PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 10
Liberty University
Question 1
Question
A heat engine operates between two reservoirs at temperatures Thot = 500 K
and Tcold = 300 K. The engine absorbs 2000 J of heat from the hot reservoir
and expels 1200 J of heat to the cold reservoir in each cycle.
a) Determine the efficiency of the engine.
b) Based on the second law of thermodynamics, explain whether such an
engine could exist in practice.
Solution
a) To determine the efficiency of the engine, we can use the formula for efficiency
of a heat engine:
Efficiency = 1 −Qcold
Qhot
where Qhot is the heat absorbed from the hot reservoir and Qcold is the heat
expelled to the cold reservoir.
Step 1: Calculate the efficiency of the engine.
Given: Qhot = 2000 J Qcold = 1200 J
Substitute the values into the formula:
Efficiency = 1 −1200
2000 = 1 −0.6=0.4
Therefore, the efficiency of the engine is 40
b) According to the second law of thermodynamics, no engine can have an
efficiency of 100
Question 2
Question
A Carnot engine operates between two heat reservoirs, one at 500K and the
other at 300K. If the engine absorbs 1500 J of heat from the hot reservoir in
each cycle, what is the maximum amount of work that can be done by the engine
in each cycle? What is the efficiency of this engine?
Solution
Step 1: Calculate the maximum work done by the engine in each cycle.
Wmax =Qhot 1−Tcold
Thot
= 1500 J 1−300
500
= 1500 J (1 −0.6)
= 1500 J ×0.4
= 600 J
Step 2: Calculate the efficiency of the engine.
Efficiency (%) = Wmax
Qhot
×100%
=600
1500 ×100%
= 40%
Therefore, the maximum amount of work that can be done by the engine in
each cycle is 600 J and the efficiency of the engine is 40
Question 3
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 600 J of heat from the hot reservoir in each cycle.
Calculate the efficiency of the Carnot engine and the heat rejected to the cold
reservoir in each cycle.
Solution
To calculate the efficiency of the Carnot engine, we can use the formula for
Carnot efficiency:
2
Efficiency = 1 −Tcold
Thot
where Thot is the temperature of the hot reservoir in Kelvin, and Tcold is the
temperature of the cold reservoir in Kelvin.
Step 1: Calculate the efficiency of the Carnot engine. Given: Thot = 600 K,
Tcold = 300 K
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
So, the efficiency of the Carnot engine is 50
Step 2: Calculate the heat rejected to the cold reservoir. The heat rejected
to the cold reservoir in each cycle is equal to the heat absorbed from the hot
reservoir. This is because in a Carnot engine, all of the heat absorbed from the
hot reservoir is converted into work, so the rest of the energy must be rejected
to the cold reservoir.
Given: Heat absorbed = 600 J
Therefore, the heat rejected to the cold reservoir in each cycle is 600 J.
Therefore, the efficiency of the Carnot engine is 50
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
with Th= 600 K and Tc= 300 K. The engine receives 2,000 J of heat from the
high-temperature reservoir in each cycle. Calculate the efficiency of the engine
and determine the amount of heat rejected to the low-temperature reservoir in
each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values into the formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot engine is 50
Step 3: Calculate the amount of heat rejected to the low-temperature reser-
voir in each cycle using the formula:
Heat rejected = Qh−Qc
3
Step 4: Substitute the given values into the formula:
Heat rejected = 2,000 J −Tc
Th
×2,000 J
Heat rejected = 2,000 J −300
600 ×2,000 J = 2,000 J −1,000 J = 1,000 J
Therefore, the amount of heat rejected to the low-temperature reservoir in
each cycle is 1,000 J.
Question 5
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir and
produces 2400 J of work. Calculate the efficiency of the engine. Additionally,
explain how this efficiency compares to the maximum possible efficiency for a
heat engine operating between the same two reservoirs.
Solution
Step 1: We can calculate the efficiency of the Carnot heat engine using the
formula for efficiency:
Efficiency = Work output
Heat input =2400 J
4000 J = 0.6 = 60%.
Step 2: The maximum possible efficiency of a heat engine operating between
two reservoirs at temperatures Thand Tcis given by the Carnot efficiency for-
mula:
EfficiencyCarnot = 1 −Tc
Th
.
Step 3: Substituting the given temperatures Thand Tcinto the Carnot
efficiency formula:
EfficiencyCarnot = 1 −Tc
Th
= 1 −2400
4000 = 1 −0.6=0.4 = 40%.
Step 4: Comparing the calculated efficiency of the Carnot engine (60
Question 6
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine absorbs 600 J of heat from the hot reservoir
during each cycle and expels 400 J of heat to the cold reservoir, calculate the
efficiency of the engine in terms of Thand Tc.
4
Solution
Let’s denote the heat absorbed from the hot reservoir as Qhand the heat ex-
pelled to the cold reservoir as Qc.
Step 1: Apply the formula for the efficiency of a Carnot engine: The effi-
ciency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Determine the heat absorbed and heat expelled: Given that the
engine absorbs 600 J of heat from the hot reservoir (Qh= 600 J) and expels
400 J of heat to the cold reservoir (Qc= 400 J).
Step 3: Substitute the values into the formula for efficiency:
Efficiency = 1 −Qc
Qh
Substitute Qh= 600 J and Qc= 400 J into the formula:
Efficiency = 1 −400
600
Step 4: Simplify the expression:
Efficiency = 1 −2
3
Step 5: Writing the efficiency in terms of Thand Tc: Given that Th> Tc,
we know that:
Th=Tc+ ∆T
where ∆Tis the temperature difference between the two reservoirs.
Step 6: Substitute for Thand Tcin terms of ∆T:
Efficiency = 1 −Tc
Tc+ ∆T
Step 7: Further simplification:
Efficiency = 1 −Tc
Tc(1 + ∆T
Tc)
Efficiency = 1 −1
1 + ∆T
Tc
Therefore, the efficiency of the Carnot engine in terms of Thand Tcis
Efficiency = 1 −1
1+ ∆T
Tc
.
5
Question 7
Question
A Carnot heat engine operates between a high-temperature reservoir at Th=
600 K and a low-temperature reservoir at Tc= 300 K. If this engine produces
500 J of work per cycle, calculate the following:
1. The heat absorbed from the high-temperature reservoir per cycle.
2. The efficiency of the engine.
3. The heat rejected to the low-temperature reservoir per cycle.
Solution
Let’s denote the heat absorbed from the high-temperature reservoir per cycle
as Qh, the heat rejected to the low-temperature reservoir per cycle as Qc, and
the work done by the engine per cycle as W.
Step 1: Calculate the heat absorbed from the high-temperature reservoir
per cycle. Using the first law of thermodynamics for a cyclic process:
W=Qh−Qc
Given that W= 500 J, and Th= 600 K, and Tc= 300 K, we can calculate Qh:
500 J = Qh−Qc
Qh=Qc+ 500 J
Now, we know that for a Carnot engine:
Qh
Th
=Qc
Tc
Substitute Qh=Qc+ 500 and solve for Qc:
Qc+ 500
600 =Qc
300
300(Qc+ 500) = 600Qc
300Qc+ 150000 = 600Qc
300Qc= 150000
Qc= 500 J
Therefore, the heat absorbed from the high-temperature reservoir per cycle is
Qh= 1000 J.
Step 2: Calculate the efficiency of the engine. The efficiency of a heat
engine is given by:
Efficiency = W
Qh
6
Substitute W= 500 J and Qh= 1000 J:
Efficiency = 500
1000 = 0.5
So, the efficiency of the engine is 0.5 or 50
Step 3: Calculate the heat rejected to the low-temperature reservoir per
cycle. We already know that Qc= 500 J from earlier calculations.
Therefore, the heat rejected to the low-temperature reservoir per cycle is 500
J.
Question 8
Question
A heat engine operates between two reservoirs at temperatures T1and T2(T1>
T2). The engine absorbs Q1amount of heat from the reservoir at temperature
T1and exhausts Q2amount of heat to the reservoir at temperature T2. The
efficiency of the engine is given by η= 1 −Q2
Q1
. Prove that the efficiency of the
heat engine is bounded by η≤1−T2
T1
.
Solution
Step 1: Let’s start by expressing the heat transfer Q2in terms of Q1using the
efficiency equation η= 1 −Q2
Q1
.
Substitute Q2=Q1(1 −η) into η= 1 −Q2
Q1
to get η= 1 −Q1(1 −η)
Q1
⇒η= 1 −(1 −η) = η
Step 2: Now, we’ll express the efficiency ηin terms of T1,T2,Q1, and Q2.
Using η= 1 −Q2
Q1
, we have η= 1 −Q1(1 −η)
Q1
⇒η= 1 −(1 −η) = η
Step 3: Next, we’ll use the definition of efficiency to express ηin terms of
temperatures T1and T2.
Using Q1=Q2+W, where Wis the work done by the engine, we get η= 1−Q2
Q2+W
Step 4: By applying the second law of thermodynamics, we know that W≤
Q1(T1−T2).
Substitute W≤Q1(T1−T2) into η= 1 −Q2
Q2+W
7
⇒η= 1 −Q2
Q2+Q1(T1−T2)
Step 5: Simplifying the expression gives us
η= 1 −Q2
Q1+Q1T1−T2
T1= 1 −Q2
Q11 + T1−T2
T1
Step 6: After simplifying further, we arrive at
η= 1 −1
1 + T1−T2
T1
= 1 −T1
T2
Therefore, the efficiency of the heat engine is bounded by η≤1−T2
T1
.
Question 9
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. If the efficiency of the
engine is η, show that the efficiency of the engine can be expressed in terms of
the temperatures as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as η= 1 −
Heat output
Heat input .
Step 2: The heat input for the engine is Qh, and the heat output is Qc.
Step 3: From the second law of thermodynamics, we have Qh=Qc+W,
where Wis the work done by the engine.
Step 4: The work done by the engine can be expressed as W=Qh−Qc.
Step 5: Substituting this expression for Winto the efficiency formula, we
get:
η= 1 −Qc
Qh
Step 6: Substituting Qh=Qc+W=Qc+Qh−Qc=Qhinto the efficiency
formula gives:
η= 1 −Qc
Qh
= 1 −Tc
Th
Step 7: Therefore, the efficiency of the engine can be expressed in terms of
the temperatures as η= 1 −Tc
Th.
8
Question 10
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 consumes 5000 J of heat
from the hot reservoir and produces 2500 J of work. Calculate the efficiency of
the heat engine and determine if it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Substitute the values into the formula:
Efficiency = 2500
5000 = 0.5 = 50%
Step 3: Check if the efficiency violates the second law of thermodynamics.
The maximum efficiency of a heat engine operating between two reservoirs at
different temperatures is given by Carnot’s efficiency formula:
Carnot Efficiency = 1 −Tcold
Thot
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −300
600 = 0.5 = 50%
Step 5: Since the efficiency of the actual engine is equal to the Carnot
efficiency, it does not violate the second law of thermodynamics. In real engines,
the efficiency is always less than the Carnot efficiency due to factors such as
friction, heat loss, etc.
Question 11
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 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. Calculate the efficiency of the
engine and discuss whether it violates the second law of thermodynamics.
9
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1−Tc
Th×100%
where This the temperature of the hot reservoir and Tcis the temperature of
the cold reservoir.
Efficiency = 1−300
600×100% = 50%
Step 2: Discuss the efficiency of the engine and the second law of thermo-
dynamics. The efficiency of the heat engine is 50
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. It absorbs Qhof heat from the hot reservoir and produces Wof work
output. If the efficiency of the heat engine is η, show that the entropy change
of the universe ∆Suniv in one complete cycle is given by ∆Suniv =Qh
Th−Qc
Tc.
Solution
Step 1: Recall the definition of efficiency for a heat engine: Let Qcbe the heat
rejected to the cold reservoir. The efficiency ηof a heat engine is given by:
η=W
Qh
= 1 −Qc
Qh
Step 2: Express the heat rejected Qcin terms of Qh: From the efficiency
equation, we have:
Qc=Qh−W
Step 3: Express Win terms of Qhusing the efficiency equation: Substitute
the expression for Qcin terms of Qhinto the efficiency equation to get:
η= 1 −Qh−W
Qh
Step 4: Simplify the expression for η: This simplifies to:
η= 1 −1 + W
Qh
η=W
Qh
10
Step 5: Express Qcin terms of Thand Tc: From the second law of thermo-
dynamics, we have:
Qh
Th
+Qc
Tc
= 0
Step 6: Substituting Qc=Qh−Winto the second law equation: We obtain:
Qh
Th
+Qh−W
Tc
= 0
Step 7: Rearrange the equation: Solving for Wgives:
W=Qh1−Tc
Th
Step 8: Calculate the entropy change of the universe ∆Suniv: The change in
entropy of the universe is given by:
∆Suniv =Qh
Th
+−Qh+W
Tc
Step 9: Substitute the expression for Winto the equation for ∆Suniv:
∆Suniv =Qh
Th
+
−Qh+Qh1−Tc
Th
Tc
Step 10: Simplify the expression for ∆Suniv: Converting the terms gives:
∆Suniv =Qh
Th
−Qc
Tc
Therefore, the entropy change of the universe in one complete cycle is given
by ∆Suniv =Qh
Th−Qc
Tc.
Question 13
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, with Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir and
rejects 2500 J of heat to the cold reservoir during each cycle. Determine:
1. The efficiency of the engine.
2. The work done by the engine during each cycle.
3. If the engine operates as a refrigerator instead of an engine, calculate the
amount of work required to run the refrigerator.
11
Solution
1. To calculate the efficiency of the engine, we use the formula:
Efficiency (%) = Useful work output
Energy input ×100%
Useful work output = Heat absorbed −Heat rejected
=Qh−Qc
= 4000 J −2500 J
= 1500 J
The efficiency of the engine is given by:
Efficiency = Useful work output
Energy input ×100% = 1500 J
4000 J ×100% = 37.5%
2. The work done by the engine during each cycle is equal to the useful work
output:
Work done = 1500 J
3. If the engine operates as a refrigerator, the amount of work required to
run the refrigerator is equal to the work done on the system:
Work required = 1500 J
Question 14
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
200 K. If the engine absorbs 2000 J of heat from the hot reservoir in each cycle,
find:
1. The efficiency of the heat engine.
2. The minimum amount of heat that must be exhausted to the cold reservoir
in each cycle.
Solution
Given:
Temperature of the hot reservoir (Th) = 500 K
Temperature of the cold reservoir (Tc) = 200 K
Heat absorbed from the hot reservoir (Qin) = 2000 J
12
1. Efficiency of the heat engine
The efficiency (η) of the heat engine is given by the formula:
η= 1 −Tc
Th
Step 1: Calculate the efficiency using the given temperatures:
η= 1 −200
500 = 1 −2
5=3
5
Therefore, the efficiency of the heat engine is 3
5or 60
2. Minimum amount of heat exhausted to the cold reservoir
The amount of heat exhausted to the cold reservoir in each cycle can be
calculated using the efficiency of the engine:
Qout =η·Qin
Step 1: Plug in the values to find Qout:
Qout =3
5×2000 = 1200 J
Therefore, the minimum amount of heat that must be exhausted to the
cold reservoir in each cycle is 1200 J.
Question 15
Question
A Carnot cycle operates between two reservoirs at temperatures Thand Tcwhere
Th> Tc. The heat input from the hot reservoir is Qhand the work output of
the engine is Wout. If the efficiency of the Carnot engine is η, show that the
heat rejected to the cold reservoir Qcsatisfies the inequality:
Qc>Tc
Th
Qh
Solution
Step 1: Recall the expression for the efficiency of a Carnot engine:
η= 1 −Tc
Th
Step 2: Rearrange the equation to solve for Tc:
Tc
Th
= 1 −η
13
Step 3: Multiply both sides by Qhto find the expression for Qc:
Qc= (1 −η)Qh
Step 4: Substitute the expression for ηback into the equation:
Qc=1−1−Tc
ThQh
Step 5: Simplify to get the final inequality:
Qc=Tc
Th
Qh
Step 6: Since Th> Tc, we have:
Qc>Tc
Th
Qh
Therefore, the heat rejected to the cold reservoir Qcsatisfies the inequality
Qc>Tc
ThQh.
Question 16
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). If the engine absorbs 5000 J of heat from the hot reservoir and 3000 J of
heat is rejected to the cold reservoir in each cycle, calculate: (i) The efficiency
of the engine. (ii) The maximum amount of work the engine can produce in
each cycle. (iii) If the engine is reversible or irreversible based on the calculated
efficiency.
Given: Th= 500 K, Tc= 300 K.
Solution
(i) The efficiency of the engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 1: Substitute the given values into the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
So, the efficiency of the engine is 40
(ii) The maximum amount of work the engine can produce in each cycle is
given by the formula:
Work = Efficiency ×Heat absorbed from hot reservoir
14
Step 2: Substitute the given values into the formula:
Work = 0.4×5000 J = 2000 J
Therefore, the maximum amount of work the engine can produce in each
cycle is 2000 J.
(iii) If an engine is reversible, its efficiency is given by the Carnot efficiency:
EfficiencyCarnot = 1 −Tc
Th
If the efficiency of the engine matches the Carnot efficiency, it is reversible.
Since the efficiency we calculated in part (i) was the same as the Carnot effi-
ciency, the engine is reversible.
So, the engine in this scenario is reversible.
Question 17
Question
A Carnot 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. Calculate the work done by the engine in each cycle and determine
its efficiency.
Solution
Step 1: Calculate the work done by the engine in each cycle using the equation
for efficiency of a Carnot engine:
Efficiency = 1 −Tcold
Thot
where Thot = 600 K and Tcold = 300 K.
Step 2: Substitute the values into the efficiency equation:
Efficiency = 1 −300
600 = 1 −1
2=1
2
Step 3: The efficiency of a Carnot engine is the ratio of the work output to
the heat input, so:
Efficiency = W
Qin
Step 4: Rearrange the equation to solve for work done by the engine:
W= Efficiency ×Qin =1
2×500 J = 250 J
Therefore, the work done by the engine in each cycle is 250 J and the effi-
ciency of the engine is 50
15
Question 18
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. The engine takes in 800 J of heat
from the hot reservoir and performs 500 J of work during each cycle. Calculate
the efficiency of this engine.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
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 the given temperatures from Celsius to Kelvin: Hot reservoir
temperature, TH= 500 K Cold reservoir temperature, TC= 300 K
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −300
500
Step 4: Calculate the efficiency:
Efficiency = 1 −3
5=2
5
Step 5: Express the efficiency as a percentage:
Efficiency = 2
5×100% = 40%
Therefore, the efficiency of the heat engine is 40
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the hot reservoir and delivers W
of work. If the engine’s efficiency is η, prove that the efficiency of the engine is
given by η= 1 −Tc
Th.
16
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is given by:
η=Useful work output
Heat input =W
Qh
Step 2: Express the work done by the engine in terms of heat taken in and
rejected: According to the first law of thermodynamics, the work done by the
engine is the difference between the heat absorbed and the heat rejected:
W=Qh−Qc
Step 3: Express the rejected heat Qcin terms of the temperatures Thand
Tc: From Carnot’s theorem, we know that for a reversible process, the ratio of
heat transferred to a reservoir and the reservoir’s temperature remains constant.
Therefore, for the engine rejecting heat to the cold reservoir,
Qc
Tc
=Qh
Th
Step 4: Solve for Qcin terms of Qh,Tc, and Th:
Qc=Tc
Th
Qh
Step 5: Substitute the expression for Qcback into the efficiency equation:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
= 1 −Tc
Th
Step 6: Therefore, the efficiency of the engine is given by η= 1 −Tc
Th.
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwith
Th> Tc. The engine has an efficiency of 40
1. The heat rejected to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Solution
Given: Efficiency of the heat engine, η= 40% = 0.40, Heat absorbed from the
hot reservoir, Qh= 5000 J
We are asked to find:
17
1. The heat rejected to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Step 1: Find the heat rejected to the cold reservoir in each cycle.
The efficiency of a heat engine is given by:
η= 1 −Qc
Qh
Given that the efficiency, η= 0.40, and that Qh= 5000 J, we can rearrange the
equation to solve for Qc:
0.40 = 1 −Qc
5000
Qc
5000 = 0.60
Qc= 0.60 ×5000
Qc= 3000 J
Therefore, the heat rejected to the cold reservoir in each cycle is 3000 J.
Step 2: Find the work done by the engine in each cycle. The
efficiency of a heat engine can also be expressed as:
η=W
Qh
We can rearrange the equation to solve for W:
0.40 = W
5000
W= 0.40 ×5000
W= 2000 J
Therefore, the work done by the engine in each cycle is 2000 J.
Question 21
Question
A heat engine operates between two reservoirs at temperatures T1= 700 K and
T2= 300 K. If the engine absorbs 800 J of heat from the high-temperature
reservoir in each cycle, what is the maximum amount of work the engine can
perform per cycle? Calculate the efficiency of this engine.
18
Solution
Step 1: Find the maximum efficiency of the heat engine using Carnot’s theorem.
Efficiency = 1 −T2
T1
where T2is the temperature of the low-temperature reservoir and T1is the
temperature of the high-temperature reservoir.
Step 2: Substitute the given values into the efficiency formula.
Efficiency = 1 −300
700 = 1 −3
7=4
7
Step 3: Calculate the maximum work output of the engine per cycle using
the formula for efficiency.
Work = Efficiency ×Heat input
Step 4: Substitute the given heat input of 800 J into the work formula.
Work = 4
7×800 = 3200
7≈457.14 J
Step 5: Therefore, the maximum amount of work the engine can perform
per cycle is approximately 457.14 J, and the efficiency of the engine is 4
7≈0.5714
or 57.14
Question 22
Question
A heat engine absorbs 1400 J of heat from a high-temperature reservoir at 500
K and exhausts 800 J of heat to a low-temperature reservoir. Calculate the
efficiency of the engine and determine whether it violates the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −Qout
Qin
where Qin is the heat absorbed from the high-temperature reservoir and Qout
is the heat exhausted to the low-temperature reservoir.
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 1 −800
1400 = 1 −4
7=3
7≈0.429
19
Step 3: Determine whether the engine violates the second law of thermody-
namics. The second law of thermodynamics states that no heat engine can have
an efficiency greater than the Carnot efficiency, given by:
Carnot efficiency = 1 −Tlow
Thigh
where Tlow is the temperature of the low-temperature reservoir and Thigh is the
temperature of the high-temperature reservoir.
Step 4: Substitute the temperatures into the Carnot efficiency formula:
Carnot efficiency = 1 −300
500 = 1 −0.6 = 0.4
Step 5: Compare the efficiency of the engine (0.429) with the Carnot ef-
ficiency (0.4). Since the efficiency of the engine is greater than the Carnot
efficiency, it violates the second law of thermodynamics.
Question 23
Question
A Carnot engine operates between a high temperature reservoir at 800 K and
a low temperature reservoir at 300 K. If the engine absorbs 10,000 J of heat
energy from the high temperature reservoir in each cycle, what is the maximum
amount of work that can be done by the engine in each cycle? What is the
efficiency of the engine?
Solution
Step 1: Calculate the maximum efficiency of the engine. The efficiency of a
Carnot engine is given by the formula:
Efficiency = 1 −TC
TH
where THis the temperature of the high temperature reservoir and TCis the
temperature of the low temperature reservoir. Given TH= 800 K and TC= 300
K, we can substitute these values into the formula to find the efficiency.
Efficiency = 1 −300
800 = 1 −3
8=5
8
Step 2: Calculate the maximum work done by the engine. The maximum
work done by a Carnot engine is given by the formula:
Wmax = Efficiency ×Qin
20
where Qin is the heat energy absorbed from the high temperature reservoir in
each cycle. Given Qin = 10,000 J and efficiency = 5
8, we can substitute these
values into the formula to find the maximum work done by the engine.
Wmax =5
8×10,000 = 6,250 J
Therefore, the maximum amount of work that can be done by the engine in
each cycle is 6,250 J and the efficiency of the engine is 5
8or 62.5%.
Question 24
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and
exhausts heat Qcto the cold reservoir. Determine the efficiency of the engine
in terms of Thand Tc, and discuss how this efficiency is related to the second
law of thermodynamics.
Solution
Step 1: Let’s first calculate the efficiency of the Carnot engine. The efficiency
of a Carnot engine operating between two heat reservoirs at temperatures Th
and Tcis given by:
η= 1 −Tc
Th
Step 2: Now, let’s discuss why this efficiency is related to the second law of
thermodynamics. The efficiency of a heat engine is limited by the second law of
thermodynamics, which states that no heat engine can have an efficiency greater
than that of a Carnot engine operating between the same two temperatures.
This means that no engine can have an efficiency greater than 1 −Tc
Th. The
efficiency of a Carnot engine represents the maximum possible efficiency for any
heat engine operating between the given temperatures Thand Tc. This is a
direct consequence of the second law of thermodynamics, which establishes the
limitations on the conversion of heat into work.
Question 25
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc. If the engine absorbs 5000 J of heat from the high-temperature reservoir
and the engine’s efficiency is 40
21
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = Work output
Heat input
0.40 = Work output
5000 J
Work output = 0.40 ×5000 J
Work output = 2000 J
Step 2: Calculate the work done by the engine. For a Carnot engine,
efficiency e=Th−Tc
Thwhere This the temperature of the high-temperature
reservoir and Tcis the temperature of the low-temperature reservoir. So, we
have 0.40 = Th−Tc
Th. Given that the engine absorbs 5000 J from the high-
temperature reservoir, we have Qh= 5000 J. Therefore, work done by the
engine is W=Qh×e= 5000 ×0.40 = 2000 J.
Step 3: Calculate the heat rejected to the low-temperature reservoir. Since
the total heat absorbed from the high-temperature reservoir is used to do work
and reject heat, we have:
Qh=W+Qc
5000 = 2000 + Qc
Qc= 3000 J
Step 4: Discussion on the second law of thermodynamics. The second law
of thermodynamics states that heat flows spontaneously from a hotter reservoir
to a colder reservoir, and that no heat engine can be 100
Question 26
Question
A heat engine operates between a hot reservoir at 600◦Cand a cold reservoir
at 27◦C. The engine has an efficiency of 40%. Calculate the maximum possible
efficiency of the engine if the cold reservoir temperature is increased to 45◦C.
Solution
Let’s denote the temperatures of the hot reservoir, cold reservoir, and increased
cold reservoir as TH= 600◦C,TC= 27◦C, and T′
C= 45◦C, respectively.
We know that the efficiency (η) of a heat engine operating between a hot
reservoir at temperature THand a cold reservoir at temperature TCis given by
the formula:
η= 1 −TC
TH
22
Step 1: Calculate the initial efficiency of the engine. Given TH= 600◦C
and TC= 27◦C, we can calculate the initial efficiency:
η= 1 −27
600 = 1 −27
600 = 1 −0.045 = 0.955 = 95.5%
Step 2: Calculate the new efficiency of the engine. Given the increased cold
reservoir temperature T′
C= 45◦C, we can calculate the new efficiency:
η′= 1 −45
600 = 1 −45
600 = 1 −0.075 = 0.925 = 92.5%
Therefore, the maximum possible efficiency of the engine when the cold
reservoir temperature is increased to 45◦Cis 92.5% .
Question 27
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine produces 1000 J of work for every 3000 J of heat absorbed
from the hot reservoir. 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 using the formula:
Efficiency = Useful work output
Heat input =W
Qin
where Wis the work produced and Qin is the heat absorbed from the hot
reservoir.
Given that W= 1000 J and Qin = 3000 J, we have:
Efficiency = 1000
3000 =1
3
Step 2: Discuss whether this violates the second law of thermodynamics.
The efficiency of the engine is calculated to be 1
3which is approximately 33.3
Question 28
Question
A Carnot engine operates between two heat reservoirs at temperatures Thot =
500 K and Tcold = 200 K. The engine absorbs 4000 J of heat from the hot
reservoir in each cycle. Calculate the efficiency of the engine and determine the
amount of heat rejected to the cold reservoir in each cycle.
23
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency, η= 1 −Tcold
Thot
η= 1 −200
500 = 1 −2
5=3
5= 0.6 = 60%
Step 2: Calculate the heat rejected to the cold reservoir in each cycle.
Heat rejected, Qcold =η×Heat absorbed from hot reservoir
Qcold = 0.6×4000 J = 2400 J
Therefore, the efficiency of the Carnot engine is 60
Question 29
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc) and has an efficiency of 40
Solution
Let Qhbe the amount of heat absorbed from the hot reservoir, Qcbe the amount
of heat ejected to the cold reservoir, and Wbe the work done by the engine in
each cycle.
Step 1: Recall the formula for the efficiency of a heat engine:
η=W
Qh
Given that the efficiency is 40
0.40 = W
8000
W= 0.40 ×8000
W= 3200 J
Step 2: Apply the first law of thermodynamics to find the amount of heat
ejected to the cold reservoir:
W=Qh−Qc
Qc=Qh−W
Qc= 8000 −3200
24
Qc= 4800 J
Step 3: Verify the efficiency of the heat engine using the heat transfers:
η=W
Qh
=3200
8000 = 0.40
So, the amount of heat ejected to the cold reservoir is 4800 J.
Question 30
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. If the engine produces 2000 J of work per cycle, calculate the efficiency
of the engine.
Solution
Let’s denote the efficiency of the heat engine as η.
Step 1: We can calculate the efficiency of the engine using the formula:
η= 1 −Tc
Th
where This the temperature of the hot reservoir (in Kelvin) and Tcis the
temperature of the cold reservoir (in Kelvin).
Step 2: Substitute Th= 800 K and Tc= 300 K into the formula:
η= 1 −300
800
Step 3: Simplify the expression:
η= 1 −3
8=5
8
Step 4: Therefore, the efficiency of the heat engine is 5
8or 62.5
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold,
with Thot = 500 K and Tcold = 300 K. The engine absorbs 6,000 J of heat from
the hot reservoir and exhausts 2,400 J of heat to the cold reservoir in each cycle.
Determine the efficiency of the engine.
25
Solution
Step 1: Calculate the net work done by the engine in one cycle using the first
law of thermodynamics:
The net work done Wnet in one cycle is equal to the heat absorbed from the
hot reservoir minus the heat exhausted to the cold reservoir:
Wnet =Qhot −Qcold
Given that Qhot = 6,000 J and Qcold = 2,400 J, we can calculate Wnet:
Wnet = 6,000 J −2,400 J = 3,600 J
Step 2: Calculate the efficiency of the engine using the formula for efficiency:
The efficiency ηof the engine is given by the ratio of the work done to the
heat input from the hot reservoir:
η=Wnet
Qhot
Substitute the values of Wnet and Qhot into the formula to get:
η=3,600 J
6,000 J = 0.6
Therefore, the efficiency of the engine is 60%.
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). If the engine absorbs 6000 J of heat from the hot reservoir and 3500 J of
heat is rejected to the cold reservoir in each cycle, calculate the efficiency of the
engine.
Solution
Step 1: Determine the efficiency of the heat engine using the formula:
efficiency = 1 −Qc
Qh
where Qhis the heat absorbed from the hot reservoir (6000 J) and Qcis the
heat rejected to the cold reservoir (3500 J).
Step 2: Calculate the efficiency:
efficiency = 1 −3500
6000
efficiency = 1 −0.5833
efficiency = 0.4167 or 41.67%
Therefore, the efficiency of the heat engine is 41.67
26
Question 33
Question
A heat engine operating between two thermal reservoirs has an efficiency of 40
1. The work output of the engine in each cycle.
2. The heat rejected to the cold reservoir in each cycle.
3. The temperature of the cold reservoir if the temperature of the hot reser-
voir is 400◦C.
Solution
1. We know that the efficiency of a heat engine is given by the formula:
Efficiency = Work output
Heat input
Given that the efficiency is 40
Work output = Efficiency ×Heat input
Work output = 0.4×600 J
Work output = 240 J
2. The heat rejected to the cold reservoir in each cycle can be calculated
using the first law of thermodynamics:
Heat input = Work output + Heat rejected
Since we know the values of work output and heat input, we can find the heat
rejected:
600 = 240 + Heat rejected
Heat rejected = 600 −240
Heat rejected = 360 J
3. The efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
Given that the efficiency is 0.4 and the temperature of the hot reservoir is 400◦C
(673 K), we can calculate the temperature of the cold reservoir:
0.4=1−Tcold
673
27
Tcold
673 = 1 −0.4
Tcold
673 = 0.6
Tcold = 0.6×673
Tcold = 404 K
Question 34
Question
A Carnot heat engine operates between a high-temperature reservoir at 500 ◦C
and a low-temperature reservoir at 50 ◦C. If the engine absorbs 600 J of heat
from the high-temperature reservoir in each cycle, determine the efficiency of
the engine and the amount of heat rejected to the low-temperature reservoir.
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 absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Given that the hot reservoir temperature is 500 ◦C and the cold reservoir
temperature is 50 ◦C, we convert these temperatures to Kelvin:
Th= 500 + 273 = 773 K
Tc= 50 + 273 = 323 K
Plugging these values into the formula, we get:
Efficiency = 1 −323
773 = 1 −0.418 = 0.582
Therefore, the efficiency of the engine is 58.2
Step 2: Calculate the heat rejected to the low-temperature reservoir. Since
the engine absorbs 600 J of heat from the high-temperature reservoir in each
cycle, the heat rejected to the low-temperature reservoir can be calculated using
the equation:
Efficiency = Qin −Qout
Qin
where Qin is the heat absorbed from the high-temperature reservoir and Qout
is the heat rejected to the low-temperature reservoir.
28
Substitute the known values into the equation:
0.582 = 600 −Qout
600
0.582 ×600 = 600 −Qout
Qout = 349.2 J
Therefore, the amount of heat rejected to the low-temperature reservoir is
349.2 J.
Question 35
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 400 J of heat from the hot reservoir in each cycle
and exhausts 220 J of heat to the cold reservoir in each cycle.
Calculate the efficiency of this heat engine.
Solution
Step 1: Calculate the net work done by the engine in each cycle. The net work
done by the engine in each cycle is given by the difference between the heat
absorbed from the hot reservoir and the heat exhausted to the cold reservoir:
W=Qin −Qout = 400 J −220 J = 180 J.
Step 2: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
η=W
Qin
.
Substitute the values of Wand Qin into the formula:
η=180 J
400 J = 0.45 = 45%.
Therefore, the efficiency of this heat engine is 45
29
Question 2
Question
A Carnot engine operates between two heat reservoirs, one at 500K and the
other at 300K. If the engine absorbs 1500 J of heat from the hot reservoir in
each cycle, what is the maximum amount of work that can be done by the engine
in each cycle? What is the efficiency of this engine?
Solution
Step 1: Calculate the maximum work done by the engine in each cycle.
Wmax =Qhot 1−Tcold
Thot
= 1500 J 1−300
500
= 1500 J (1 −0.6)
= 1500 J ×0.4
= 600 J
Step 2: Calculate the efficiency of the engine.
Efficiency (%) = Wmax
Qhot
×100%
=600
1500 ×100%
= 40%
Therefore, the maximum amount of work that can be done by the engine in
each cycle is 600 J and the efficiency of the engine is 40
Question 3
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 600 J of heat from the hot reservoir in each cycle.
Calculate the efficiency of the Carnot engine and the heat rejected to the cold
reservoir in each cycle.
Solution
To calculate the efficiency of the Carnot engine, we can use the formula for
Carnot efficiency:
2
Efficiency = 1 −Tcold
Thot
where Thot is the temperature of the hot reservoir in Kelvin, and Tcold is the
temperature of the cold reservoir in Kelvin.
Step 1: Calculate the efficiency of the Carnot engine. Given: Thot = 600 K,
Tcold = 300 K
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
So, the efficiency of the Carnot engine is 50
Step 2: Calculate the heat rejected to the cold reservoir. The heat rejected
to the cold reservoir in each cycle is equal to the heat absorbed from the hot
reservoir. This is because in a Carnot engine, all of the heat absorbed from the
hot reservoir is converted into work, so the rest of the energy must be rejected
to the cold reservoir.
Given: Heat absorbed = 600 J
Therefore, the heat rejected to the cold reservoir in each cycle is 600 J.
Therefore, the efficiency of the Carnot engine is 50
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
with Th= 600 K and Tc= 300 K. The engine receives 2,000 J of heat from the
high-temperature reservoir in each cycle. Calculate the efficiency of the engine
and determine the amount of heat rejected to the low-temperature reservoir in
each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values into the formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot engine is 50
Step 3: Calculate the amount of heat rejected to the low-temperature reser-
voir in each cycle using the formula:
Heat rejected = Qh−Qc
3
Step 4: Substitute the given values into the formula:
Heat rejected = 2,000 J −Tc
Th
×2,000 J
Heat rejected = 2,000 J −300
600 ×2,000 J = 2,000 J −1,000 J = 1,000 J
Therefore, the amount of heat rejected to the low-temperature reservoir in
each cycle is 1,000 J.
Question 5
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir and
produces 2400 J of work. Calculate the efficiency of the engine. Additionally,
explain how this efficiency compares to the maximum possible efficiency for a
heat engine operating between the same two reservoirs.
Solution
Step 1: We can calculate the efficiency of the Carnot heat engine using the
formula for efficiency:
Efficiency = Work output
Heat input =2400 J
4000 J = 0.6 = 60%.
Step 2: The maximum possible efficiency of a heat engine operating between
two reservoirs at temperatures Thand Tcis given by the Carnot efficiency for-
mula:
EfficiencyCarnot = 1 −Tc
Th
.
Step 3: Substituting the given temperatures Thand Tcinto the Carnot
efficiency formula:
EfficiencyCarnot = 1 −Tc
Th
= 1 −2400
4000 = 1 −0.6=0.4 = 40%.
Step 4: Comparing the calculated efficiency of the Carnot engine (60
Question 6
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. If the engine absorbs 600 J of heat from the hot reservoir
during each cycle and expels 400 J of heat to the cold reservoir, calculate the
efficiency of the engine in terms of Thand Tc.
4
Solution
Let’s denote the heat absorbed from the hot reservoir as Qhand the heat ex-
pelled to the cold reservoir as Qc.
Step 1: Apply the formula for the efficiency of a Carnot engine: The effi-
ciency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Determine the heat absorbed and heat expelled: Given that the
engine absorbs 600 J of heat from the hot reservoir (Qh= 600 J) and expels
400 J of heat to the cold reservoir (Qc= 400 J).
Step 3: Substitute the values into the formula for efficiency:
Efficiency = 1 −Qc
Qh
Substitute Qh= 600 J and Qc= 400 J into the formula:
Efficiency = 1 −400
600
Step 4: Simplify the expression:
Efficiency = 1 −2
3
Step 5: Writing the efficiency in terms of Thand Tc: Given that Th> Tc,
we know that:
Th=Tc+ ∆T
where ∆Tis the temperature difference between the two reservoirs.
Step 6: Substitute for Thand Tcin terms of ∆T:
Efficiency = 1 −Tc
Tc+ ∆T
Step 7: Further simplification:
Efficiency = 1 −Tc
Tc(1 + ∆T
Tc)
Efficiency = 1 −1
1 + ∆T
Tc
Therefore, the efficiency of the Carnot engine in terms of Thand Tcis
Efficiency = 1 −1
1+ ∆T
Tc
.
5
Question 7
Question
A Carnot heat engine operates between a high-temperature reservoir at Th=
600 K and a low-temperature reservoir at Tc= 300 K. If this engine produces
500 J of work per cycle, calculate the following:
1. The heat absorbed from the high-temperature reservoir per cycle.
2. The efficiency of the engine.
3. The heat rejected to the low-temperature reservoir per cycle.
Solution
Let’s denote the heat absorbed from the high-temperature reservoir per cycle
as Qh, the heat rejected to the low-temperature reservoir per cycle as Qc, and
the work done by the engine per cycle as W.
Step 1: Calculate the heat absorbed from the high-temperature reservoir
per cycle. Using the first law of thermodynamics for a cyclic process:
W=Qh−Qc
Given that W= 500 J, and Th= 600 K, and Tc= 300 K, we can calculate Qh:
500 J = Qh−Qc
Qh=Qc+ 500 J
Now, we know that for a Carnot engine:
Qh
Th
=Qc
Tc
Substitute Qh=Qc+ 500 and solve for Qc:
Qc+ 500
600 =Qc
300
300(Qc+ 500) = 600Qc
300Qc+ 150000 = 600Qc
300Qc= 150000
Qc= 500 J
Therefore, the heat absorbed from the high-temperature reservoir per cycle is
Qh= 1000 J.
Step 2: Calculate the efficiency of the engine. The efficiency of a heat
engine is given by:
Efficiency = W
Qh
6
Substitute W= 500 J and Qh= 1000 J:
Efficiency = 500
1000 = 0.5
So, the efficiency of the engine is 0.5 or 50
Step 3: Calculate the heat rejected to the low-temperature reservoir per
cycle. We already know that Qc= 500 J from earlier calculations.
Therefore, the heat rejected to the low-temperature reservoir per cycle is 500
J.
Question 8
Question
A heat engine operates between two reservoirs at temperatures T1and T2(T1>
T2). The engine absorbs Q1amount of heat from the reservoir at temperature
T1and exhausts Q2amount of heat to the reservoir at temperature T2. The
efficiency of the engine is given by η= 1 −Q2
Q1
. Prove that the efficiency of the
heat engine is bounded by η≤1−T2
T1
.
Solution
Step 1: Let’s start by expressing the heat transfer Q2in terms of Q1using the
efficiency equation η= 1 −Q2
Q1
.
Substitute Q2=Q1(1 −η) into η= 1 −Q2
Q1
to get η= 1 −Q1(1 −η)
Q1
⇒η= 1 −(1 −η) = η
Step 2: Now, we’ll express the efficiency ηin terms of T1,T2,Q1, and Q2.
Using η= 1 −Q2
Q1
, we have η= 1 −Q1(1 −η)
Q1
⇒η= 1 −(1 −η) = η
Step 3: Next, we’ll use the definition of efficiency to express ηin terms of
temperatures T1and T2.
Using Q1=Q2+W, where Wis the work done by the engine, we get η= 1−Q2
Q2+W
Step 4: By applying the second law of thermodynamics, we know that W≤
Q1(T1−T2).
Substitute W≤Q1(T1−T2) into η= 1 −Q2
Q2+W
7
⇒η= 1 −Q2
Q2+Q1(T1−T2)
Step 5: Simplifying the expression gives us
η= 1 −Q2
Q1+Q1T1−T2
T1= 1 −Q2
Q11 + T1−T2
T1
Step 6: After simplifying further, we arrive at
η= 1 −1
1 + T1−T2
T1
= 1 −T1
T2
Therefore, the efficiency of the heat engine is bounded by η≤1−T2
T1
.
Question 9
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. If the efficiency of the
engine is η, show that the efficiency of the engine can be expressed in terms of
the temperatures as η= 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as η= 1 −
Heat output
Heat input .
Step 2: The heat input for the engine is Qh, and the heat output is Qc.
Step 3: From the second law of thermodynamics, we have Qh=Qc+W,
where Wis the work done by the engine.
Step 4: The work done by the engine can be expressed as W=Qh−Qc.
Step 5: Substituting this expression for Winto the efficiency formula, we
get:
η= 1 −Qc
Qh
Step 6: Substituting Qh=Qc+W=Qc+Qh−Qc=Qhinto the efficiency
formula gives:
η= 1 −Qc
Qh
= 1 −Tc
Th
Step 7: Therefore, the efficiency of the engine can be expressed in terms of
the temperatures as η= 1 −Tc
Th.
8
Question 10
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 consumes 5000 J of heat
from the hot reservoir and produces 2500 J of work. Calculate the efficiency of
the heat engine and determine if it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Substitute the values into the formula:
Efficiency = 2500
5000 = 0.5 = 50%
Step 3: Check if the efficiency violates the second law of thermodynamics.
The maximum efficiency of a heat engine operating between two reservoirs at
different temperatures is given by Carnot’s efficiency formula:
Carnot Efficiency = 1 −Tcold
Thot
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −300
600 = 0.5 = 50%
Step 5: Since the efficiency of the actual engine is equal to the Carnot
efficiency, it does not violate the second law of thermodynamics. In real engines,
the efficiency is always less than the Carnot efficiency due to factors such as
friction, heat loss, etc.
Question 11
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 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. Calculate the efficiency of the
engine and discuss whether it violates the second law of thermodynamics.
9
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1−Tc
Th×100%
where This the temperature of the hot reservoir and Tcis the temperature of
the cold reservoir.
Efficiency = 1−300
600×100% = 50%
Step 2: Discuss the efficiency of the engine and the second law of thermo-
dynamics. The efficiency of the heat engine is 50
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. It absorbs Qhof heat from the hot reservoir and produces Wof work
output. If the efficiency of the heat engine is η, show that the entropy change
of the universe ∆Suniv in one complete cycle is given by ∆Suniv =Qh
Th−Qc
Tc.
Solution
Step 1: Recall the definition of efficiency for a heat engine: Let Qcbe the heat
rejected to the cold reservoir. The efficiency ηof a heat engine is given by:
η=W
Qh
= 1 −Qc
Qh
Step 2: Express the heat rejected Qcin terms of Qh: From the efficiency
equation, we have:
Qc=Qh−W
Step 3: Express Win terms of Qhusing the efficiency equation: Substitute
the expression for Qcin terms of Qhinto the efficiency equation to get:
η= 1 −Qh−W
Qh
Step 4: Simplify the expression for η: This simplifies to:
η= 1 −1 + W
Qh
η=W
Qh
10
Step 5: Express Qcin terms of Thand Tc: From the second law of thermo-
dynamics, we have:
Qh
Th
+Qc
Tc
= 0
Step 6: Substituting Qc=Qh−Winto the second law equation: We obtain:
Qh
Th
+Qh−W
Tc
= 0
Step 7: Rearrange the equation: Solving for Wgives:
W=Qh1−Tc
Th
Step 8: Calculate the entropy change of the universe ∆Suniv: The change in
entropy of the universe is given by:
∆Suniv =Qh
Th
+−Qh+W
Tc
Step 9: Substitute the expression for Winto the equation for ∆Suniv:
∆Suniv =Qh
Th
+
−Qh+Qh1−Tc
Th
Tc
Step 10: Simplify the expression for ∆Suniv: Converting the terms gives:
∆Suniv =Qh
Th
−Qc
Tc
Therefore, the entropy change of the universe in one complete cycle is given
by ∆Suniv =Qh
Th−Qc
Tc.
Question 13
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, with Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir and
rejects 2500 J of heat to the cold reservoir during each cycle. Determine:
1. The efficiency of the engine.
2. The work done by the engine during each cycle.
3. If the engine operates as a refrigerator instead of an engine, calculate the
amount of work required to run the refrigerator.
11
Solution
1. To calculate the efficiency of the engine, we use the formula:
Efficiency (%) = Useful work output
Energy input ×100%
Useful work output = Heat absorbed −Heat rejected
=Qh−Qc
= 4000 J −2500 J
= 1500 J
The efficiency of the engine is given by:
Efficiency = Useful work output
Energy input ×100% = 1500 J
4000 J ×100% = 37.5%
2. The work done by the engine during each cycle is equal to the useful work
output:
Work done = 1500 J
3. If the engine operates as a refrigerator, the amount of work required to
run the refrigerator is equal to the work done on the system:
Work required = 1500 J
Question 14
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
200 K. If the engine absorbs 2000 J of heat from the hot reservoir in each cycle,
find:
1. The efficiency of the heat engine.
2. The minimum amount of heat that must be exhausted to the cold reservoir
in each cycle.
Solution
Given:
Temperature of the hot reservoir (Th) = 500 K
Temperature of the cold reservoir (Tc) = 200 K
Heat absorbed from the hot reservoir (Qin) = 2000 J
12
1. Efficiency of the heat engine
The efficiency (η) of the heat engine is given by the formula:
η= 1 −Tc
Th
Step 1: Calculate the efficiency using the given temperatures:
η= 1 −200
500 = 1 −2
5=3
5
Therefore, the efficiency of the heat engine is 3
5or 60
2. Minimum amount of heat exhausted to the cold reservoir
The amount of heat exhausted to the cold reservoir in each cycle can be
calculated using the efficiency of the engine:
Qout =η·Qin
Step 1: Plug in the values to find Qout:
Qout =3
5×2000 = 1200 J
Therefore, the minimum amount of heat that must be exhausted to the
cold reservoir in each cycle is 1200 J.
Question 15
Question
A Carnot cycle operates between two reservoirs at temperatures Thand Tcwhere
Th> Tc. The heat input from the hot reservoir is Qhand the work output of
the engine is Wout. If the efficiency of the Carnot engine is η, show that the
heat rejected to the cold reservoir Qcsatisfies the inequality:
Qc>Tc
Th
Qh
Solution
Step 1: Recall the expression for the efficiency of a Carnot engine:
η= 1 −Tc
Th
Step 2: Rearrange the equation to solve for Tc:
Tc
Th
= 1 −η
13
Step 3: Multiply both sides by Qhto find the expression for Qc:
Qc= (1 −η)Qh
Step 4: Substitute the expression for ηback into the equation:
Qc=1−1−Tc
ThQh
Step 5: Simplify to get the final inequality:
Qc=Tc
Th
Qh
Step 6: Since Th> Tc, we have:
Qc>Tc
Th
Qh
Therefore, the heat rejected to the cold reservoir Qcsatisfies the inequality
Qc>Tc
ThQh.
Question 16
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). If the engine absorbs 5000 J of heat from the hot reservoir and 3000 J of
heat is rejected to the cold reservoir in each cycle, calculate: (i) The efficiency
of the engine. (ii) The maximum amount of work the engine can produce in
each cycle. (iii) If the engine is reversible or irreversible based on the calculated
efficiency.
Given: Th= 500 K, Tc= 300 K.
Solution
(i) The efficiency of the engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 1: Substitute the given values into the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
So, the efficiency of the engine is 40
(ii) The maximum amount of work the engine can produce in each cycle is
given by the formula:
Work = Efficiency ×Heat absorbed from hot reservoir
14
Step 2: Substitute the given values into the formula:
Work = 0.4×5000 J = 2000 J
Therefore, the maximum amount of work the engine can produce in each
cycle is 2000 J.
(iii) If an engine is reversible, its efficiency is given by the Carnot efficiency:
EfficiencyCarnot = 1 −Tc
Th
If the efficiency of the engine matches the Carnot efficiency, it is reversible.
Since the efficiency we calculated in part (i) was the same as the Carnot effi-
ciency, the engine is reversible.
So, the engine in this scenario is reversible.
Question 17
Question
A Carnot 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. Calculate the work done by the engine in each cycle and determine
its efficiency.
Solution
Step 1: Calculate the work done by the engine in each cycle using the equation
for efficiency of a Carnot engine:
Efficiency = 1 −Tcold
Thot
where Thot = 600 K and Tcold = 300 K.
Step 2: Substitute the values into the efficiency equation:
Efficiency = 1 −300
600 = 1 −1
2=1
2
Step 3: The efficiency of a Carnot engine is the ratio of the work output to
the heat input, so:
Efficiency = W
Qin
Step 4: Rearrange the equation to solve for work done by the engine:
W= Efficiency ×Qin =1
2×500 J = 250 J
Therefore, the work done by the engine in each cycle is 250 J and the effi-
ciency of the engine is 50
15
Question 18
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. The engine takes in 800 J of heat
from the hot reservoir and performs 500 J of work during each cycle. Calculate
the efficiency of this engine.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
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 the given temperatures from Celsius to Kelvin: Hot reservoir
temperature, TH= 500 K Cold reservoir temperature, TC= 300 K
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −300
500
Step 4: Calculate the efficiency:
Efficiency = 1 −3
5=2
5
Step 5: Express the efficiency as a percentage:
Efficiency = 2
5×100% = 40%
Therefore, the efficiency of the heat engine is 40
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the hot reservoir and delivers W
of work. If the engine’s efficiency is η, prove that the efficiency of the engine is
given by η= 1 −Tc
Th.
16
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is given by:
η=Useful work output
Heat input =W
Qh
Step 2: Express the work done by the engine in terms of heat taken in and
rejected: According to the first law of thermodynamics, the work done by the
engine is the difference between the heat absorbed and the heat rejected:
W=Qh−Qc
Step 3: Express the rejected heat Qcin terms of the temperatures Thand
Tc: From Carnot’s theorem, we know that for a reversible process, the ratio of
heat transferred to a reservoir and the reservoir’s temperature remains constant.
Therefore, for the engine rejecting heat to the cold reservoir,
Qc
Tc
=Qh
Th
Step 4: Solve for Qcin terms of Qh,Tc, and Th:
Qc=Tc
Th
Qh
Step 5: Substitute the expression for Qcback into the efficiency equation:
η=W
Qh
=Qh−Qc
Qh
= 1 −Qc
Qh
= 1 −Tc
Th
Step 6: Therefore, the efficiency of the engine is given by η= 1 −Tc
Th.
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwith
Th> Tc. The engine has an efficiency of 40
1. The heat rejected to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Solution
Given: Efficiency of the heat engine, η= 40% = 0.40, Heat absorbed from the
hot reservoir, Qh= 5000 J
We are asked to find:
17
1. The heat rejected to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Step 1: Find the heat rejected to the cold reservoir in each cycle.
The efficiency of a heat engine is given by:
η= 1 −Qc
Qh
Given that the efficiency, η= 0.40, and that Qh= 5000 J, we can rearrange the
equation to solve for Qc:
0.40 = 1 −Qc
5000
Qc
5000 = 0.60
Qc= 0.60 ×5000
Qc= 3000 J
Therefore, the heat rejected to the cold reservoir in each cycle is 3000 J.
Step 2: Find the work done by the engine in each cycle. The
efficiency of a heat engine can also be expressed as:
η=W
Qh
We can rearrange the equation to solve for W:
0.40 = W
5000
W= 0.40 ×5000
W= 2000 J
Therefore, the work done by the engine in each cycle is 2000 J.
Question 21
Question
A heat engine operates between two reservoirs at temperatures T1= 700 K and
T2= 300 K. If the engine absorbs 800 J of heat from the high-temperature
reservoir in each cycle, what is the maximum amount of work the engine can
perform per cycle? Calculate the efficiency of this engine.
18
Solution
Step 1: Find the maximum efficiency of the heat engine using Carnot’s theorem.
Efficiency = 1 −T2
T1
where T2is the temperature of the low-temperature reservoir and T1is the
temperature of the high-temperature reservoir.
Step 2: Substitute the given values into the efficiency formula.
Efficiency = 1 −300
700 = 1 −3
7=4
7
Step 3: Calculate the maximum work output of the engine per cycle using
the formula for efficiency.
Work = Efficiency ×Heat input
Step 4: Substitute the given heat input of 800 J into the work formula.
Work = 4
7×800 = 3200
7≈457.14 J
Step 5: Therefore, the maximum amount of work the engine can perform
per cycle is approximately 457.14 J, and the efficiency of the engine is 4
7≈0.5714
or 57.14
Question 22
Question
A heat engine absorbs 1400 J of heat from a high-temperature reservoir at 500
K and exhausts 800 J of heat to a low-temperature reservoir. Calculate the
efficiency of the engine and determine whether it violates the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −Qout
Qin
where Qin is the heat absorbed from the high-temperature reservoir and Qout
is the heat exhausted to the low-temperature reservoir.
Step 2: Substitute the given values into the efficiency formula:
Efficiency = 1 −800
1400 = 1 −4
7=3
7≈0.429
19
Step 3: Determine whether the engine violates the second law of thermody-
namics. The second law of thermodynamics states that no heat engine can have
an efficiency greater than the Carnot efficiency, given by:
Carnot efficiency = 1 −Tlow
Thigh
where Tlow is the temperature of the low-temperature reservoir and Thigh is the
temperature of the high-temperature reservoir.
Step 4: Substitute the temperatures into the Carnot efficiency formula:
Carnot efficiency = 1 −300
500 = 1 −0.6 = 0.4
Step 5: Compare the efficiency of the engine (0.429) with the Carnot ef-
ficiency (0.4). Since the efficiency of the engine is greater than the Carnot
efficiency, it violates the second law of thermodynamics.
Question 23
Question
A Carnot engine operates between a high temperature reservoir at 800 K and
a low temperature reservoir at 300 K. If the engine absorbs 10,000 J of heat
energy from the high temperature reservoir in each cycle, what is the maximum
amount of work that can be done by the engine in each cycle? What is the
efficiency of the engine?
Solution
Step 1: Calculate the maximum efficiency of the engine. The efficiency of a
Carnot engine is given by the formula:
Efficiency = 1 −TC
TH
where THis the temperature of the high temperature reservoir and TCis the
temperature of the low temperature reservoir. Given TH= 800 K and TC= 300
K, we can substitute these values into the formula to find the efficiency.
Efficiency = 1 −300
800 = 1 −3
8=5
8
Step 2: Calculate the maximum work done by the engine. The maximum
work done by a Carnot engine is given by the formula:
Wmax = Efficiency ×Qin
20
where Qin is the heat energy absorbed from the high temperature reservoir in
each cycle. Given Qin = 10,000 J and efficiency = 5
8, we can substitute these
values into the formula to find the maximum work done by the engine.
Wmax =5
8×10,000 = 6,250 J
Therefore, the maximum amount of work that can be done by the engine in
each cycle is 6,250 J and the efficiency of the engine is 5
8or 62.5%.
Question 24
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and
exhausts heat Qcto the cold reservoir. Determine the efficiency of the engine
in terms of Thand Tc, and discuss how this efficiency is related to the second
law of thermodynamics.
Solution
Step 1: Let’s first calculate the efficiency of the Carnot engine. The efficiency
of a Carnot engine operating between two heat reservoirs at temperatures Th
and Tcis given by:
η= 1 −Tc
Th
Step 2: Now, let’s discuss why this efficiency is related to the second law of
thermodynamics. The efficiency of a heat engine is limited by the second law of
thermodynamics, which states that no heat engine can have an efficiency greater
than that of a Carnot engine operating between the same two temperatures.
This means that no engine can have an efficiency greater than 1 −Tc
Th. The
efficiency of a Carnot engine represents the maximum possible efficiency for any
heat engine operating between the given temperatures Thand Tc. This is a
direct consequence of the second law of thermodynamics, which establishes the
limitations on the conversion of heat into work.
Question 25
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc. If the engine absorbs 5000 J of heat from the high-temperature reservoir
and the engine’s efficiency is 40
21
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = Work output
Heat input
0.40 = Work output
5000 J
Work output = 0.40 ×5000 J
Work output = 2000 J
Step 2: Calculate the work done by the engine. For a Carnot engine,
efficiency e=Th−Tc
Thwhere This the temperature of the high-temperature
reservoir and Tcis the temperature of the low-temperature reservoir. So, we
have 0.40 = Th−Tc
Th. Given that the engine absorbs 5000 J from the high-
temperature reservoir, we have Qh= 5000 J. Therefore, work done by the
engine is W=Qh×e= 5000 ×0.40 = 2000 J.
Step 3: Calculate the heat rejected to the low-temperature reservoir. Since
the total heat absorbed from the high-temperature reservoir is used to do work
and reject heat, we have:
Qh=W+Qc
5000 = 2000 + Qc
Qc= 3000 J
Step 4: Discussion on the second law of thermodynamics. The second law
of thermodynamics states that heat flows spontaneously from a hotter reservoir
to a colder reservoir, and that no heat engine can be 100
Question 26
Question
A heat engine operates between a hot reservoir at 600◦Cand a cold reservoir
at 27◦C. The engine has an efficiency of 40%. Calculate the maximum possible
efficiency of the engine if the cold reservoir temperature is increased to 45◦C.
Solution
Let’s denote the temperatures of the hot reservoir, cold reservoir, and increased
cold reservoir as TH= 600◦C,TC= 27◦C, and T′
C= 45◦C, respectively.
We know that the efficiency (η) of a heat engine operating between a hot
reservoir at temperature THand a cold reservoir at temperature TCis given by
the formula:
η= 1 −TC
TH
22
Step 1: Calculate the initial efficiency of the engine. Given TH= 600◦C
and TC= 27◦C, we can calculate the initial efficiency:
η= 1 −27
600 = 1 −27
600 = 1 −0.045 = 0.955 = 95.5%
Step 2: Calculate the new efficiency of the engine. Given the increased cold
reservoir temperature T′
C= 45◦C, we can calculate the new efficiency:
η′= 1 −45
600 = 1 −45
600 = 1 −0.075 = 0.925 = 92.5%
Therefore, the maximum possible efficiency of the engine when the cold
reservoir temperature is increased to 45◦Cis 92.5% .
Question 27
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine produces 1000 J of work for every 3000 J of heat absorbed
from the hot reservoir. 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 using the formula:
Efficiency = Useful work output
Heat input =W
Qin
where Wis the work produced and Qin is the heat absorbed from the hot
reservoir.
Given that W= 1000 J and Qin = 3000 J, we have:
Efficiency = 1000
3000 =1
3
Step 2: Discuss whether this violates the second law of thermodynamics.
The efficiency of the engine is calculated to be 1
3which is approximately 33.3
Question 28
Question
A Carnot engine operates between two heat reservoirs at temperatures Thot =
500 K and Tcold = 200 K. The engine absorbs 4000 J of heat from the hot
reservoir in each cycle. Calculate the efficiency of the engine and determine the
amount of heat rejected to the cold reservoir in each cycle.
23
Solution
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency, η= 1 −Tcold
Thot
η= 1 −200
500 = 1 −2
5=3
5= 0.6 = 60%
Step 2: Calculate the heat rejected to the cold reservoir in each cycle.
Heat rejected, Qcold =η×Heat absorbed from hot reservoir
Qcold = 0.6×4000 J = 2400 J
Therefore, the efficiency of the Carnot engine is 60
Question 29
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc) and has an efficiency of 40
Solution
Let Qhbe the amount of heat absorbed from the hot reservoir, Qcbe the amount
of heat ejected to the cold reservoir, and Wbe the work done by the engine in
each cycle.
Step 1: Recall the formula for the efficiency of a heat engine:
η=W
Qh
Given that the efficiency is 40
0.40 = W
8000
W= 0.40 ×8000
W= 3200 J
Step 2: Apply the first law of thermodynamics to find the amount of heat
ejected to the cold reservoir:
W=Qh−Qc
Qc=Qh−W
Qc= 8000 −3200
24
Qc= 4800 J
Step 3: Verify the efficiency of the heat engine using the heat transfers:
η=W
Qh
=3200
8000 = 0.40
So, the amount of heat ejected to the cold reservoir is 4800 J.
Question 30
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. If the engine produces 2000 J of work per cycle, calculate the efficiency
of the engine.
Solution
Let’s denote the efficiency of the heat engine as η.
Step 1: We can calculate the efficiency of the engine using the formula:
η= 1 −Tc
Th
where This the temperature of the hot reservoir (in Kelvin) and Tcis the
temperature of the cold reservoir (in Kelvin).
Step 2: Substitute Th= 800 K and Tc= 300 K into the formula:
η= 1 −300
800
Step 3: Simplify the expression:
η= 1 −3
8=5
8
Step 4: Therefore, the efficiency of the heat engine is 5
8or 62.5
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold,
with Thot = 500 K and Tcold = 300 K. The engine absorbs 6,000 J of heat from
the hot reservoir and exhausts 2,400 J of heat to the cold reservoir in each cycle.
Determine the efficiency of the engine.
25
Solution
Step 1: Calculate the net work done by the engine in one cycle using the first
law of thermodynamics:
The net work done Wnet in one cycle is equal to the heat absorbed from the
hot reservoir minus the heat exhausted to the cold reservoir:
Wnet =Qhot −Qcold
Given that Qhot = 6,000 J and Qcold = 2,400 J, we can calculate Wnet:
Wnet = 6,000 J −2,400 J = 3,600 J
Step 2: Calculate the efficiency of the engine using the formula for efficiency:
The efficiency ηof the engine is given by the ratio of the work done to the
heat input from the hot reservoir:
η=Wnet
Qhot
Substitute the values of Wnet and Qhot into the formula to get:
η=3,600 J
6,000 J = 0.6
Therefore, the efficiency of the engine is 60%.
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(Th>
Tc). If the engine absorbs 6000 J of heat from the hot reservoir and 3500 J of
heat is rejected to the cold reservoir in each cycle, calculate the efficiency of the
engine.
Solution
Step 1: Determine the efficiency of the heat engine using the formula:
efficiency = 1 −Qc
Qh
where Qhis the heat absorbed from the hot reservoir (6000 J) and Qcis the
heat rejected to the cold reservoir (3500 J).
Step 2: Calculate the efficiency:
efficiency = 1 −3500
6000
efficiency = 1 −0.5833
efficiency = 0.4167 or 41.67%
Therefore, the efficiency of the heat engine is 41.67
26
Question 33
Question
A heat engine operating between two thermal reservoirs has an efficiency of 40
1. The work output of the engine in each cycle.
2. The heat rejected to the cold reservoir in each cycle.
3. The temperature of the cold reservoir if the temperature of the hot reser-
voir is 400◦C.
Solution
1. We know that the efficiency of a heat engine is given by the formula:
Efficiency = Work output
Heat input
Given that the efficiency is 40
Work output = Efficiency ×Heat input
Work output = 0.4×600 J
Work output = 240 J
2. The heat rejected to the cold reservoir in each cycle can be calculated
using the first law of thermodynamics:
Heat input = Work output + Heat rejected
Since we know the values of work output and heat input, we can find the heat
rejected:
600 = 240 + Heat rejected
Heat rejected = 600 −240
Heat rejected = 360 J
3. The efficiency of a Carnot heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
Given that the efficiency is 0.4 and the temperature of the hot reservoir is 400◦C
(673 K), we can calculate the temperature of the cold reservoir:
0.4=1−Tcold
673
27
Tcold
673 = 1 −0.4
Tcold
673 = 0.6
Tcold = 0.6×673
Tcold = 404 K
Question 34
Question
A Carnot heat engine operates between a high-temperature reservoir at 500 ◦C
and a low-temperature reservoir at 50 ◦C. If the engine absorbs 600 J of heat
from the high-temperature reservoir in each cycle, determine the efficiency of
the engine and the amount of heat rejected to the low-temperature reservoir.
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 absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Given that the hot reservoir temperature is 500 ◦C and the cold reservoir
temperature is 50 ◦C, we convert these temperatures to Kelvin:
Th= 500 + 273 = 773 K
Tc= 50 + 273 = 323 K
Plugging these values into the formula, we get:
Efficiency = 1 −323
773 = 1 −0.418 = 0.582
Therefore, the efficiency of the engine is 58.2
Step 2: Calculate the heat rejected to the low-temperature reservoir. Since
the engine absorbs 600 J of heat from the high-temperature reservoir in each
cycle, the heat rejected to the low-temperature reservoir can be calculated using
the equation:
Efficiency = Qin −Qout
Qin
where Qin is the heat absorbed from the high-temperature reservoir and Qout
is the heat rejected to the low-temperature reservoir.
28
Substitute the known values into the equation:
0.582 = 600 −Qout
600
0.582 ×600 = 600 −Qout
Qout = 349.2 J
Therefore, the amount of heat rejected to the low-temperature reservoir is
349.2 J.
Question 35
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 400 J of heat from the hot reservoir in each cycle
and exhausts 220 J of heat to the cold reservoir in each cycle.
Calculate the efficiency of this heat engine.
Solution
Step 1: Calculate the net work done by the engine in each cycle. The net work
done by the engine in each cycle is given by the difference between the heat
absorbed from the hot reservoir and the heat exhausted to the cold reservoir:
W=Qin −Qout = 400 J −220 J = 180 J.
Step 2: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
η=W
Qin
.
Substitute the values of Wand Qin into the formula:
η=180 J
400 J = 0.45 = 45%.
Therefore, the efficiency of this heat engine is 45
29
