiency equation: Substitute
Qh≤Th
TcQcinto the efficiency equation:
η= 1 −Qc
Qh
≥1−Tc
Th
Step 8: Conclusion: Therefore, we have shown that the efficiency of the heat
engine is given by η= 1 −Tc
Th. This expression indicates that the efficiency of a
heat engine is limited by the temperature difference between the hot and cold
reservoirs, in accordance with the second law of thermodynamics.
Question 24
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhamount of heat from the high-
temperature reservoir and expels Qcamount of heat to the low-temperature
reservoir. Calculate the efficiency of the Carnot engine in terms of Qhand Th.
20
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: We can express the efficiency in terms of the heat absorbed from the
high-temperature reservoir Qhby using the relation Qh= (1 −Tc
Th)Qh.
Step 3: Simplify the expression:
Efficiency = 1 −Tc
Th
= 1 −Qc
Qh
Therefore, the efficiency of the Carnot engine in terms of Qhand This
1−Qc
Qh
.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc), where Qhis the heat absorbed from the hot reservoir and Wis the
work done by the engine. Prove that the efficiency of the engine is given by
η= 1 −Tc
Th
.
Solution
Step 1: Recall that the First Law of Thermodynamics states that the net work
done by a heat engine is equal to the heat absorbed from the hot reservoir minus
the heat rejected to the cold reservoir:
W=Qh−Qc.
Step 2: The efficiency of a heat engine is defined as the ratio of the work
done by the engine to the heat absorbed from the hot reservoir:
η=W
Qh
.
Step 3: Substitute W=Qh−Qcinto the efficiency formula:
η=Qh−Qc
Qh
.
Step 4: Rearrange terms and factor out Qhto simplify the expression:
η= 1 −Qc
Qh
.
21
Step 5: Recall the definition of efficiency in terms of temperatures:
η= 1 −Tc
Th
.
Step 6: Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 26
Question
A Carnot engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc. The engine absorbs 6000 J of heat from the hot
reservoir and discharges 4000 J to the cold reservoir in each cycle. Calculate
the efficiency of the engine in terms of Thand Tc.
Solution
Step 1: Recall that the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: First, let’s find the value of Tc
Th. Given that the engine absorbs 6000
J from the hot reservoir and discharges 4000 J to the cold reservoir in each cycle,
we have: Qh
Qc
=6000
4000 =Th
Tc
Step 3: Simplify the expression to find Tc
Th:
Tc
Th
=4000
6000 =2
3
Step 4: Substitute Tc
Thinto the efficiency formula:
Efficiency = 1 −2
3=1
3
Therefore, the efficiency of the engine in terms of Thand Tcis 1
3.
Question 27
Question
A Carnot engine operating between two reservoirs produces 500 J of work for
every 1000 J of heat absorbed. What is the efficiency of this engine? In addition,
if the colder reservoir is at 50◦C, what is the minimum temperature of the hotter
reservoir?
22
Solution
Step 1: Recall the formula for efficiency of a heat engine:
Efficiency = Useful work output
Heat input
Step 2: Given that the Carnot engine produces 500 J of work for every 1000
J of heat absorbed, we have:
Efficiency = 500 J
1000 J = 0.5 = 50%
Therefore, the efficiency of the engine is 50
Step 3: The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the colder reservoir and This the abso-
lute temperature of the hotter reservoir.
Step 4: We can rearrange the formula to solve for Th:
Tc
Th
= 1 −Efficiency
Th=Tc
1−Efficiency
Step 5: Convert the temperature in Celsius to Kelvin:
Tc= 50◦C + 273 = 323 K
Step 6: Substitute the values into the formula:
Th=323
1−0.5= 646 K
Therefore, the minimum temperature of the hotter reservoir is 646 K.
Question 28
Question
A heat engine operates between a hot reservoir at 700 K and a cold reservoir at
300 K. The engine absorbs 2000 J of heat from the hot reservoir in each cycle
and exhausts 1200 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine. Is this engine in violation of the second law of thermodynamics?
23
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Substitute the values for heat input and heat output into the equa-
tion above:
Efficiency = 1 −1200
2000
Step 3: Perform the calculation to find the efficiency of the engine:
Efficiency = 1 −1200
2000 = 1 −0.6=0.4
Step 4: Therefore, the efficiency of the engine is 40
Step 5: The efficiency of the engine is less than 100
Step 6: The second law of thermodynamics states that heat always flows
from a hotter body to a colder body, and no heat engine can be 100
Carnot efficiency = 1 −Tcold
Thot
Step 7: For this case, the Carnot efficiency would be:
Carnot efficiency = 1 −300
700 = 1 −0.4286 = 0.5714 = 57.14%
Step 8: Comparing the efficiency of the engine (40
Question 29
Question
A heat engine operates between two reservoirs at temperatures Thand Tc. The
engine absorbs Qhof heat from the high-temperature reservoir and does Wof
work. The efficiency of the engine is defined as η=W
Qh. Show that the maximum
possible efficiency of any heat engine operating between two reservoirs is given
by ηmax = 1−Tc
Th, where This the absolute temperature of the high-temperature
reservoir and Tcis the absolute temperature of the low-temperature reservoir.
Solution
Step 1: Let’s start with the First Law of Thermodynamics:
∆U=Q−W
where ∆Uis the change in internal energy, Qis the heat absorbed, and Wis
the work done by the engine.
24
Step 2: For a heat engine, the net work done by the engine is the difference
between the heat absorbed from the high-temperature reservoir and the heat
rejected to the low-temperature reservoir:
W=Qh−Qc
where Qhis the heat absorbed from the high-temperature reservoir and Qcis
the heat rejected to the low-temperature reservoir.
Step 3: The efficiency of the heat engine is defined as:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: By the Second Law of Thermodynamics, the heat engine cannot
operate at 100
Step 5: From the definition of temperature in the Carnot cycle:
Qc
Tc
=Qh
Th
Dividing through by Qh:Qc
Qh
=Tc
Th
Step 6: Substituting into the efficiency expression:
η≤1−Tc
Th
Hence, the maximum possible efficiency of any heat engine operating between
two reservoirs is ηmax = 1 −Tc
Th.
Question 30
Question
A heat 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 3500 J of heat
from the hot reservoir and delivers 2000 J of work output during each cycle.
Calculate: a) The efficiency of the engine. b) The amount of heat rejected to
the cold reservoir during each cycle. c) Discuss whether the engine adheres to
the second law of thermodynamics.
Solution
a) The efficiency of the engine is given by the formula:
Efficiency = Work Output
Heat Input
25
Step 1: Calculate the efficiency of the engine. The work output is 2000 J
and the heat input is 3500 J.
Efficiency = 2000 J
3500 J = 0.5714 or 57.14%
b) The amount of heat rejected to the cold reservoir during each cycle is
given by:
Heat Rejected = Heat Input −Work Output
Step 2: Calculate the amount of heat rejected to the cold reservoir. The
heat input is 3500 J and the work output is 2000 J.
Heat Rejected = 3500 J −2000 J = 1500 J
c) A heat engine operates by converting heat into work, but it cannot convert
all the heat it absorbs into work. According to the second law of thermodynam-
ics, the efficiency of a heat engine is never 100Therefore, the engine adheres to
the second law of thermodynamics as it cannot convert all the input heat into
work and some heat is always rejected to the cold reservoir.
Question 31
Question
A heat engine operates between a high temperature reservoir at 500 K and a low
temperature reservoir at 300 K. The engine absorbs 4000 J of heat from the high
temperature reservoir in each cycle and produces 2000 J of work. Calculate the
efficiency of the engine and discuss whether the operation of this engine violates
the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency (%) = Work output
Heat input ×100%
Step 2: Substitute the given values into the formula:
Efficiency (%) = 2000
4000 ×100% = 50%
Therefore, the efficiency of the heat engine is 50%.
Step 3: According to the second law of thermodynamics, no heat engine
operating in a cycle can absorb heat from a single reservoir and convert it
completely into work without producing any other effect.
Step 4: In this case, the engine violates the second law of thermodynamics
because it is operating at 100
Therefore, the operation of this engine violates the second law of thermody-
namics.
26
Question 32
Question
A heat engine operates between a source temperature of 600 K and a sink
temperature of 300 K. The engine absorbs 2000 J of heat from the source in
each cycle. Calculate the efficiency of the engine. Is this engine violating the
second law of thermodynamics?
Solution
Step 1: Calculate the work done by the engine during each cycle using the
formula for efficiency.
Efficiency = 1 −Heat rejected
Heat absorbed
Efficiency = 1 −Tsink
Tsource
Step 2: Substitute the source and sink temperatures into the formula.
Efficiency = 1 −300
600
Efficiency = 1 −0.5
Efficiency = 0.5
Step 3: Determine if the engine is violating the second law of thermodynam-
ics. According to the second law of thermodynamics, the efficiency of a heat
engine cannot be 100
Therefore, the efficiency of the engine is 50
Question 33
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
releases heat Qcto the low-temperature reservoir. If the efficiency of the engine
is η, show that the efficiency of the engine can be expressed in terms of the
temperatures of the reservoirs as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
η= 1 −|Qc|
|Qh|.
Step 2: We know that the work output of the engine is W=|Qh|−|Qc|.
27
Step 3: Using the first law of thermodynamics, we have W=|Qh|−|Qc|=
|Qh|(1 −|Qc|
|Qh|).
Step 4: Now, we can rewrite the efficiency in terms of the temperatures of
the reservoirs. The heat absorbed Qhis related to Thand Tcby the equation
Qh=ThS, where Sis the entropy change of the hot reservoir.
Step 5: Similarly, the heat rejected Qcis related to Thand Tcby the equation
Qc=TcS, where Sis the entropy change of the cold reservoir.
Step 6: Plugging these expressions back into the efficiency formula, we get
η= 1 −TcS
ThS= 1 −Tc
Th.
Step 7: Therefore, the efficiency of the heat engine can be expressed in terms
of the temperatures of the reservoirs as η= 1 −Tc
Th.
Question 34
Question
A heat engine operates between a high-temperature reservoir at 600 K and a
low-temperature reservoir at 300 K. The engine absorbs 600 J of heat from the
high-temperature reservoir in each cycle.
1. Calculate the maximum efficiency of this engine.
2. If the engine releases 300 J of heat to the low-temperature reservoir in
each cycle, calculate the work done by the engine in each cycle.
Solution
1. To find the maximum efficiency of the heat engine, we can use the formula
for the efficiency of a Carnot engine:
η= 1 −Tlow
Thigh
where Tlow and Thigh are the absolute temperatures of the low-temperature and
high-temperature reservoirs, respectively.
Given Thigh = 600 K and Tlow = 300 K, we can plug in these values to find
the efficiency:
η= 1 −300
600 = 1 −0.5 = 0.5 = 50%
Therefore, the maximum efficiency of this engine is 50%.
2. The work done by the engine in each cycle can be calculated using the
formula:
Work done = Heat absorbed −Heat released
Given that the engine absorbs 600 J from the high-temperature reservoir and
releases 300 J to the low-temperature reservoir, we can substitute these values
into the formula:
Work done = 600 J −300 J = 300 J
28
Therefore, the work done by the engine in each cycle is 300 J.
Question 35
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine takes in heat energy QHfrom the hot reservoir
and produces work output W. If the efficiency of the engine is ηand the engine
follows the Carnot cycle, prove that the efficiency of the engine can be expressed
as η= 1 −TC
TH.
Solution
Step 1: Recall the definition of efficiency for a heat engine:
η=Useful work output
Heat energy input
Step 2: For the Carnot engine, the efficiency is given by:
η= 1 −TC
TH
Step 3: The efficiency of a Carnot engine is based on the fact that the work
done by the engine is the difference between the heat energy input from the hot
reservoir and the heat energy rejected to the cold reservoir. Mathematically,
this can be expressed as:
W=QH−QC
Step 4: The efficiency can then be expressed as:
η=W
QH
Step 5: Substituting QC=QH−Winto the efficiency equation:
η=QH−(QH−W)
QH
Step 6: Simplifying the equation, we get:
η=W
QH
=W
W+QC
Step 7: Using the equation QC=TC(S2−S1) and the definition of entropy,
QC=TC∆S, where ∆Sis the change in entropy, we can write the equation in
terms of temperature:
QC=TC∆S=TC(S2−S1)
29
Step 8: Substitute QC=TC(S2−S1) into the efficiency equation:
η=W
W+TC(S2−S1)
Step 9: Since the engine follows the Carnot cycle, the entropy change is zero,
∆S= 0. This means that S2=S1and the equation simplifies to:
η=W
W
Step 10: Finally, the efficiency of a Carnot engine is given by:
η= 1 −TC
TH
Therefore, the efficiency of the engine can be expressed as η= 1 −TC
TH.
30
