PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 9
Liberty University
Question 1
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and releases heat
Qcto the cold reservoir. The engine does 3000 J of work while absorbing 5000
J of heat from the hot reservoir. Determine the efficiency of the engine and
discuss how it relates to the second law of thermodynamics.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
Efficiency = Work done by the engine
Heat absorbed from the hot reservoir
Step 2: In this case, the work done by the engine is 3000 J and the heat
absorbed from the hot reservoir is 5000 J. Substituting these values into the
formula, we get:
Efficiency = 3000 J
5000 J
Step 3: Simplifying the expression, we find:
Efficiency = 0.6
Step 4: The efficiency of the engine is 0.6 or 60
Step 5: The second law of thermodynamics states that no heat engine can
be 100
Question 2
Question
A heat engine operates between two reservoirs at temperatures T1and T2, with
T1> T2. The engine absorbs heat Q1from the reservoir at temperature T1and
exhausts heat Q2to the reservoir at temperature T2. Prove that the efficiency
of the engine is given by η= 1 −T2
T1.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
η= 1 −|Q2|
|Q1|.
Step 2: Using the fact that the net work output of the engine is W=
|Q1|−|Q2|, rewrite the efficiency formula as η= 1 −|Q2|
|Q1|= 1 −|Q1|−|W|
|Q1|.
Step 3: Substitute the expressions for |Q1|and |Q2|in terms of the reservoir
temperatures and heat transferred formula: |Q1|=Q1=T1∆Sand |Q2|=
Q2=T2∆S.
Step 4: Now, plug in the expressions for |Q1|and |Q2|and simplify to find
the formula for efficiency:
η= 1 −T1∆S−(T1−T2)∆S
T1∆S= 1 −T2
T1
.
Step 5: Therefore, the efficiency of the engine is given by η= 1 −T2
T1, as
required.
Question 3
Question
A Carnot heat engine operates between a high temperature reservoir at 450◦C
and a low temperature reservoir at 150◦C. The engine absorbs 5000 J of heat
from the high temperature reservoir per cycle. Calculate the efficiency of the
engine and determine the maximum amount of work that can be produced in a
cycle.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
Thigh = 450 + 273.15 = 723.15 K
Tlow = 150 + 273.15 = 423.15 K
2
Step 2: Calculate the efficiency of the engine using the formula for a Carnot
engine, η= 1 −Tlow
Thigh .
η= 1 −423.15
723.15
= 1 −0.5854
= 0.4146 or 41.46%
Step 3: Determine the maximum amount of work that can be produced in
a cycle using the formula Wmax =Qhigh ×η.
Wmax = 5000 J ×0.4146
= 2073 J
Therefore, the efficiency of the engine is 41.46
Question 4
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 4000 J of heat from the hot reservoir and has an
efficiency of 25
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
Efficiency = Work output
Heat input ×100%
Substitute the given values into the formula:
25% = Work output
4000 J
Work output = 0.25 ×4000 J = 1000 J
Step 2: The amount of heat rejected to the cold reservoir can be found using
the first law of thermodynamics:
Heat input = Work output + Heat rejected
Substitute the known values:
4000 J = 1000 J + Heat rejected
Heat rejected = 4000 J −1000 J = 3000 J
Therefore, the engine performs 1000 J of work and rejects 3000 J of heat to
the cold reservoir.
3
Question 5
Question
A heat engine operates between two reservoirs at temperatures Th= 800 K and
Tc= 300 K. The engine has an efficiency of 40%. Calculate the maximum theo-
retical efficiency of the engine if it were operating between these two reservoirs.
Solution
Let Qhbe the heat absorbed from the hot reservoir at temperature Thand Qc
be the heat rejected to the cold reservoir at temperature Tc. The efficiency of a
heat engine is given by:
Efficiency = Work done by the engine
Heat absorbed from the hot reservoir =W
Qh
Given that the efficiency is 40% or 0.40, we have:
0.40 = W
Qh
W= 0.40 ·Qh
From the first law of thermodynamics, we have:
W=Qh−Qc
Therefore:
Qh=W+Qc= 0.40 ·Qh+Qc
0.60 ·Qh=Qc
Now, we can calculate the maximum theoretical efficiency of the engine using
Carnot’s efficiency formula:
Carnot Efficiency = 1 −Tc
Th
Substitute Th= 800 K and Tc= 300 K into the formula:
Carnot Efficiency = 1 −300
800 = 1 −0.375 = 0.625
Therefore, the maximum theoretical efficiency of the engine operating be-
tween these two reservoirs is 62.5%.
4
Question 6
Question
A heat engine operates between two reservoirs at temperatures T1= 500 K and
T2= 300 K. The engine absorbs 5000 J of heat from the hot reservoir in each
cycle and exhausts 3000 J of heat to the cold reservoir in each cycle. Calculate
the efficiency of the engine. Is the engine operating in accordance with the
second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Qout
Qin
where Qin is the heat absorbed from the hot reservoir and Qout is the heat
exhausted to the cold reservoir.
Step 2: Substitute the values into the formula:
Efficiency = 1 −3000 J
5000 J
Step 3: Calculate the efficiency:
Efficiency = 1 −0.6=0.4 = 40%
Step 4: Analyze the efficiency: The efficiency of the engine is 40
Step 5: Check the second law of thermodynamics: The second law of ther-
modynamics states that no engine can have an efficiency of 100
Question 7
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and a
cold reservoir at a temperature of 200 K. The engine absorbs 4000 J of heat from
the hot reservoir in each cycle. (a) Calculate the maximum theoretical efficiency
of this heat engine. (b) If the engine actually operates with an efficiency of 40%
of the maximum theoretical efficiency determined in part (a), calculate the heat
rejected to the cold reservoir in each cycle.
Solution
(a) Let’s denote the efficiency of the heat engine as ηand the temperatures of the
hot and cold reservoirs as THand TC, respectively. The maximum theoretical
5
efficiency of a heat engine is given by the Carnot efficiency, which is determined
by the formula
ηmax = 1 −TC
TH
.
Step 1: Calculate the maximum theoretical efficiency.
ηmax = 1 −200
500 = 1 −2
5=3
5= 0.6.
Therefore, the maximum theoretical efficiency of this heat engine is 60%.
(b) The actual efficiency of the engine is 40% of the maximum theoretical
efficiency, so the actual efficiency is calculated as
η= 0.4×0.6 = 0.24.
The heat input to the engine in each cycle is QH= 4000 J. Using the formula
for efficiency η=W
QHand rearranging, we can find the work output W:
W=η×QH= 0.24 ×4000 = 960 J.
The heat rejected to the cold reservoir can be calculated using the conser-
vation of energy:
QC=QH−W= 4000 −960 = 3040 J.
Therefore, the heat rejected to the cold reservoir in each cycle is 3040 J.
Question 8
Question
A Carnot engine operates between two reservoirs at temperatures THand TC,
with TH> TC. The engine absorbs 6000 J of heat from the high-temperature
reservoir in each cycle and exhausts 3600 J to the low-temperature reservoir.
Calculate the efficiency of the engine and determine the heat rejected to the
environment in each cycle.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −TC
TH
Step 2: We are given that the high-temperature reservoir supplies 6000 J
of heat to the engine in each cycle (QH= 6000 J) and the low-temperature
reservoir absorbs 3600 J of heat from the engine in each cycle (QC= 3600 J).
6
Step 3: Using the efficiency formula, we can calculate the efficiency:
Efficiency = 1 −3600
6000 = 1 −3
5=2
5
Step 4: Therefore, the efficiency of the engine is 2
5or 40
Step 5: Next, we can determine the heat rejected to the environment in each
cycle using the conservation of energy:
QC=QH−W
where Wis the work done by the engine.
Step 6: Since this is a Carnot engine, the work done is given by
W= (1 −Efficiency) ×QH
Step 7: Substituting the given values and efficiency calculated earlier, we
have
W=3
5×6000 = 3600 J
Step 8: Therefore, the heat rejected to the environment in each cycle is
QC=QH−W= 6000 −3600 = 2400 J
Step 9: Hence, the efficiency of the engine is 40
Question 9
Question
A heat engine operates between two reservoirs at temperatures T1and T2(with
T1> T2). The engine absorbs Q1heat from the reservoir at temperature T1and
exhausts Q2heat to the reservoir at temperature T2. Show that the efficiency
of the engine, η, is given by
η= 1 −T2
T1
and explain how this result is related to the second law of thermodynamics.
Solution
Step 1: Recall the definition of efficiency for a heat engine, η:
η=Useful work output
Heat input
Step 2: The useful work output of the engine is equal to the difference
between the heat absorbed from the high-temperature reservoir and the heat
exhausted to the low-temperature reservoir:
Useful work output = Q1−Q2
7
Step 3: Given that the total heat input Q1is consumed by the engine to
perform work Wand reject heat Q2, we have the equation:
Q1=W+Q2
Step 4: Substituting the expression for useful work output and the relation
between Q1and Q2into the efficiency equation, we find:
η=Q1−Q2
Q1
= 1 −Q2
Q1
Step 5: Using the fact that Q1and Q2are related to the temperatures T1
and T2by the Carnot efficiency, we have:
Q2
Q1
=T2
T1
Step 6: Substituting this expression back into the efficiency equation, we
find the efficiency of the engine:
η= 1 −T2
T1
Step 7: The result obtained, η= 1 −T2
T1, shows that the efficiency of a heat
engine is limited by the temperature of the reservoirs it operates between. This
result is related to the second law of thermodynamics, which states that no
engine operating between two reservoirs can have an efficiency of 100
Question 10
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc. If
the efficiency of the engine is η, show that the efficiency of a reversible engine
operating between the same two reservoirs is given by ηrev = 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a Carnot engine is given by η= 1 −Tc
Th,
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir.
Step 2: Let’s consider a reversible engine operating between the same two
reservoirs. By the second law of thermodynamics, the efficiency of a reversible
engine is less than or equal to the efficiency of a Carnot engine operating between
the same two reservoirs. Therefore, ηrev ≤η.
Step 3: We need to show that ηrev = 1 −Tc
Th. Let’s assume that ηrev is equal
to 1 −kTc
Th, where kis a positive constant.
8
Step 4: Since the engine is reversible, the net work done by the engine is
equal to the change in internal energy of the system. Therefore, we have:
Wrev =Qh−Qc= (Th−Tc)S
Step 5: The efficiency of the reversible engine is given by:
ηrev =Wrev
Qh
=(Th−Tc)S
ThS= 1 −Tc
Th
Step 6: Therefore, we have shown that the efficiency of a reversible engine
operating between the same two reservoirs is given by ηrev = 1 −Tc
Th.
Question 11
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 600 J of heat from the hot reservoir and delivers
250 J of work output. Calculate the efficiency of the engine and discuss how it
relates to the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful Work Output
Heat Absorbed
Step 2: Substitute the given values into the formula. Step 3: Use the second law
of thermodynamics to discuss the maximum possible efficiency of the engine.
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful Work Output
Heat Absorbed
Step 2: Substitute the given values into the formula:
Efficiency = 250 J
600 J = 0.4167 or 41.67%
Step 3: According to the second law of thermodynamics, no engine operat-
ing between two reservoirs can be more efficient than a Carnot engine operating
between the same reservoirs. The maximum efficiency of a Carnot engine is
given by:
EfficiencyCarnot = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively. Substituting the values into the formula:
EfficiencyCarnot = 1 −300 K
800 K = 0.625 or 62.5%
Thus, the efficiency of the actual engine (41.67
9
Question 12
Question
A heat engine operates between two reservoirs at temperatures Th= 600 K and
Tc= 300 K. The engine absorbs 1500 J of heat from the hot reservoir in each
cycle. Find the maximum efficiency of this engine.
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency of
a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures Th= 600 K and Tc= 300 K into
the equation:
Efficiency = 1 −300
600
Step 3: Simplify the expression:
Efficiency = 1 −1
2=1
2= 50%
Step 4: The maximum efficiency of this heat engine operating between the
given temperatures is 50
Question 13
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. If the engine absorbs 600 J of heat from the hot reservoir in each cycle,
calculate the maximum possible efficiency of this heat engine. Additionally,
discuss whether this heat engine violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using 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: Th= 600 K, Tc= 300 K, and heat absorbed Qh= 600 J.
Plugging in the values:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
10
Step 2: Discussion of the second law of thermodynamics: The maximum
possible efficiency of a heat engine operating between two reservoirs at temper-
atures Thand Tc, respectively, is given by 1 −Tc
Th. In this case, the efficiency
comes out to be 50
This means that the maximum efficiency that can be achieved by this heat
engine is 50
Therefore, the heat engine described does not violate the second law of
thermodynamics.
Question 14
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 400 K. The engine produces 5000 J of work in each cycle. Determine 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 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Given that Th= 800 K and Tc= 400 K, we have:
Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Thus, the efficiency of the engine is 50
Step 2: Discuss whether this violates the second law of thermodynamics
which states that the efficiency of a heat engine operating between two temper-
ature reservoirs cannot exceed:
EfficiencyCarnot = 1 −Tc
Th
For this engine, the maximum theoretical efficiency (Carnot efficiency) is:
EfficiencyCarnot = 1 −400
800 =1
2= 50%
Since the efficiency of the engine is equal to the Carnot efficiency, it does
not violate the second law of thermodynamics. This means that the engine is
operating at the maximum possible efficiency for this temperature difference.
11
Question 15
Question
A heat engine operates between a hot reservoir at 127◦C and a cold reservoir
at 27◦C. If the engine has an efficiency of 40
Solution
Let’s denote the temperature of the hot reservoir as THand the temperature of
the cold reservoir as TC. The efficiency of a heat engine is given by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the maximum possible efficiency of the first heat engine
operating between 127◦C and 27◦C. Given:
TH= 127◦C + 273 = 400K
TC= 27◦C + 273 = 300K
The efficiency of the first heat engine is:
Efficiency1= 1 −300
400 = 1 −3
4=1
4= 25%
Step 2: Calculate the maximum possible efficiency of the second heat engine
operating between 227◦C and 127◦C. Given:
TH= 227◦C + 273 = 500K
TC= 27◦C + 273 = 300K
The efficiency of the second heat engine is:
Efficiency2= 1 −300
500 = 1 −3
5=2
5= 40%
Therefore, the maximum possible efficiency of a heat engine operating be-
tween reservoirs at 227◦C and 127◦C is 40
Question 16
Question
A heat engine operates between two heat reservoirs at temperatures Thand Tc
(with Th> Tc). The engine absorbs Qhof heat from the hot reservoir and
expels Qcof heat to the cold reservoir in one complete cycle. If the engine is
found to have an efficiency of 30
12
Solution
Let’s denote the efficiency of the heat engine as η(expressed as a decimal). The
efficiency of a heat engine is given by the formula:
η= 1 −Tc
Th
Step 1: We are given that the efficiency of the heat engine is 30
η= 0.30
Step 2: Substitute the known efficiency into the efficiency formula:
0.30 = 1 −Tc
Th
Step 3: Rearrange the equation to solve for Tc/Th:
Tc
Th
= 1 −0.30 = 0.70
Step 4: We are also given that Th> Tc, so Th/Tc>1. Since Tc
Th= 0.70, we
have Th
Tc=1
0.70 = 1.4286.
Therefore, the ratio of Th/Tcis 1.4286.
Question 17
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and
releases heat Qcto the cold reservoir. Let ηbe the efficiency of the engine.
Prove that the efficiency of a Carnot engine is given by η= 1 −Tc
Th.
Solution
Step 1: Recall the Carnot efficiency formula: Recall that the efficiency of a
Carnot engine is given by η= 1 −Qc
Qh.
Step 2: Express Qcand Qhin terms of Tcand Thusing the Carnot engine.
Since the Carnot engine is reversible, by the first law of thermodynamics, we
have Qh
Th
=Qc
Tc
=Wnet
where Wnet is the net work done by the engine.
Step 3: Express Wnet using Tcand Th. Since the Carnot engine is operating
between two temperatures Tcand Th, the net work done is given by
Wnet =Qh−Qc=Qh1−Tc
Th
13
Step 4: Substitute Wnet into the efficiency formula. Substitute the expression
for the net work done into the efficiency formula:
η= 1 −Qc
Qh
= 1 −Qc
Qh
= 1 −Tc
Th
Step 5: Therefore, we have proved that the efficiency of a Carnot engine is
given by η= 1 −Tc
Th.
Question 18
Question
A heat engine operates between a hot reservoir at 500◦C and a cold reservoir
at 25◦C. The engine takes in 600 J of heat from the hot reservoir in each cycle
and exhausts 375 J to the cold reservoir. Calculate the efficiency of the engine
and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Find the efficiency of the engine using the formula for efficiency:
Efficiency = Useful work output
Heat input =Qin −Qout
Qin
where Qin is the heat input and Qout is the heat output.
Step 2: Calculate the heat input and output: Given: Qin = 600 J (heat from
hot reservoir) Qout = 375 J (heat to cold reservoir)
Substitute these values into the formula to find the efficiency:
Efficiency = 600 −375
600
Step 3: Calculate the efficiency:
Efficiency = 225
600 = 0.375 = 37.5%
Step 4: The efficiency of the engine is 37.5
EfficiencyCarnot = 1 −Tcold
Thot
where Thot and Tcold are the temperatures of the hot and cold reservoirs in
Kelvin.
Step 5: Convert the temperatures to Kelvin: Hot reservoir temperature,
Thot = 500 + 273 = 773 K Cold reservoir temperature, Tcold = 25 + 273 = 298 K
Step 6: Calculate the Carnot efficiency:
EfficiencyCarnot = 1 −298
773 ≈0.614
Step 7: Since the efficiency of the engine (37.5
14
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
The engine takes in 1000 J of heat from the hot reservoir and exhausts 600 J to
the cold reservoir during each cycle. Calculate the efficiency of the engine and
discuss whether it violates the second law of thermodynamics.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
Efficiency = Work output
Heat input
Step 2: We are not given the work output directly, but we can calculate it
using the first law of thermodynamics. The first law of thermodynamics states
that the net heat input must equal the net work output. Therefore, the work
output can be found as:
Work output = Heat input −Heat output
Step 3: Given that the engine takes in 1000 J of heat from the hot reservoir
and exhausts 600 J to the cold reservoir during each cycle, we have:
Work output = 1000 J −600 J = 400 J
Step 4: Now we can calculate the efficiency of the engine:
Efficiency = 400 J
1000 J ×100% = 40%
Step 5: The efficiency of the engine is 40
Question 20
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 exhausts Qc
of heat into the cold reservoir. Given that the efficiency of the engine is η, prove
that the efficiency of a reversible engine operating between the same reservoirs
is greater than the given efficiency, η > 1−Tc
Th.
15
Solution
Suppose we have a reversible engine operating between the same reservoirs at
temperatures Thand Tc. Let Q′
hbe the amount of heat absorbed from the hot
reservoir and Q′
cbe the amount of heat exhausted into the cold reservoir by the
reversible engine.
Step 1: Derive the efficiency of the reversible engine. The efficiency
of the reversible engine is given by the Carnot efficiency formula:
η′= 1 −Tc
Th
Step 2: Compare the two efficiencies. We want to show that η > η′.
To do this, we compare the two expressions:
η > η′⇒1−Qc
Qh
>1−Tc
Th
Step 3: Use the definition of efficiency. Recall that the efficiency ηof
a heat engine is given by:
η=Work output
Heat input =Qh−Qc
Qh
Step 4: Substitute the expression for efficiency into the inequality.
Substitute the definition of efficiency into the inequality we derived in step 2:
Qh−Qc
Qh
>1−Tc
Th
Step 5: Simplify the inequality. Simplify the inequality:
Qh
Qh
−Qc
Qh
>1−Tc
Th
Step 6:Further simplify the inequality. Further simplify the inequality:
1−Qc
Qh
>1−Tc
Th
Step 7: Conclusion. From the above steps, we have shown that the
efficiency of a reversible engine operating between the same reservoirs is greater
than the given efficiency, η > 1−Tc
Th.
Question 21
Question
A heat engine operates between a high-temperature reservoir at 500◦Cand a
low-temperature reservoir at 100◦C. The engine absorbs 800 J of heat from
the high-temperature reservoir in each cycle and exhausts 500 J to the low-
temperature reservoir. What is the efficiency of the engine? Is this engine
operating in accordance with the second law of thermodynamics?
16
Solution
Step 1: Determine the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency = 1 −Heat rejected
Heat absorbed
Step 2: Calculate the efficiency of the heat engine using the given values:
Efficiency = 1 −500 J
800 J = 1 −5
8=3
8≈0.375
Therefore, the efficiency of the heat engine is 37.5%.
Step 3: Evaluate whether the engine is operating in accordance with the
second law of thermodynamics. A heat engine’s efficiency is limited by the
Carnot efficiency, which is given by:
Carnot Efficiency = 1 −Tlow
Thigh
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −373 K
773 K ≈0.517
Step 5: Compare the actual efficiency of the heat engine with the Carnot
efficiency to determine if it is operating in accordance with the second law of
thermodynamics. Since the actual efficiency is lower than the Carnot efficiency,
the engine operates in accordance with the second law of thermodynamics.
Question 22
Question
A heat engine operates between two reservoirs at temperatures of 600 K and 300
K. The engine consumes 250 J of heat from the high-temperature reservoir in
each cycle. Calculate the maximum 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 engine using the formula:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir (300 K) and This the temper-
ature of the hot reservoir (600 K).
Step 2: Substitute the values in the efficiency formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
17
Step 3: Comment on whether the operation of this engine violates the second
law of thermodynamics. The efficiency of the engine is 50
Question 23
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 600 J of heat from the hot reservoir per cycle and
exhausts 400 J of heat to the cold reservoir per cycle. Calculate the efficiency
of the engine. Is this engine violating the second law of thermodynamics?
Solution
Let’s denote the heat absorbed from the hot reservoir as QH= 600 J and the
heat exhausted to the cold reservoir as QC= 400 J. The efficiency of the engine
can be calculated using the formula:
Efficiency = 1 −QC
QH
Step 1: Calculate the efficiency of the engine.
Efficiency = 1 −QC
QH
= 1 −400
600
= 1 −2
3
=1
3
Therefore, the efficiency of the engine is 1
3or approximately 33.33
Step 2: Determine whether the engine is violating the second law of ther-
modynamics. The efficiency of the engine is given by Efficiency = 1 −QC
QH,
which implies that the efficiency is less than 1. This means that the engine is
not violating the second law of thermodynamics, as it is impossible to have 100
Therefore, the efficiency of the engine is 1
3and it is not violating the second
law of thermodynamics.
Question 24
Question
A heat engine operates between two reservoirs at temperatures Th= 600 K
and Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir and
exhausts 1200 J to the cold reservoir in each cycle. Determine the efficiency of
this engine and discuss whether it violates the second law of thermodynamics.
18
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for efficiency
(η= 1 −Qc
Qh).
Efficiency, η = 1 −Qc
Qh
Qh= 2000 J
Qc= 1200 J
η= 1 −1200
2000
η= 1 −0.6
η= 0.4
Step 2: Determine whether the efficiency of the engine violates the second
law of thermodynamics. For a heat engine operating between two reservoirs,
the maximum possible efficiency is given by Carnot’s theorem:
ηmax = 1 −Tc
Th
ηmax = 1 −300
600
ηmax = 1 −0.5
ηmax = 0.5
Since the efficiency of the engine we calculated (0.4) is less than the maximum
possible efficiency based on Carnot’s theorem (0.5), the engine does not violate
the second law of thermodynamics. The efficiency is lower than the theoretical
maximum due to irreversibilities in the engine.
Question 25
Question
A Carnot engine operates between a hot reservoir at 500 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 amount of heat rejected
to the cold reservoir in each cycle.
19
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
Carnot efficiency:
Efficiency = 1 −Tcold
Thot
Given:
Thot = 500 K
Tcold = 300 K
Plugging in the values:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Therefore, the efficiency of the Carnot engine is 40%.
Step 2: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
Heat rejected = Heat absorbed −Work done
Since the Carnot engine is reversible, we can use the formula for work done:
Work done = Heat absorbed ·Efficiency
Given:
Heat absorbed = 600 J
Efficiency = 0.4
Plugging in the values:
Work done = 600 J ·0.4 = 240 J
Now, calculate the heat rejected:
Heat rejected = 600 J −240 J = 360 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
360 J.
Question 26
Question
A heat engine operates between a high-temperature reservoir at 600◦Cand a
low-temperature reservoir at 50◦C. The engine has an efficiency of 40
20
Solution
Let’s denote the high-temperature reservoir temperature as TH= 600 K and
the low-temperature reservoir temperature as TL= 50 K. The efficiency of a
heat engine is given by the formula:
Efficiency = 1 −TL
TH
Given that the efficiency of the engine is 40
0.40 = 1 −50
600
Step 1: Calculate the efficiency using the given temperatures.
0.40 = 1 −1
12
1
12 = 1 −0.40 = 0.60
TL
TH
= 0.60
TL= 0.60 ×TH
Step 2: Determine the maximum efficiency based on the second law of
thermodynamics. For a Carnot heat engine, the maximum efficiency is given
by:
Efficiencymax = 1 −TL
TH
Step 3: Substitute the calculated values into the formula for maximum
efficiency.
Efficiencymax = 1 −0.60 ×TH
TH
Efficiencymax = 1 −0.60
Efficiencymax = 0.40
Therefore, the maximum possible efficiency of this engine according to the
second law of thermodynamics is 40
Question 27
Question
A Carnot heat engine operates between two reservoirs at temperatures Th=
500 K and Tc= 300 K. The engine absorbs 500 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.
21
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
where This the temperature of the hot reservoir and Tcis the temperature of
the cold reservoir.
Step 2: Substitute the given values:
Efficiency = 1 −300 K
500 K = 1 −3
5=2
5= 0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Step 4: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
Heat rejected = Efficiency ×Heat absorbed from hot reservoir
Step 5: Substitute the known values:
Heat rejected = 0.4×500 J = 200 J
Step 6: Thus, the amount of heat rejected to the cold reservoir in each cycle
is 200 J.
Question 28
Question
A heat engine operates between two reservoirs at temperatures of 600 K and
300 K. The engine absorbs 10,000 J of heat from the hot reservoir in each cycle
and exhausts 5,000 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine.
Solution
Step 1: Calculate the net work done by the engine.
Net Work = Heat Input −Heat Output
Net Work = 10,000 J −5,000 J = 5,000 J
Step 2: Calculate the efficiency of the engine.
Efficiency = Net Work
Heat Input
Efficiency = 5,000 J
10,000 J = 0.5 = 50%
Therefore, the efficiency of the engine is 50
22
Question 29
Question
A heat engine operates between two reservoirs at temperatures TH= 500 K and
TL= 300 K. The engine absorbs 600 J of heat from the hot reservoir in each
cycle and exhausts 300 J of heat to the cold reservoir in each cycle. Calculate the
efficiency of the engine and determine whether this engine violates the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −TL
TH
where THis the temperature of the hot reservoir and TLis the temperature of
the cold reservoir.
Step 2: Plug in the given values:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
So, the efficiency of the engine is 40
Step 3: Check if the engine violates the second law of thermodynamics.
According to the second law of thermodynamics, the efficiency of a heat engine
cannot be more than the Carnot efficiency, which is given by:
Carnot Efficiency = 1 −TL
TH
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Step 5: Since the efficiency of the engine (40
Question 30
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Determine the maximum efficiency
that this engine could have and comment on the engine’s capability to achieve
this efficiency.
23
Solution
Given: Hot reservoir temperature, Th= 800 K
Cold reservoir temperature, Tc= 300 K
Efficiency of the engine, η= 40% = 0.4
According to the second law of thermodynamics, the maximum efficiency of
a heat engine operating between two reservoirs is given by the Carnot efficiency:
ηmax = 1 −Tc
Th
Step 1: Calculate the Carnot efficiency.
ηmax = 1 −300
800 = 1 −0.375 = 0.625
Step 2: Compare the engine’s actual efficiency with the maximum efficiency.
The engine’s efficiency, η= 0.4, is less than the maximum efficiency, ηmax =
0.625. This means that the engine is not operating at its maximum potential
efficiency.
Step 3: Comment on the engine’s capability to achieve the maximum ef-
ficiency. Since the engine’s efficiency is lower than the Carnot efficiency, it
indicates that the engine is not operating reversibly and is not achieving the
maximum potential efficiency. The engine could potentially improve its effi-
ciency by operating closer to the idealized Carnot cycle, which is only achievable
in theory for reversible processes.
Question 31
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine takes in 5000 J of heat from the hot reservoir
and rejects 3000 J of heat to the cold reservoir in each cycle. Calculate the
efficiency of this Carnot engine.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −Tc
Th
Step 2: Given that the engine takes in 5000 J of heat from the hot reservoir,
the efficiency can be expressed as:
Efficiency = 1 −3000
5000
24
Step 3: Simplify the expression to find the efficiency:
Efficiency = 1 −3
5=2
5
Step 4: Therefore, the efficiency of the Carnot engine is 2
5or 40
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 800 J of heat from the hot reservoir per cycle
and rejects 450 J of heat to the cold reservoir per cycle, calculate the efficiency
of the engine. Is this engine operating in violation of the second law of thermo-
dynamics?
Solution
Step 1: Recall that the efficiency (η) of a heat engine is defined as the ratio of
the work output to the heat input:
η= 1 −Qc
Qh
where Qcis the heat rejected to the cold reservoir and Qhis the heat absorbed
from the hot reservoir.
Step 2: Given Qh= 800 J and Qc= 450 J, we can substitute these values
into the formula to find the efficiency:
η= 1 −450
800
Step 3: Calculate the efficiency:
η= 1 −450
800 = 1 −0.5625 = 0.4375
So, the efficiency of the engine is 43.75
Step 4: The efficiency of the engine is less than 100
Therefore, the efficiency of the engine is 43.75
Question 33
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 1200 J of heat from the hot reservoir in each cycle.
Calculate the maximum possible efficiency of this engine. Is this efficiency
achievable according to the second law of thermodynamics?
25
Solution
Step 1: Calculate the net work done by the engine in each cycle using the first
law of thermodynamics:
Given that the engine absorbs 1200 J of heat from the hot reservoir, ∆Qh=
1200 J. Since the engine operates in a cycle, the net work done by the engine
(Wnet) is the difference between the heat absorbed from the hot reservoir and
the heat rejected to the cold reservoir:
Wnet = ∆Qh−∆Qc
Step 2: Calculate the heat rejected to the cold reservoir:
Using the efficiency of the engine (η), we can express the heat rejected to
the cold reservoir ∆Qcin terms of the heat absorbed from the hot reservoir:
η=Wnet
∆Qh
= 1 −∆Qc
∆Qh
From this expression, we can solve for ∆Qcin terms of ∆Qhand the effi-
ciency.
Step 3: Calculate the maximum possible efficiency of the engine:
The efficiency of a heat engine is given by:
η= 1 −Tc
Th
Substitute the temperatures of the hot and cold reservoirs into the efficiency
equation to find the maximum possible efficiency of the engine.
Step 4: Determine if the calculated efficiency is achievable according to the
second law of thermodynamics:
The efficiency calculated in step 3 represents the maximum possible efficiency
of the engine based on the second law of thermodynamics. If the calculated
efficiency is less than the maximum possible efficiency, then it is achievable.
Otherwise, it is not achievable.
Question 34
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs Qhof heat from the hot reservoir and expels Qc
of heat to the cold reservoir. Given that Qc= 4Qh, calculate the efficiency of
the engine in terms of Thand Tc.
26
Solution
Step 1: Recall the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Heat input =Qh−Qc
Qh
Step 2: Substituting Qc= 4Qhinto the efficiency formula, we have:
Efficiency = Qh−4Qh
Qh
=−3Qh
Qh
=−3
Step 3: The negative sign in the efficiency calculation means that the engine
is not operating efficiently. It indicates that more work is required to operate
the engine than the work produced by the engine.
Therefore, the efficiency of the engine in terms of Thand Tcis −3 .
Question 35
Question
A heat engine operates between two reservoirs at temperatures THand TC, with
TH> TC. The engine has an efficiency of 40
1. The amount of heat expelled to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Solution
Given:
Efficiency = W
QH= 40% = 0.4
Heat absorbed QH= 6000 J
We are asked to find:
1. Heat expelled QC
2. Work done W
Step 1: Find the Heat Expelled QC
Efficiency = W
QH
0.4 = W
6000
W= 0.4×6000
W= 2400 J
27
Question 2
Question
A heat engine operates between two reservoirs at temperatures T1and T2, with
T1> T2. The engine absorbs heat Q1from the reservoir at temperature T1and
exhausts heat Q2to the reservoir at temperature T2. Prove that the efficiency
of the engine is given by η= 1 −T2
T1.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
η= 1 −|Q2|
|Q1|.
Step 2: Using the fact that the net work output of the engine is W=
|Q1|−|Q2|, rewrite the efficiency formula as η= 1 −|Q2|
|Q1|= 1 −|Q1|−|W|
|Q1|.
Step 3: Substitute the expressions for |Q1|and |Q2|in terms of the reservoir
temperatures and heat transferred formula: |Q1|=Q1=T1∆Sand |Q2|=
Q2=T2∆S.
Step 4: Now, plug in the expressions for |Q1|and |Q2|and simplify to find
the formula for efficiency:
η= 1 −T1∆S−(T1−T2)∆S
T1∆S= 1 −T2
T1
.
Step 5: Therefore, the efficiency of the engine is given by η= 1 −T2
T1, as
required.
Question 3
Question
A Carnot heat engine operates between a high temperature reservoir at 450◦C
and a low temperature reservoir at 150◦C. The engine absorbs 5000 J of heat
from the high temperature reservoir per cycle. Calculate the efficiency of the
engine and determine the maximum amount of work that can be produced in a
cycle.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
Thigh = 450 + 273.15 = 723.15 K
Tlow = 150 + 273.15 = 423.15 K
2
Step 2: Calculate the efficiency of the engine using the formula for a Carnot
engine, η= 1 −Tlow
Thigh .
η= 1 −423.15
723.15
= 1 −0.5854
= 0.4146 or 41.46%
Step 3: Determine the maximum amount of work that can be produced in
a cycle using the formula Wmax =Qhigh ×η.
Wmax = 5000 J ×0.4146
= 2073 J
Therefore, the efficiency of the engine is 41.46
Question 4
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 4000 J of heat from the hot reservoir and has an
efficiency of 25
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
Efficiency = Work output
Heat input ×100%
Substitute the given values into the formula:
25% = Work output
4000 J
Work output = 0.25 ×4000 J = 1000 J
Step 2: The amount of heat rejected to the cold reservoir can be found using
the first law of thermodynamics:
Heat input = Work output + Heat rejected
Substitute the known values:
4000 J = 1000 J + Heat rejected
Heat rejected = 4000 J −1000 J = 3000 J
Therefore, the engine performs 1000 J of work and rejects 3000 J of heat to
the cold reservoir.
3
Question 5
Question
A heat engine operates between two reservoirs at temperatures Th= 800 K and
Tc= 300 K. The engine has an efficiency of 40%. Calculate the maximum theo-
retical efficiency of the engine if it were operating between these two reservoirs.
Solution
Let Qhbe the heat absorbed from the hot reservoir at temperature Thand Qc
be the heat rejected to the cold reservoir at temperature Tc. The efficiency of a
heat engine is given by:
Efficiency = Work done by the engine
Heat absorbed from the hot reservoir =W
Qh
Given that the efficiency is 40% or 0.40, we have:
0.40 = W
Qh
W= 0.40 ·Qh
From the first law of thermodynamics, we have:
W=Qh−Qc
Therefore:
Qh=W+Qc= 0.40 ·Qh+Qc
0.60 ·Qh=Qc
Now, we can calculate the maximum theoretical efficiency of the engine using
Carnot’s efficiency formula:
Carnot Efficiency = 1 −Tc
Th
Substitute Th= 800 K and Tc= 300 K into the formula:
Carnot Efficiency = 1 −300
800 = 1 −0.375 = 0.625
Therefore, the maximum theoretical efficiency of the engine operating be-
tween these two reservoirs is 62.5%.
4
Question 6
Question
A heat engine operates between two reservoirs at temperatures T1= 500 K and
T2= 300 K. The engine absorbs 5000 J of heat from the hot reservoir in each
cycle and exhausts 3000 J of heat to the cold reservoir in each cycle. Calculate
the efficiency of the engine. Is the engine operating in accordance with the
second law of thermodynamics?
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Qout
Qin
where Qin is the heat absorbed from the hot reservoir and Qout is the heat
exhausted to the cold reservoir.
Step 2: Substitute the values into the formula:
Efficiency = 1 −3000 J
5000 J
Step 3: Calculate the efficiency:
Efficiency = 1 −0.6=0.4 = 40%
Step 4: Analyze the efficiency: The efficiency of the engine is 40
Step 5: Check the second law of thermodynamics: The second law of ther-
modynamics states that no engine can have an efficiency of 100
Question 7
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and a
cold reservoir at a temperature of 200 K. The engine absorbs 4000 J of heat from
the hot reservoir in each cycle. (a) Calculate the maximum theoretical efficiency
of this heat engine. (b) If the engine actually operates with an efficiency of 40%
of the maximum theoretical efficiency determined in part (a), calculate the heat
rejected to the cold reservoir in each cycle.
Solution
(a) Let’s denote the efficiency of the heat engine as ηand the temperatures of the
hot and cold reservoirs as THand TC, respectively. The maximum theoretical
5
efficiency of a heat engine is given by the Carnot efficiency, which is determined
by the formula
ηmax = 1 −TC
TH
.
Step 1: Calculate the maximum theoretical efficiency.
ηmax = 1 −200
500 = 1 −2
5=3
5= 0.6.
Therefore, the maximum theoretical efficiency of this heat engine is 60%.
(b) The actual efficiency of the engine is 40% of the maximum theoretical
efficiency, so the actual efficiency is calculated as
η= 0.4×0.6 = 0.24.
The heat input to the engine in each cycle is QH= 4000 J. Using the formula
for efficiency η=W
QHand rearranging, we can find the work output W:
W=η×QH= 0.24 ×4000 = 960 J.
The heat rejected to the cold reservoir can be calculated using the conser-
vation of energy:
QC=QH−W= 4000 −960 = 3040 J.
Therefore, the heat rejected to the cold reservoir in each cycle is 3040 J.
Question 8
Question
A Carnot engine operates between two reservoirs at temperatures THand TC,
with TH> TC. The engine absorbs 6000 J of heat from the high-temperature
reservoir in each cycle and exhausts 3600 J to the low-temperature reservoir.
Calculate the efficiency of the engine and determine the heat rejected to the
environment in each cycle.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −TC
TH
Step 2: We are given that the high-temperature reservoir supplies 6000 J
of heat to the engine in each cycle (QH= 6000 J) and the low-temperature
reservoir absorbs 3600 J of heat from the engine in each cycle (QC= 3600 J).
6
Step 3: Using the efficiency formula, we can calculate the efficiency:
Efficiency = 1 −3600
6000 = 1 −3
5=2
5
Step 4: Therefore, the efficiency of the engine is 2
5or 40
Step 5: Next, we can determine the heat rejected to the environment in each
cycle using the conservation of energy:
QC=QH−W
where Wis the work done by the engine.
Step 6: Since this is a Carnot engine, the work done is given by
W= (1 −Efficiency) ×QH
Step 7: Substituting the given values and efficiency calculated earlier, we
have
W=3
5×6000 = 3600 J
Step 8: Therefore, the heat rejected to the environment in each cycle is
QC=QH−W= 6000 −3600 = 2400 J
Step 9: Hence, the efficiency of the engine is 40
Question 9
Question
A heat engine operates between two reservoirs at temperatures T1and T2(with
T1> T2). The engine absorbs Q1heat from the reservoir at temperature T1and
exhausts Q2heat to the reservoir at temperature T2. Show that the efficiency
of the engine, η, is given by
η= 1 −T2
T1
and explain how this result is related to the second law of thermodynamics.
Solution
Step 1: Recall the definition of efficiency for a heat engine, η:
η=Useful work output
Heat input
Step 2: The useful work output of the engine is equal to the difference
between the heat absorbed from the high-temperature reservoir and the heat
exhausted to the low-temperature reservoir:
Useful work output = Q1−Q2
7
Step 3: Given that the total heat input Q1is consumed by the engine to
perform work Wand reject heat Q2, we have the equation:
Q1=W+Q2
Step 4: Substituting the expression for useful work output and the relation
between Q1and Q2into the efficiency equation, we find:
η=Q1−Q2
Q1
= 1 −Q2
Q1
Step 5: Using the fact that Q1and Q2are related to the temperatures T1
and T2by the Carnot efficiency, we have:
Q2
Q1
=T2
T1
Step 6: Substituting this expression back into the efficiency equation, we
find the efficiency of the engine:
η= 1 −T2
T1
Step 7: The result obtained, η= 1 −T2
T1, shows that the efficiency of a heat
engine is limited by the temperature of the reservoirs it operates between. This
result is related to the second law of thermodynamics, which states that no
engine operating between two reservoirs can have an efficiency of 100
Question 10
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc. If
the efficiency of the engine is η, show that the efficiency of a reversible engine
operating between the same two reservoirs is given by ηrev = 1 −Tc
Th.
Solution
Step 1: Recall that the efficiency of a Carnot engine is given by η= 1 −Tc
Th,
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir.
Step 2: Let’s consider a reversible engine operating between the same two
reservoirs. By the second law of thermodynamics, the efficiency of a reversible
engine is less than or equal to the efficiency of a Carnot engine operating between
the same two reservoirs. Therefore, ηrev ≤η.
Step 3: We need to show that ηrev = 1 −Tc
Th. Let’s assume that ηrev is equal
to 1 −kTc
Th, where kis a positive constant.
8
Step 4: Since the engine is reversible, the net work done by the engine is
equal to the change in internal energy of the system. Therefore, we have:
Wrev =Qh−Qc= (Th−Tc)S
Step 5: The efficiency of the reversible engine is given by:
ηrev =Wrev
Qh
=(Th−Tc)S
ThS= 1 −Tc
Th
Step 6: Therefore, we have shown that the efficiency of a reversible engine
operating between the same two reservoirs is given by ηrev = 1 −Tc
Th.
Question 11
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 600 J of heat from the hot reservoir and delivers
250 J of work output. Calculate the efficiency of the engine and discuss how it
relates to the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful Work Output
Heat Absorbed
Step 2: Substitute the given values into the formula. Step 3: Use the second law
of thermodynamics to discuss the maximum possible efficiency of the engine.
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = Useful Work Output
Heat Absorbed
Step 2: Substitute the given values into the formula:
Efficiency = 250 J
600 J = 0.4167 or 41.67%
Step 3: According to the second law of thermodynamics, no engine operat-
ing between two reservoirs can be more efficient than a Carnot engine operating
between the same reservoirs. The maximum efficiency of a Carnot engine is
given by:
EfficiencyCarnot = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively. Substituting the values into the formula:
EfficiencyCarnot = 1 −300 K
800 K = 0.625 or 62.5%
Thus, the efficiency of the actual engine (41.67
9
Question 12
Question
A heat engine operates between two reservoirs at temperatures Th= 600 K and
Tc= 300 K. The engine absorbs 1500 J of heat from the hot reservoir in each
cycle. Find the maximum efficiency of this engine.
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency of
a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures Th= 600 K and Tc= 300 K into
the equation:
Efficiency = 1 −300
600
Step 3: Simplify the expression:
Efficiency = 1 −1
2=1
2= 50%
Step 4: The maximum efficiency of this heat engine operating between the
given temperatures is 50
Question 13
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. If the engine absorbs 600 J of heat from the hot reservoir in each cycle,
calculate the maximum possible efficiency of this heat engine. Additionally,
discuss whether this heat engine violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the heat engine using 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: Th= 600 K, Tc= 300 K, and heat absorbed Qh= 600 J.
Plugging in the values:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
10
Step 2: Discussion of the second law of thermodynamics: The maximum
possible efficiency of a heat engine operating between two reservoirs at temper-
atures Thand Tc, respectively, is given by 1 −Tc
Th. In this case, the efficiency
comes out to be 50
This means that the maximum efficiency that can be achieved by this heat
engine is 50
Therefore, the heat engine described does not violate the second law of
thermodynamics.
Question 14
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 400 K. The engine produces 5000 J of work in each cycle. Determine 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 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Given that Th= 800 K and Tc= 400 K, we have:
Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Thus, the efficiency of the engine is 50
Step 2: Discuss whether this violates the second law of thermodynamics
which states that the efficiency of a heat engine operating between two temper-
ature reservoirs cannot exceed:
EfficiencyCarnot = 1 −Tc
Th
For this engine, the maximum theoretical efficiency (Carnot efficiency) is:
EfficiencyCarnot = 1 −400
800 =1
2= 50%
Since the efficiency of the engine is equal to the Carnot efficiency, it does
not violate the second law of thermodynamics. This means that the engine is
operating at the maximum possible efficiency for this temperature difference.
11
Question 15
Question
A heat engine operates between a hot reservoir at 127◦C and a cold reservoir
at 27◦C. If the engine has an efficiency of 40
Solution
Let’s denote the temperature of the hot reservoir as THand the temperature of
the cold reservoir as TC. The efficiency of a heat engine is given by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the maximum possible efficiency of the first heat engine
operating between 127◦C and 27◦C. Given:
TH= 127◦C + 273 = 400K
TC= 27◦C + 273 = 300K
The efficiency of the first heat engine is:
Efficiency1= 1 −300
400 = 1 −3
4=1
4= 25%
Step 2: Calculate the maximum possible efficiency of the second heat engine
operating between 227◦C and 127◦C. Given:
TH= 227◦C + 273 = 500K
TC= 27◦C + 273 = 300K
The efficiency of the second heat engine is:
Efficiency2= 1 −300
500 = 1 −3
5=2
5= 40%
Therefore, the maximum possible efficiency of a heat engine operating be-
tween reservoirs at 227◦C and 127◦C is 40
Question 16
Question
A heat engine operates between two heat reservoirs at temperatures Thand Tc
(with Th> Tc). The engine absorbs Qhof heat from the hot reservoir and
expels Qcof heat to the cold reservoir in one complete cycle. If the engine is
found to have an efficiency of 30
12
Solution
Let’s denote the efficiency of the heat engine as η(expressed as a decimal). The
efficiency of a heat engine is given by the formula:
η= 1 −Tc
Th
Step 1: We are given that the efficiency of the heat engine is 30
η= 0.30
Step 2: Substitute the known efficiency into the efficiency formula:
0.30 = 1 −Tc
Th
Step 3: Rearrange the equation to solve for Tc/Th:
Tc
Th
= 1 −0.30 = 0.70
Step 4: We are also given that Th> Tc, so Th/Tc>1. Since Tc
Th= 0.70, we
have Th
Tc=1
0.70 = 1.4286.
Therefore, the ratio of Th/Tcis 1.4286.
Question 17
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and
releases heat Qcto the cold reservoir. Let ηbe the efficiency of the engine.
Prove that the efficiency of a Carnot engine is given by η= 1 −Tc
Th.
Solution
Step 1: Recall the Carnot efficiency formula: Recall that the efficiency of a
Carnot engine is given by η= 1 −Qc
Qh.
Step 2: Express Qcand Qhin terms of Tcand Thusing the Carnot engine.
Since the Carnot engine is reversible, by the first law of thermodynamics, we
have Qh
Th
=Qc
Tc
=Wnet
where Wnet is the net work done by the engine.
Step 3: Express Wnet using Tcand Th. Since the Carnot engine is operating
between two temperatures Tcand Th, the net work done is given by
Wnet =Qh−Qc=Qh1−Tc
Th
13
Step 4: Substitute Wnet into the efficiency formula. Substitute the expression
for the net work done into the efficiency formula:
η= 1 −Qc
Qh
= 1 −Qc
Qh
= 1 −Tc
Th
Step 5: Therefore, we have proved that the efficiency of a Carnot engine is
given by η= 1 −Tc
Th.
Question 18
Question
A heat engine operates between a hot reservoir at 500◦C and a cold reservoir
at 25◦C. The engine takes in 600 J of heat from the hot reservoir in each cycle
and exhausts 375 J to the cold reservoir. Calculate the efficiency of the engine
and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Find the efficiency of the engine using the formula for efficiency:
Efficiency = Useful work output
Heat input =Qin −Qout
Qin
where Qin is the heat input and Qout is the heat output.
Step 2: Calculate the heat input and output: Given: Qin = 600 J (heat from
hot reservoir) Qout = 375 J (heat to cold reservoir)
Substitute these values into the formula to find the efficiency:
Efficiency = 600 −375
600
Step 3: Calculate the efficiency:
Efficiency = 225
600 = 0.375 = 37.5%
Step 4: The efficiency of the engine is 37.5
EfficiencyCarnot = 1 −Tcold
Thot
where Thot and Tcold are the temperatures of the hot and cold reservoirs in
Kelvin.
Step 5: Convert the temperatures to Kelvin: Hot reservoir temperature,
Thot = 500 + 273 = 773 K Cold reservoir temperature, Tcold = 25 + 273 = 298 K
Step 6: Calculate the Carnot efficiency:
EfficiencyCarnot = 1 −298
773 ≈0.614
Step 7: Since the efficiency of the engine (37.5
14
Question 19
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
The engine takes in 1000 J of heat from the hot reservoir and exhausts 600 J to
the cold reservoir during each cycle. Calculate the efficiency of the engine and
discuss whether it violates the second law of thermodynamics.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
Efficiency = Work output
Heat input
Step 2: We are not given the work output directly, but we can calculate it
using the first law of thermodynamics. The first law of thermodynamics states
that the net heat input must equal the net work output. Therefore, the work
output can be found as:
Work output = Heat input −Heat output
Step 3: Given that the engine takes in 1000 J of heat from the hot reservoir
and exhausts 600 J to the cold reservoir during each cycle, we have:
Work output = 1000 J −600 J = 400 J
Step 4: Now we can calculate the efficiency of the engine:
Efficiency = 400 J
1000 J ×100% = 40%
Step 5: The efficiency of the engine is 40
Question 20
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 exhausts Qc
of heat into the cold reservoir. Given that the efficiency of the engine is η, prove
that the efficiency of a reversible engine operating between the same reservoirs
is greater than the given efficiency, η > 1−Tc
Th.
15
Solution
Suppose we have a reversible engine operating between the same reservoirs at
temperatures Thand Tc. Let Q′
hbe the amount of heat absorbed from the hot
reservoir and Q′
cbe the amount of heat exhausted into the cold reservoir by the
reversible engine.
Step 1: Derive the efficiency of the reversible engine. The efficiency
of the reversible engine is given by the Carnot efficiency formula:
η′= 1 −Tc
Th
Step 2: Compare the two efficiencies. We want to show that η > η′.
To do this, we compare the two expressions:
η > η′⇒1−Qc
Qh
>1−Tc
Th
Step 3: Use the definition of efficiency. Recall that the efficiency ηof
a heat engine is given by:
η=Work output
Heat input =Qh−Qc
Qh
Step 4: Substitute the expression for efficiency into the inequality.
Substitute the definition of efficiency into the inequality we derived in step 2:
Qh−Qc
Qh
>1−Tc
Th
Step 5: Simplify the inequality. Simplify the inequality:
Qh
Qh
−Qc
Qh
>1−Tc
Th
Step 6:Further simplify the inequality. Further simplify the inequality:
1−Qc
Qh
>1−Tc
Th
Step 7: Conclusion. From the above steps, we have shown that the
efficiency of a reversible engine operating between the same reservoirs is greater
than the given efficiency, η > 1−Tc
Th.
Question 21
Question
A heat engine operates between a high-temperature reservoir at 500◦Cand a
low-temperature reservoir at 100◦C. The engine absorbs 800 J of heat from
the high-temperature reservoir in each cycle and exhausts 500 J to the low-
temperature reservoir. What is the efficiency of the engine? Is this engine
operating in accordance with the second law of thermodynamics?
16
Solution
Step 1: Determine the efficiency of the heat engine using the formula for effi-
ciency:
Efficiency = 1 −Heat rejected
Heat absorbed
Step 2: Calculate the efficiency of the heat engine using the given values:
Efficiency = 1 −500 J
800 J = 1 −5
8=3
8≈0.375
Therefore, the efficiency of the heat engine is 37.5%.
Step 3: Evaluate whether the engine is operating in accordance with the
second law of thermodynamics. A heat engine’s efficiency is limited by the
Carnot efficiency, which is given by:
Carnot Efficiency = 1 −Tlow
Thigh
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −373 K
773 K ≈0.517
Step 5: Compare the actual efficiency of the heat engine with the Carnot
efficiency to determine if it is operating in accordance with the second law of
thermodynamics. Since the actual efficiency is lower than the Carnot efficiency,
the engine operates in accordance with the second law of thermodynamics.
Question 22
Question
A heat engine operates between two reservoirs at temperatures of 600 K and 300
K. The engine consumes 250 J of heat from the high-temperature reservoir in
each cycle. Calculate the maximum 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 engine using the formula:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir (300 K) and This the temper-
ature of the hot reservoir (600 K).
Step 2: Substitute the values in the efficiency formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
17
Step 3: Comment on whether the operation of this engine violates the second
law of thermodynamics. The efficiency of the engine is 50
Question 23
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 600 J of heat from the hot reservoir per cycle and
exhausts 400 J of heat to the cold reservoir per cycle. Calculate the efficiency
of the engine. Is this engine violating the second law of thermodynamics?
Solution
Let’s denote the heat absorbed from the hot reservoir as QH= 600 J and the
heat exhausted to the cold reservoir as QC= 400 J. The efficiency of the engine
can be calculated using the formula:
Efficiency = 1 −QC
QH
Step 1: Calculate the efficiency of the engine.
Efficiency = 1 −QC
QH
= 1 −400
600
= 1 −2
3
=1
3
Therefore, the efficiency of the engine is 1
3or approximately 33.33
Step 2: Determine whether the engine is violating the second law of ther-
modynamics. The efficiency of the engine is given by Efficiency = 1 −QC
QH,
which implies that the efficiency is less than 1. This means that the engine is
not violating the second law of thermodynamics, as it is impossible to have 100
Therefore, the efficiency of the engine is 1
3and it is not violating the second
law of thermodynamics.
Question 24
Question
A heat engine operates between two reservoirs at temperatures Th= 600 K
and Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir and
exhausts 1200 J to the cold reservoir in each cycle. Determine the efficiency of
this engine and discuss whether it violates the second law of thermodynamics.
18
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for efficiency
(η= 1 −Qc
Qh).
Efficiency, η = 1 −Qc
Qh
Qh= 2000 J
Qc= 1200 J
η= 1 −1200
2000
η= 1 −0.6
η= 0.4
Step 2: Determine whether the efficiency of the engine violates the second
law of thermodynamics. For a heat engine operating between two reservoirs,
the maximum possible efficiency is given by Carnot’s theorem:
ηmax = 1 −Tc
Th
ηmax = 1 −300
600
ηmax = 1 −0.5
ηmax = 0.5
Since the efficiency of the engine we calculated (0.4) is less than the maximum
possible efficiency based on Carnot’s theorem (0.5), the engine does not violate
the second law of thermodynamics. The efficiency is lower than the theoretical
maximum due to irreversibilities in the engine.
Question 25
Question
A Carnot engine operates between a hot reservoir at 500 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 amount of heat rejected
to the cold reservoir in each cycle.
19
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
Carnot efficiency:
Efficiency = 1 −Tcold
Thot
Given:
Thot = 500 K
Tcold = 300 K
Plugging in the values:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Therefore, the efficiency of the Carnot engine is 40%.
Step 2: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
Heat rejected = Heat absorbed −Work done
Since the Carnot engine is reversible, we can use the formula for work done:
Work done = Heat absorbed ·Efficiency
Given:
Heat absorbed = 600 J
Efficiency = 0.4
Plugging in the values:
Work done = 600 J ·0.4 = 240 J
Now, calculate the heat rejected:
Heat rejected = 600 J −240 J = 360 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
360 J.
Question 26
Question
A heat engine operates between a high-temperature reservoir at 600◦Cand a
low-temperature reservoir at 50◦C. The engine has an efficiency of 40
20
Solution
Let’s denote the high-temperature reservoir temperature as TH= 600 K and
the low-temperature reservoir temperature as TL= 50 K. The efficiency of a
heat engine is given by the formula:
Efficiency = 1 −TL
TH
Given that the efficiency of the engine is 40
0.40 = 1 −50
600
Step 1: Calculate the efficiency using the given temperatures.
0.40 = 1 −1
12
1
12 = 1 −0.40 = 0.60
TL
TH
= 0.60
TL= 0.60 ×TH
Step 2: Determine the maximum efficiency based on the second law of
thermodynamics. For a Carnot heat engine, the maximum efficiency is given
by:
Efficiencymax = 1 −TL
TH
Step 3: Substitute the calculated values into the formula for maximum
efficiency.
Efficiencymax = 1 −0.60 ×TH
TH
Efficiencymax = 1 −0.60
Efficiencymax = 0.40
Therefore, the maximum possible efficiency of this engine according to the
second law of thermodynamics is 40
Question 27
Question
A Carnot heat engine operates between two reservoirs at temperatures Th=
500 K and Tc= 300 K. The engine absorbs 500 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.
21
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −Tc
Th
where This the temperature of the hot reservoir and Tcis the temperature of
the cold reservoir.
Step 2: Substitute the given values:
Efficiency = 1 −300 K
500 K = 1 −3
5=2
5= 0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Step 4: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
Heat rejected = Efficiency ×Heat absorbed from hot reservoir
Step 5: Substitute the known values:
Heat rejected = 0.4×500 J = 200 J
Step 6: Thus, the amount of heat rejected to the cold reservoir in each cycle
is 200 J.
Question 28
Question
A heat engine operates between two reservoirs at temperatures of 600 K and
300 K. The engine absorbs 10,000 J of heat from the hot reservoir in each cycle
and exhausts 5,000 J to the cold reservoir in each cycle. Calculate the efficiency
of the engine.
Solution
Step 1: Calculate the net work done by the engine.
Net Work = Heat Input −Heat Output
Net Work = 10,000 J −5,000 J = 5,000 J
Step 2: Calculate the efficiency of the engine.
Efficiency = Net Work
Heat Input
Efficiency = 5,000 J
10,000 J = 0.5 = 50%
Therefore, the efficiency of the engine is 50
22
Question 29
Question
A heat engine operates between two reservoirs at temperatures TH= 500 K and
TL= 300 K. The engine absorbs 600 J of heat from the hot reservoir in each
cycle and exhausts 300 J of heat to the cold reservoir in each cycle. Calculate the
efficiency of the engine and determine whether this engine violates the second
law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −TL
TH
where THis the temperature of the hot reservoir and TLis the temperature of
the cold reservoir.
Step 2: Plug in the given values:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
So, the efficiency of the engine is 40
Step 3: Check if the engine violates the second law of thermodynamics.
According to the second law of thermodynamics, the efficiency of a heat engine
cannot be more than the Carnot efficiency, which is given by:
Carnot Efficiency = 1 −TL
TH
Step 4: Calculate the Carnot efficiency for the given temperatures:
Carnot Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Step 5: Since the efficiency of the engine (40
Question 30
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Determine the maximum efficiency
that this engine could have and comment on the engine’s capability to achieve
this efficiency.
23
Solution
Given: Hot reservoir temperature, Th= 800 K
Cold reservoir temperature, Tc= 300 K
Efficiency of the engine, η= 40% = 0.4
According to the second law of thermodynamics, the maximum efficiency of
a heat engine operating between two reservoirs is given by the Carnot efficiency:
ηmax = 1 −Tc
Th
Step 1: Calculate the Carnot efficiency.
ηmax = 1 −300
800 = 1 −0.375 = 0.625
Step 2: Compare the engine’s actual efficiency with the maximum efficiency.
The engine’s efficiency, η= 0.4, is less than the maximum efficiency, ηmax =
0.625. This means that the engine is not operating at its maximum potential
efficiency.
Step 3: Comment on the engine’s capability to achieve the maximum ef-
ficiency. Since the engine’s efficiency is lower than the Carnot efficiency, it
indicates that the engine is not operating reversibly and is not achieving the
maximum potential efficiency. The engine could potentially improve its effi-
ciency by operating closer to the idealized Carnot cycle, which is only achievable
in theory for reversible processes.
Question 31
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine takes in 5000 J of heat from the hot reservoir
and rejects 3000 J of heat to the cold reservoir in each cycle. Calculate the
efficiency of this Carnot engine.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −Tc
Th
Step 2: Given that the engine takes in 5000 J of heat from the hot reservoir,
the efficiency can be expressed as:
Efficiency = 1 −3000
5000
24
Step 3: Simplify the expression to find the efficiency:
Efficiency = 1 −3
5=2
5
Step 4: Therefore, the efficiency of the Carnot engine is 2
5or 40
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 800 J of heat from the hot reservoir per cycle
and rejects 450 J of heat to the cold reservoir per cycle, calculate the efficiency
of the engine. Is this engine operating in violation of the second law of thermo-
dynamics?
Solution
Step 1: Recall that the efficiency (η) of a heat engine is defined as the ratio of
the work output to the heat input:
η= 1 −Qc
Qh
where Qcis the heat rejected to the cold reservoir and Qhis the heat absorbed
from the hot reservoir.
Step 2: Given Qh= 800 J and Qc= 450 J, we can substitute these values
into the formula to find the efficiency:
η= 1 −450
800
Step 3: Calculate the efficiency:
η= 1 −450
800 = 1 −0.5625 = 0.4375
So, the efficiency of the engine is 43.75
Step 4: The efficiency of the engine is less than 100
Therefore, the efficiency of the engine is 43.75
Question 33
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine absorbs 1200 J of heat from the hot reservoir in each cycle.
Calculate the maximum possible efficiency of this engine. Is this efficiency
achievable according to the second law of thermodynamics?
25
Solution
Step 1: Calculate the net work done by the engine in each cycle using the first
law of thermodynamics:
Given that the engine absorbs 1200 J of heat from the hot reservoir, ∆Qh=
1200 J. Since the engine operates in a cycle, the net work done by the engine
(Wnet) is the difference between the heat absorbed from the hot reservoir and
the heat rejected to the cold reservoir:
Wnet = ∆Qh−∆Qc
Step 2: Calculate the heat rejected to the cold reservoir:
Using the efficiency of the engine (η), we can express the heat rejected to
the cold reservoir ∆Qcin terms of the heat absorbed from the hot reservoir:
η=Wnet
∆Qh
= 1 −∆Qc
∆Qh
From this expression, we can solve for ∆Qcin terms of ∆Qhand the effi-
ciency.
Step 3: Calculate the maximum possible efficiency of the engine:
The efficiency of a heat engine is given by:
η= 1 −Tc
Th
Substitute the temperatures of the hot and cold reservoirs into the efficiency
equation to find the maximum possible efficiency of the engine.
Step 4: Determine if the calculated efficiency is achievable according to the
second law of thermodynamics:
The efficiency calculated in step 3 represents the maximum possible efficiency
of the engine based on the second law of thermodynamics. If the calculated
efficiency is less than the maximum possible efficiency, then it is achievable.
Otherwise, it is not achievable.
Question 34
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs Qhof heat from the hot reservoir and expels Qc
of heat to the cold reservoir. Given that Qc= 4Qh, calculate the efficiency of
the engine in terms of Thand Tc.
26
Solution
Step 1: Recall the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Heat input =Qh−Qc
Qh
Step 2: Substituting Qc= 4Qhinto the efficiency formula, we have:
Efficiency = Qh−4Qh
Qh
=−3Qh
Qh
=−3
Step 3: The negative sign in the efficiency calculation means that the engine
is not operating efficiently. It indicates that more work is required to operate
the engine than the work produced by the engine.
Therefore, the efficiency of the engine in terms of Thand Tcis −3 .
Question 35
Question
A heat engine operates between two reservoirs at temperatures THand TC, with
TH> TC. The engine has an efficiency of 40
1. The amount of heat expelled to the cold reservoir in each cycle.
2. The work done by the engine in each cycle.
Solution
Given:
Efficiency = W
QH= 40% = 0.4
Heat absorbed QH= 6000 J
We are asked to find:
1. Heat expelled QC
2. Work done W
Step 1: Find the Heat Expelled QC
Efficiency = W
QH
0.4 = W
6000
W= 0.4×6000
W= 2400 J
27
Since efficiency is the ratio of work done to heat absorbed, the rest of the
heat is expelled:
W+QC=QH
2400 + QC= 6000
QC= 6000 −2400
QC= 3600 J
Therefore, the amount of heat expelled to the cold reservoir in each cycle is
3600 J.
Step 2: Find the Work Done WWe have already found the work done
by the engine:
W= 2400 J
Therefore, the work done by the engine in each cycle is 2400 J.
28
