PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 8
Liberty University
Question 1
Question
A heat engine operates between a hot reservoir at 127◦C and a cold reservoir
at 27◦C. The engine follows the Carnot cycle and has an efficiency of 25
Solution
Step 1: Let’s first express the efficiency of the Carnot engine in terms of the
reservoir temperatures. The efficiency of a Carnot engine is given by:
Efficiency = 1 −TC
TH
where TCis the absolute temperature of the cold reservoir and THis the absolute
temperature of the hot reservoir.
Step 2: Given that the efficiency of the first engine is 25
0.25 = 1 −TC
TH
Step 3: Substitute the temperatures in Celsius to absolute temperatures
using the relation T(Kelvin) = T(Celsius) + 273.15. For the first engine:
0.25 = 1 −27 + 273.15
127 + 273.15
Step 4: Solve for TC/THto find the ratio of the cold to hot reservoir tem-
peratures for the first engine.
Step 5: Next, let’s determine the cold reservoir temperature for the second
engine operating between 227◦C and 27◦C in terms of absolute temperature in
Kelvin.
Step 6: Use the distinction of efficiency between the two engines to calculate
the efficiency of the second Carnot engine using the formula:
Efficiency = 1 −T′
C
T′
H
where T′
Cand T′
Hare the absolute temperatures of the cold and hot reservoirs
for the second engine respectively.
Step 7: Substitute T′
Cand T′
Hwith the respective temperatures in Kelvin
and the previously determined TC/THratio.
Step 8: Calculate the efficiency of the second Carnot engine.
Question 2
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine produces 4000 J of work per cycle. Calculate the efficiency
of the engine and determine if it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency:
Efficiency = 1 −Heat output
Heat input
Step 2: Calculate the heat input using the work output and the efficiency
formula:
Work output = Heat input −Heat output
4000 J = Heat input −Heat output
Step 3: Calculate the heat input using the temperatures of the reservoirs
and the principles of heat engines:
Heat input = QH=TC
TH−TC
×Heat output
Heat input = QH=300
600 −300 ×4000 = 4000 J
Step 4: Calculate the efficiency of the engine:
Efficiency = 1 −4000
4000 = 0
2
Step 5: Discuss the efficiency and the second law of thermodynamics: The
efficiency of the engine is 0, which indicates that it violates the second law of
thermodynamics. According to the second law, no heat engine can have an
efficiency of 100
Question 3
Question
A heat engine operates between a hot reservoir at a temperature of 500◦C and
a cold reservoir at a temperature of 20◦C. If the engine has an efficiency of 40
Solution
Step 1: Convert the temperatures to Kelvin.
Hot reservoir temperature: T1= 500 + 273 = 773 K
Cold reservoir temperature: T2= 20 + 273 = 293 K
Step 2: Calculate the efficiency of the engine.
Given efficiency, η=W
QH, where Wis the work done per cycle and QHis
the heat absorbed from the hot reservoir.
Efficiency is given as 40
Therefore, 0.40 = 800
QH.
Step 3: Calculate the heat absorbed from the hot reservoir.
Using the efficiency equation, QH=800
0.40 = 2000 J
Step 4: Determine the heat expelled to the cold reservoir.
Since the engine is reversible, the heat expelled to the cold reservoir is
given by QC=QH−W.
Therefore, QC= 2000 −800 = 1200 J
Step 5: Express the efficiency in terms of the temperatures.
Using the Carnot efficiency formula, η= 1 −T2
T1.
Substitute the temperatures in Kelvin: 0.40 = 1 −293
773 .
Step 6: Solve for the unknown temperature.
Rearranging the equation gives: 293
773 = 0.60.
Finally, solve for the hot reservoir temperature T1:T1=293
0.60 = 488 K
3
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Th= 500 K
and Tc= 200 K. If the engine absorbs 3000 J of heat from the hot reservoir
in each cycle, calculate the following: a) The efficiency of the engine b) The
amount of heat rejected to the cold reservoir in each cycle
Solution
a) To find the efficiency of the engine, we can use the formula for efficiency of a
Carnot engine:
Efficiency = 1 −Tc
Th
Step 1: Calculate the efficiency using the given temperatures:
Efficiency = 1 −200
500 = 1 −2
5=3
5= 0.6
Therefore, the efficiency of the engine is 60
b) The amount of heat rejected to the cold reservoir can be found using the
formula:
Amount of heat rejected = Qc= Efficiency×Amount of heat absorbed from the hot reservoir
Step 2: Calculate the amount of heat rejected:
Qc= 0.6×3000 = 1800 J
Therefore, the engine rejects 1800 J of heat to the cold reservoir in each
cycle.
Question 5
Question
A heat engine operates between a hot reservoir at 500◦Cand a cold reservoir
at 20◦C. If the engine has an efficiency of 35
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(C) +
273.15. Hot reservoir temperature: Thot = 500 + 273.15 = 773.15 K
Cold reservoir temperature: Tcold = 20 + 273.15 = 293.15 K
Step 2: Calculate the efficiency of the engine using the Carnot efficiency
formula:
Efficiency = 1 −Tcold
Thot
4
0.35 = 1 −293.15
773.15
Step 3: Calculate the heat input Qin per 1000 kJ: Let Qin be the amount of
heat input from the hot reservoir per 1000 kJ of heat:
Qin =1000
Efficiency
Step 4: Calculate the work output Wout per 1000 kJ:
Wout = Efficiency ×Qin
Step 5: Calculate the minimum possible amount of work Wmin that the
engine can perform per 1000 kJ of heat input:
Wmin =Wout −Qin
Calculate Wmin using the values calculated in the previous steps.
Question 6
Question
A heat engine operates between two reservoirs at temperatures T1and T2(with
T1> T2) with an efficiency of η. Prove that the efficiency of the engine cannot
be 100
Solution
Let Q1be the heat absorbed from the hot reservoir at temperature T1and Q2
be the heat rejected to the cold reservoir at temperature T2.
Step 1: Write the expression for efficiency η. The efficiency of a heat engine
is defined as
η= 1 −Q2
Q1
Step 2: Apply the second law of thermodynamics. According to the sec-
ond law of thermodynamics, the net work output of a heat engine (in absolute
magnitude) cannot exceed the net heat input. Mathematically:
|W|=|Q1−Q2|≤|Q1|
Step 3: Derive an expression for the efficiency in terms of T1and T2. Using
the definition of efficiency and the fact that Q1=|W|+Q2, we have:
η= 1 −Q2
W+Q2
= 1 −Q2
Q1
≤1−Q2
W+Q2
η≤1−Q2
W+Q2
≤1−Q2
Q2
= 1
Step 4: Conclude that the efficiency cannot be 100Since η≤1, the efficiency
of the engine cannot be 100
5
Question 7
Question
A Carnot engine operating between two heat reservoirs absorbs 600 J of heat
from the high-temperature reservoir and exhausts 400 J to the low-temperature
reservoir in each cycle. Calculate the efficiency of this engine. If the low-
temperature reservoir is at a temperature of 300 K, determine the temperature
of the high-temperature reservoir.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine: The efficiency
(η) of a Carnot engine is given by:
η= 1 −TL
TH
where TLis the absolute temperature of the low-temperature reservoir and TH
is the absolute temperature of the high-temperature reservoir.
Step 2: Calculate the efficiency of the engine: Given that the engine absorbs
600 J of heat from the high-temperature reservoir and exhausts 400 J to the
low-temperature reservoir in each cycle, the net work done by the engine in each
cycle is:
W=QH−QC= 600 −400 = 200 J
The efficiency can be calculated as:
η=W
QH
=200
600 =1
3= 0.3333
Step 3: Set up the equation to determine the temperature of the high-
temperature reservoir: Substitute the values into the efficiency formula:
0.3333 = 1 −300
TH
Step 4: Solve for the temperature of the high-temperature reservoir:
300
TH
= 1 −0.3333
300
TH
= 0.6667
TH=300
0.6667 = 450 K
Therefore, the efficiency of the engine is 0.3333 or 33.33
6
Question 8
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the reservoir at tem-
perature Thand exhausts Qcof heat to the reservoir at temperature Tc. Prove
that the efficiency of the Carnot engine is given by η= 1 −Tc
Thusing the second
law of thermodynamics.
Solution
Step 1: Let’s start by using the first law of thermodynamics which states that
for any cyclic process, the net heat absorbed must equal the net work done by
the engine.
According to the first law of thermodynamics:
Qh=W+Qc
Step 2: Next, we’ll use the definition of efficiency for a heat engine. The
efficiency, η, of a heat engine is defined as the ratio of the work done by the
engine to the heat absorbed from the hot reservoir. Mathematically, this is
given by:
η=W
Qh
Step 3: Since the Carnot engine is the most efficient heat engine possible,
its efficiency is equal to the efficiency of a Carnot engine, which is given by:
ηCarnot = 1 −Tc
Th
Step 4: Now, we’ll substitute the values of work and heat into the efficiency
equation. From Step 1, W=Qh−Qc, so:
η=Qh−Qc
Qh
Step 5: We can simplify the expression for efficiency by substituting Qc=
Qh−W. This gives:
η=Qh−(Qh−W)
Qh
Step 6: Further simplifying, we get:
η=W
Qh
= 1 −Qc
Qh
Step 7: Finally, we express Qcand Qhin terms of Tcand Th. From the
second law of thermodynamics, we know that for a reversible process:
Qh
Th
=Qc
Tc
7
Step 8: Solving for Qc, we get:
Qc=Tc
Th
Qh
Step 9: Substituting this expression back into the efficiency equation, we
get:
η= 1 −Tc
Th
Therefore, the efficiency of the Carnot engine is given by η= 1 −Tc
Thusing
the second law of thermodynamics.
Question 9
Question
A Carnot heat engine operates between a high-temperature reservoir at 600 K
and a low-temperature reservoir at 300 K. If the engine absorbs 1000 J of heat
from the high-temperature reservoir in each cycle, calculate the efficiency of the
engine. Is this efficiency greater or less than the efficiency of a Carnot engine
operating between reservoirs at 400 K and 300 K?
Solution
Step 1: Determine the efficiency of the Carnot heat engine operating between
600 K and 300 K. Given temperatures TH= 600 K and TC= 300 K, we can
calculate the efficiency using the formula for Carnot efficiency:
η= 1 −TC
TH
η= 1 −300
600
η= 1 −0.5
η= 0.5
Step 2: Calculate the work done by the engine. Since the efficiency of the
engine is 0.5, the work done by the engine is half the heat absorbed from the
high-temperature reservoir:
W=ηQH
W= 0.5×1000 J
W= 500 J
Step 3: Compare the efficiencies of the two Carnot engines. For the Carnot
engine operating between 400 K and 300 K, the efficiency would be:
η′= 1 −TC
TH
8
η′= 1 −300
400
η′= 1 −0.75
η′= 0.25
Since 0.25 <0.5, the efficiency of the Carnot engine operating between 600 K
and 300 K is greater than the efficiency of the engine operating between 400 K
and 300 K.
Question 10
Question
A heat engine operates between two reservoirs at 400◦C and 50◦C. 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. We are given that TH= 400◦C and TC= 50◦C.
We are also given that the efficiency of the engine is 40
Efficiency = 1 −TC
TH
Step 1: Calculate the efficiency of the engine with the given temperatures.
Substitute TH= 400◦C and TC= 50◦C into the efficiency equation:
Efficiency = 1 −50 + 273
400 + 273 = 1 −323
673 ≈0.5201
So, the efficiency of the engine with the given temperatures is 52.01
Step 2: Calculate the maximum possible efficiency of the engine if the reser-
voir temperatures were reversed. In this case, the new hot reservoir temperature
T′
H= 50◦C and the new cold reservoir temperature T′
C= 400◦C.
The maximum possible efficiency of the engine with the reversed tempera-
tures is given by
Efficiency = 1 −T′
C
T′
H
= 1 −400 + 273
50 + 273 = 1 −673
323 ≈0.5867
Therefore, the maximum possible efficiency of the engine if the reservoir
temperatures were reversed is approximately 58.67
9
Question 11
Question
A Carnot heat engine operates between reservoirs at temperatures of 500 K
and 300 K. If the engine absorbs 600 J of heat from the hot reservoir in each
cycle, what is the efficiency of the engine? Also, explain how the second law of
thermodynamics applies to this situation.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir (300 K) and THis the tem-
perature of the hot reservoir (500 K).
Step 2: Substitute the values into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Step 4: The second law of thermodynamics states that it is impossible for a
heat engine to have an efficiency of 100
Step 5: In this case, even though the Carnot engine is the most efficient heat
engine possible, its efficiency is limited by the temperature difference between
the hot and cold reservoirs as well as the irreversible processes inside the engine.
Step 6: The second law of thermodynamics also implies that no process is
possible in which the sole result would be the transfer of heat from a cooler
to a hotter body without any other change. This principle sets limits on the
efficiency of all heat engines.
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(where
Th> Tc). If the heat engine has an efficiency of 25
Solution
Let’s denote the ratio Th
Tcas r, the efficiency of the heat engine as η= 0.25, the
heat input as Qh, and the heat output as Qc. According to the second law of
thermodynamics for a heat engine, the efficiency can be expressed as:
η=W
Qh
= 1 −Qc
Qh
10
where Wis the work output of the engine. We also know that
Qc
Qh
=Tc
Th
Combining these equations, we have:
0.25 = 1 −Tc
Th
Solving for Tc
Thwill give us the ratio we are looking for.
Step 1: Simplify the equation to solve for Tc
Th.
Tc
Th
= 1 −0.25 = 0.75
Step 2: Therefore, the ratio Tc
This 0.75. To find the ratio Th
Tc, we take the
reciprocal:
r=1
0.75 =4
3
Step 3: Thus, the required ratio Th
Tcis 4
3.
Question 13
Question
A certain heat engine takes in 500 J of heat from a high-temperature reservoir
at 400 K and exhausts 300 J of heat to a low-temperature reservoir at 200 K.
Calculate the efficiency of the 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
Given that the engine takes in 500 J of heat and exhausts 300 J of heat, the
useful work output can be calculated as:
Useful Work Output = Heat Input −Heat Output
Useful Work Output = 500 J −300 J = 200 J
Therefore, the efficiency of the engine is:
Efficiency = 200 J
500 J = 0.4 = 40%
11
Step 2: Interpretation of efficiency: An efficiency of 40
Step 3: Analyzing the violation of the second law of thermodynamics: The
second law of thermodynamics states that no heat engine can be more effi-
cient than a Carnot engine working between the same two temperatures. The
theoretical maximum efficiency of a Carnot engine can be calculated using the
formula:
EfficiencyCarnot = 1 −Tlow
Thigh
Given that Tlow = 200 K and Thigh = 400 K, the maximum efficiency of a
Carnot engine operating between these temperatures would be:
EfficiencyCarnot = 1 −200
400 = 0.5 = 50%
Since the actual efficiency of the engine (40
Question 14
Question
A Carnot heat engine operates between a hot reservoir at 500
°
C and a cold
reservoir at 200
°
C. The engine absorbs 5000 J of heat energy from the hot
reservoir in each cycle. Calculate: (a) The efficiency of the engine (b) The
amount of heat energy rejected to the cold reservoir (c) Determine if this engine
violates the second law of thermodynamics.
Solution
(a) To find the efficiency of the engine, we can use the formula for the efficiency
of a Carnot engine:
Efficiency = 1 −Tcold
Thot
Step 1: Convert the temperatures to kelvin.
Thot = 500 + 273 = 773 K
Tcold = 200 + 273 = 473 K
Step 2: Substitute the values into the efficiency formula.
Efficiency = 1 −473
773 = 1 −0.611 = 0.389 = 38.9%
So, the efficiency of the engine is 38.9
(b) The amount of heat energy rejected to the cold reservoir can be found
using the formula:
Heat rejected = Heat absorbed −Work done
12
Since this is a Carnot engine, the work done can be determined using:
Work done = Efficiency ×Heat absorbed
Step 3: Calculate the work done.
Work done = 0.389 ×5000 = 1945 J
Step 4: Calculate the heat rejected.
Heat rejected = 5000 −1945 = 3055 J
So, the amount of heat energy rejected to the cold reservoir is 3055 J.
(c) This engine does not violate the second law of thermodynamics because
the efficiency is less than 100
Question 15
Question
A Carnot 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
during each cycle and expels 2400 J of heat to the cold reservoir during each
cycle, calculate the efficiency of the engine. Also, discuss how the efficiency of
this engine compares to the maximum possible efficiency for any heat engine
operating between the same two reservoirs.
Solution
Step 1: Determine the efficiency of the Carnot engine. Let Qhbe the heat
absorbed from the hot reservoir and Qcbe the heat expelled to the cold reservoir
during each cycle. The efficiency (η) of a Carnot engine is given by the formula:
η= 1 −Tc
Th
Given that Qh= 4000 J and Qc= 2400 J, we know that the efficiency can also
be expressed as:
η= 1 −Qc
Qh
Plugging in the values, we get:
η= 1 −2400
4000 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot engine is 40
13
Step 2: Comparison with maximum possible efficiency. The maximum pos-
sible efficiency for any heat engine operating between the same two reservoirs
can be found using the formula:
Max efficiency = 1 −Tc
Th
Plugging in the given temperatures, we get:
Max efficiency = 1 −Tc
Th
= 1 −2400
4000 = 1 −0.6=0.4
Comparing the efficiency of the Carnot engine with the maximum possi-
ble efficiency, we see that the Carnot engine achieves the maximum possible
efficiency for a heat engine operating between the given two reservoirs.
Question 16
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 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.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency for a Carnot engine:
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 600 K and the cold reservoir
temperature is 300 K, we have:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot engine is 50%.
Step 2: Calculate the amount of heat rejected to the cold reservoir. Since
the engine absorbs 4000 J of heat from the hot reservoir in each cycle, by
conservation of energy, the amount of heat rejected to the cold reservoir must
also be 4000 J.
Hence, the amount of heat rejected to the cold reservoir is 4000 J.
14
Question 17
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, with
Th> Tc. The engine absorbs Qhamount of heat from the hot reservoir and
discards Qcamount of heat to the cold reservoir. Calculate the efficiency of this
heat engine in terms of Th,Tc,Qh, and Qc.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency of a
heat engine is given by the formula:
Efficiency = 1 −Qc
Qh
Step 2: Determine the expression for Qcin terms of Th,Tc, and Qh: Ac-
cording to the second law of thermodynamics, we have:
Qc
Tc
=Qh
Th
Solving for Qc, we get:
Qc=Tc
Th
Qh
Step 3: Substitute the expression for Qcinto the efficiency formula:
Efficiency = 1 −Tc
Th
Therefore, the efficiency of the heat engine in terms of Th,Tc,Qh, and Qc
is 1 −Tc
Th.
Question 18
Question
A heat engine operates between two heat reservoirs at temperatures TH= 500 K
and TC= 300 K. The engine consumes 500 J of heat from the hot reservoir in
each cycle. Calculate the maximum efficiency of the engine according to the
second law of thermodynamics.
Solution
Step 1: Determine the maximum efficiency of the engine using Carnot’s theo-
rem. According to Carnot’s theorem, the maximum efficiency of a heat engine
operating between two reservoirs at temperatures THand TCis given by:
ηmax = 1 −TC
TH
15
Step 2: Substitute the given values into the efficiency formula. Given TH=
500 K and TC= 300 K, we can calculate the maximum efficiency:
ηmax = 1 −300
500 = 1 −0.6=0.4
Step 3: Convert the efficiency to a percentage. To express the efficiency as
a percentage, we multiply by 100:
ηmax = 0.4×100% = 40%
Therefore, the maximum efficiency of the heat engine operating between the
two given heat reservoirs is 40
Question 19
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Determine the maximum possible
efficiency for a heat engine operating between reservoirs at these temperatures.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
η= 1 −Tcold
Thot
Step 2: Substitute the given temperatures into the equation:
η= 1 −300
600
Step 3: Simplify the expression:
η= 1 −1
2=1
2= 50%
Answer: The maximum possible efficiency for a heat engine operating be-
tween reservoirs at 600 K and 300 K is 50%.
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir. If the efficiency of the engine is η, prove that the
efficiency can be expressed as η= 1 −Tc
Th.
16
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency (η)
of a heat engine is defined as the ratio of the work done by the engine to the
heat absorbed from the hot reservoir:
η=W
Qh
Step 2: Apply the first law of thermodynamics for the engine: The first law
of thermodynamics states that the net work done by the engine is equal to the
difference between the heat absorbed and the heat exhausted by the engine:
W=Qh−Qc
Step 3: Substitute the expression for work done from Step 2 into the effi-
ciency equation from Step 1:
η=Qh−Qc
Qh
Step 4: Rearrange the expression using algebraic manipulation:
η= 1 −Qc
Qh
Step 5: Apply the definition of efficiency from the Carnot cycle: For a Carnot
engine, the efficiency is given by the formula:
η= 1 −Tc
Th
Step 6: Equate the equations for efficiency from Step 4 and Step 5:
1−Qc
Qh
= 1 −Tc
Th
Step 7: Simplify the equation to obtain the desired expression for efficiency:
η= 1 −Tc
Th
Therefore, the efficiency of the heat engine can be expressed as η= 1 −Tc
Th.
Question 21
Question
A heat engine operates between a hot reservoir at 500K and a cold reservoir
at 300K. If the engine absorbs 2000J of heat from the hot reservoir during
each cycle, what is the maximum efficiency of the engine? Also, determine the
amount of heat rejected to the cold reservoir during each cycle.
17
Solution
Let’s denote the heat absorbed from the hot reservoir as Qh= 2000 J, the
temperature of the hot reservoir as Th= 500 K, and the temperature of the
cold reservoir as Tc= 300 K.
Step 1: Find the efficiency of the engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Tc
Th
Substitute the given values:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Therefore, the maximum efficiency of the engine is 40
Step 2: Find the amount of heat rejected to the cold reservoir. The
amount of heat rejected to the cold reservoir during each cycle can be calculated
using the formula:
Qc=Qh−W
where Wis the work done by the engine during each cycle. Since the engine is
working in a cycle, the work done is equal to the difference between the heat
absorbed and the heat rejected.
W=Qh−Qc
Substitute the given values and solve for the heat rejected:
Qc= 2000 −(2000 ×0.4) = 2000 −800 = 1200 J
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 1200 J.
Question 22
Question
A Carnot engine operates between a hot reservoir at a temperature of 500 K
and a cold reservoir at a temperature of 300 K. If it absorbs 500 J of heat from
the hot reservoir in each cycle, determine:
1. The efficiency of the engine.
2. The amount of heat rejected to the cold reservoir in each cycle.
18
Solution
1. To find the efficiency of the Carnot engine, we use the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
Step 1: Given Thot = 500 K and Tcold = 300 K, we plug these values into
the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4
2. To find the amount of heat rejected to the cold reservoir in each cycle,
we can use the fact that for a Carnot engine, the net work output in each cycle
is equal to the difference in the heat absorbed from the hot reservoir and the
heat rejected to the cold reservoir. Therefore:
Qcold =Tcold
Thot
·Qhot
where Qcold is the heat rejected to the cold reservoir and Qhot is the heat
absorbed from the hot reservoir.
Step 2: Given Qhot = 500 J, we calculate the heat rejected to the cold
reservoir:
Qcold =300
500 ·500 = 300 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
300 J .
Question 23
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold,
where Thot = 500 K and Tcold = 300 K. The engine absorbs 500 J of heat
from the hot reservoir in each cycle and exhausts 300 J to the cold reservoir.
Calculate the efficiency of the engine.
Solution
Step 1: Calculate the efficiency of the engine.
The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
19
Given that Thot = 500 K and Tcold = 300 K, we can substitute these values
into the formula:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4
Therefore, the efficiency of the engine is 40
Question 24
Question
A heat engine operates between a hot reservoir at 600◦Cand a cold reservoir
at 20◦C. The engine has an efficiency of 40
Solution
Step 1: Convert the temperatures to Kelvin. Given: Hot reservoir temperature,
Thot = 600◦C= 600 + 273 = 873 K
Cold reservoir temperature, Tcold = 20◦C= 20 + 273 = 293 K
Step 2: Calculate the maximum efficiency using Carnot efficiency formula.
The Carnot efficiency is given by:
Maximum efficiency, ηCarnot = 1 −Tcold
Thot
Substitute the values:
ηCarnot = 1 −293
873
ηCarnot = 1 −293
873
ηCarnot = 1 −293
873
ηCarnot = 1 −0.3360
ηCarnot = 0.6640 or 66.40%
Step 3: Analyze the obtained result. The maximum efficiency for this engine
(66.40
Question 25
Question
A heat engine operates between two reservoirs at temperatures T1= 400 K and
T2= 100 K. If the engine extracts 2000 J of heat from the hot reservoir in each
cycle, calculate the maximum possible efficiency of the engine. Also, discuss
whether this efficiency violates the second law of thermodynamics.
20
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −T2
T1
Step 2: Substitute the given values into the formula:
Efficiency = 1 −100
400 = 1 −1
4=3
4= 75%
Step 3: Discuss whether this efficiency violates the second law of thermody-
namics. According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Question 26
Question
A heat engine operates with a hot reservoir temperature of 500 K and a cold
reservoir temperature of 300 K. The engine absorbs 2000 J of heat from the hot
reservoir in each cycle and exhausts 1400 J to the cold reservoir.
a) Calculate the efficiency of the heat engine. b) Determine the maximum
possible efficiency of a heat engine operating between these two reservoir tem-
peratures. c) Explain how the second law of thermodynamics relates to the
efficiency of heat engines.
Solution
a) To calculate the efficiency of the heat engine, we can use the formula:
Efficiency = 1 −Qc
Qh
Where Qcis the heat expelled to the cold reservoir and Qhis the heat
absorbed from the hot reservoir.
Step 1: Calculate the efficiency using the given values:
Efficiency = 1 −1400
2000
Efficiency = 1 −0.7 = 0.3
Therefore, the efficiency of the heat engine is 30
b) The maximum possible efficiency of a heat engine operating between two
reservoir temperatures Thand Tcis given by Carnot efficiency:
EfficiencyCarnot = 1 −Tc
Th
21
Step 2: Calculate the Carnot efficiency using the given temperatures:
EfficiencyCarnot = 1 −300
500
EfficiencyCarnot = 1 −0.6=0.4
Therefore, the maximum possible efficiency of a heat engine operating be-
tween 500 K and 300 K is 40
c) The second law of thermodynamics states that it is impossible for any
heat engine to be 100
Question 27
Question
A heat engine operating between a hot reservoir at 600 K and a cold reservoir
at 300 K has an efficiency of 40%. If the engine absorbs 2000 J of energy from
the hot reservoir in each cycle, calculate:
1. The work done by the engine in each cycle.
2. The heat energy rejected to the cold reservoir in each cycle.
Solution
Let’s denote the efficiency of the heat engine as η, the heat absorbed from the
hot reservoir as Qh, the work done by the engine as W, and the heat rejected
to the cold reservoir as Qc. We know that the efficiency is given by:
η=W
Qh
Step 1: Calculate the work done by the engine in each cycle. Given that
the efficiency of the engine is 40% or 0.4, we have:
η= 0.4 = W
Qh
0.4 = W
2000 J
W= 0.4×2000 J
W= 800 J
Therefore, the work done by the engine in each cycle is 800 J.
Step 2: Calculate the heat energy rejected to the cold reservoir in each
cycle. Since the efficiency of the engine is given by:
η=Qh−Qc
Qh
22
We can rearrange this equation to find Qc:
Qc= (1 −η)×Qh
Qc= (1 −0.4) ×2000 J
Qc= 0.6×2000 J
Qc= 1200 J
Therefore, the heat energy rejected to the cold reservoir in each cycle is 1200
J.
Question 28
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. If the engine produces work W, prove that the efficiency
of the engine is given by η= 1 −Tc
Th.
Solution
Step 1: According to the first law of thermodynamics, the net work done by
the engine is equal to the difference between the heat absorbed and the heat
released:
W=Qh−Qc
Step 2: The efficiency of the engine is defined as the ratio of the work done
by the engine to the heat absorbed from the hot reservoir:
η=W
Qh
Step 3: Substituting the expression for Wfrom Step 1 into the efficiency
formula:
η=Qh−Qc
Qh
Step 4: Simplifying the expression:
η= 1 −Qc
Qh
Step 5: We can relate Qcand Qhto the temperatures Tcand Thusing the
Carnot efficiency:
ηC= 1 −Tc
Th
=Qh−Qc
Qh
23
Step 6: If we compare the Carnot efficiency with the efficiency of the engine,
we see that they are equal:
η=ηC= 1 −Tc
Th
Therefore, the efficiency of the engine is given by η= 1 −Tc
Th.
Question 29
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc(Th> Tc). The engine absorbs 600 J of heat from the hot reservoir and
exhausts 400 J to the cold reservoir for each cycle. Find the efficiency of the
engine.
Solution
Let’s denote the efficiency of the Carnot engine as η. We know that the efficiency
of a Carnot engine is given by the formula:
η= 1 −Tc
Th
Step 1: Calculate the efficiency using the given values of heat exchange:
η= 1 −Tc
Th
We are given that the engine absorbs 600 J of heat (Qh= 600 J) and exhausts
400 J (Qc= 400 J) for each cycle. By the first law of thermodynamics, the net
work done by the engine in each cycle is given by:
W=Qh−Qc
This work output is used to express the efficiency in terms of the heat ex-
change:
η= 1 −Qc
Qh
Substitute the given values of heat exchange:
η= 1 −400
600
η= 1 −2
3
η=1
3
24
Step 2: Express the efficiency in percentage:
Efficiency(%) = η×100%
Substitute the value of η:
Efficiency(%) = 1
3×100%
Efficiency(%) = 33.33%
Therefore, the efficiency of the Carnot heat engine is 33.33
Question 30
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and has an
efficiency of η. Prove that the efficiency of any heat engine operating between the
same two temperatures is less than that of a Carnot engine operating between
the same two temperatures.
Solution
Let’s consider a Carnot engine operating between the same two reservoirs at
temperatures Thand Tc.
Step 1: Identify the efficiency of the Carnot engine. The efficiency of a
Carnot engine is given by:
ηCarnot = 1 −Tc
Th
Step 2: Identify the efficiency of the given engine. The efficiency of the
given engine is ηand the heat absorbed from the hot reservoir is Qh. Therefore,
the work output of the engine is given by:
Wgiven =ηQh
and the heat rejected to the cold reservoir is:
Qc=Qh−Wgiven =Qh(1 −η)
Also, we have Qc=Tc∆S, where ∆Sis the change in entropy.
Step 3: Compare the efficiency of the given engine with that of the Carnot
engine. From step 2, we have:
Qh(1 −η) = Tc∆S
25
Since η < 1, we have 1 −η > 0. Thus, ∆S > 0. Now, the heat absorbed
from the hot reservoir for the Carnot engine is Qh. Using the second law of
thermodynamics, we have:
∆SCarnot =Qh
Th
−Qc
Tc
≥0
Substitute the values of Qh,Qc, and ∆SCarnot:
Th
Th
−Tc
Tc
= 1 −Tc
Th
≥0
Therefore, 1 −Tc
Th≥0, which implies that η < ηCarnot. Hence, the efficiency of
any heat engine operating between the same two temperatures is less than that
of a Carnot engine operating between the same two temperatures.
Question 31
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir in each cycle. If the efficiency of the
engine is η, show that η= 1 −Tc
Th.
Solution
Step 1: Recall the efficiency of a Carnot heat 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 denote the net work done by the engine in each cycle as W.
By the first law of thermodynamics, we have
W=Qh−Qc
Step 3: Since the engine is operating between two heat reservoirs, the heat
absorbed from the hot reservoir must equal the work done plus the heat rejected
to the cold reservoir. Therefore, we have
Qh=W+Qc
Step 4: The efficiency of the engine is defined as the ratio of the work done
to the heat input from the hot reservoir. Hence, we can write
η=W
Qh
26
Step 5: Substituting W=Qh−Qcinto the equation for efficiency, we get
η=Qh−Qc
Qh
Step 6: Simplifying the expression, we find
η= 1 −Qc
Qh
Step 7: We know that for a Carnot engine, the efficiency is given by
η= 1 −Tc
Th
Step 8: Equating the two expressions for efficiency, we have
1−Tc
Th
= 1 −Qc
Qh
Step 9: Since Qc
Qh=Tc
Th, we have shown that
η= 1 −Tc
Th
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(where
Th> Tc). The heat engine consumes 4000 J of heat from the hot reservoir and
produces 2000 J of work output. Determine the efficiency of the heat engine
and discuss whether this violates the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = Useful work output
Heat input
Step 2: Given that the heat input to the engine is 4000 J and the work output
is 2000 J, we can plug these values into the formula to find the efficiency:
Efficiency = 2000
4000 = 0.5
Step 3: The efficiency of the heat engine is 50
27
Step 4: According to the second law of thermodynamics, the efficiency of a
heat engine is limited by the Carnot efficiency, which is given by:
Carnot efficiency = 1 −Tc
Th
Step 5: Calculate the Carnot efficiency using the given temperatures:
Carnot efficiency = 1 −Tc
Th
= 1 −2000
4000 = 0.5
Step 6: The Carnot efficiency of the heat engine operating between the
temperatures Thand Tcis 50
Step 7: Since the actual efficiency of the heat engine (50
Question 33
Question
A Carnot heat engine operates between two reservoirs at 500 K and 300 K. If the
heat engine absorbs 1000 J of heat energy from the high-temperature reservoir
in each cycle, calculate:
1. The efficiency of the Carnot engine.
2. The heat energy rejected by the engine 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
where TCis the temperature of the low-temperature reservoir and THis the
temperature of the high-temperature reservoir.
Given that TC= 300 K and TH= 500 K, we can substitute these values into
the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot engine is 40%.
Step 2: Calculate the heat energy rejected by the engine to the low-
temperature reservoir in each cycle. We can use the efficiency formula to find
the heat energy rejected:
Efficiency = Work done by engine
Heat energy absorbed from high-temperature reservoir
28
Since the efficiency is 0.4 and the heat energy absorbed from the high-
temperature reservoir is 1000 J, we can rearrange the formula to find the work
done by the engine:
Work done by engine = Efficiency ×Heat energy absorbed
Work done by engine = 0.4×1000 = 400 J
Since the total energy input is 1000 J and the work done by the engine is
400 J, the heat energy rejected to the low-temperature reservoir is:
Heat energy rejected = Total energy input −Work done by engine
Heat energy rejected = 1000 −400 = 600 J
Therefore, the heat energy rejected by the engine to the low-temperature
reservoir in each cycle is 600 J.
Question 34
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tcwhere Th> Tc. If the engine absorbs 600 J of heat energy from the high-
temperature reservoir and exhausts 375 J to the low-temperature reservoir dur-
ing each cycle, what is the efficiency of the engine? Explain your answer in
terms of the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a Carnot heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values into the formula. The engine absorbs
600 J of heat energy from the high-temperature reservoir (Th) and exhausts 375
J to the low-temperature reservoir (Tc). Therefore, Th= 600 J and Tc= 375 J.
Step 3: Calculate the efficiency using the formula:
Efficiency = 1 −375
600
Step 4: Simplify the expression to find the efficiency:
Efficiency = 1 −375
600 = 1 −5
8=3
8= 0.375
Step 5: The efficiency of the engine is 0.375 or 37.5
29
Question 35
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine is able to extract 4000 J of heat from the hot reservoir
and delivers 2500 J of work per cycle. Calculate the efficiency of the engine and
determine the minimum value of Tcif Th= 500 K.
Solution
Step 1: Calculate the efficiency of the engine. The efficiency of a heat engine is
given by the formula:
Efficiency = Useful work output
Heat input
Given that the engine delivers 2500 J of work and extracts 4000 J of heat, the
efficiency is:
Efficiency = 2500 J
4000 J ×100% = 62.5%
Step 2: Determine the minimum value of Tc. The efficiency of a heat engine
operating between two reservoirs is given by:
Efficiency = 1 −Tc
Th
Given that the efficiency is 62.5
0.625 = 1 −Tc
500
Tc
500 = 0.375
Tc= 0.375 ×500 = 187.5 K
Therefore, the efficiency of the engine is 62.5
30
Step 4: Solve for TC/THto find the ratio of the cold to hot reservoir tem-
peratures for the first engine.
Step 5: Next, let’s determine the cold reservoir temperature for the second
engine operating between 227◦C and 27◦C in terms of absolute temperature in
Kelvin.
Step 6: Use the distinction of efficiency between the two engines to calculate
the efficiency of the second Carnot engine using the formula:
Efficiency = 1 −T′
C
T′
H
where T′
Cand T′
Hare the absolute temperatures of the cold and hot reservoirs
for the second engine respectively.
Step 7: Substitute T′
Cand T′
Hwith the respective temperatures in Kelvin
and the previously determined TC/THratio.
Step 8: Calculate the efficiency of the second Carnot engine.
Question 2
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine produces 4000 J of work per cycle. Calculate the efficiency
of the engine and determine if it violates the second law of thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula for efficiency:
Efficiency = 1 −Heat output
Heat input
Step 2: Calculate the heat input using the work output and the efficiency
formula:
Work output = Heat input −Heat output
4000 J = Heat input −Heat output
Step 3: Calculate the heat input using the temperatures of the reservoirs
and the principles of heat engines:
Heat input = QH=TC
TH−TC
×Heat output
Heat input = QH=300
600 −300 ×4000 = 4000 J
Step 4: Calculate the efficiency of the engine:
Efficiency = 1 −4000
4000 = 0
2
Step 5: Discuss the efficiency and the second law of thermodynamics: The
efficiency of the engine is 0, which indicates that it violates the second law of
thermodynamics. According to the second law, no heat engine can have an
efficiency of 100
Question 3
Question
A heat engine operates between a hot reservoir at a temperature of 500◦C and
a cold reservoir at a temperature of 20◦C. If the engine has an efficiency of 40
Solution
Step 1: Convert the temperatures to Kelvin.
Hot reservoir temperature: T1= 500 + 273 = 773 K
Cold reservoir temperature: T2= 20 + 273 = 293 K
Step 2: Calculate the efficiency of the engine.
Given efficiency, η=W
QH, where Wis the work done per cycle and QHis
the heat absorbed from the hot reservoir.
Efficiency is given as 40
Therefore, 0.40 = 800
QH.
Step 3: Calculate the heat absorbed from the hot reservoir.
Using the efficiency equation, QH=800
0.40 = 2000 J
Step 4: Determine the heat expelled to the cold reservoir.
Since the engine is reversible, the heat expelled to the cold reservoir is
given by QC=QH−W.
Therefore, QC= 2000 −800 = 1200 J
Step 5: Express the efficiency in terms of the temperatures.
Using the Carnot efficiency formula, η= 1 −T2
T1.
Substitute the temperatures in Kelvin: 0.40 = 1 −293
773 .
Step 6: Solve for the unknown temperature.
Rearranging the equation gives: 293
773 = 0.60.
Finally, solve for the hot reservoir temperature T1:T1=293
0.60 = 488 K
3
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Th= 500 K
and Tc= 200 K. If the engine absorbs 3000 J of heat from the hot reservoir
in each cycle, calculate the following: a) The efficiency of the engine b) The
amount of heat rejected to the cold reservoir in each cycle
Solution
a) To find the efficiency of the engine, we can use the formula for efficiency of a
Carnot engine:
Efficiency = 1 −Tc
Th
Step 1: Calculate the efficiency using the given temperatures:
Efficiency = 1 −200
500 = 1 −2
5=3
5= 0.6
Therefore, the efficiency of the engine is 60
b) The amount of heat rejected to the cold reservoir can be found using the
formula:
Amount of heat rejected = Qc= Efficiency×Amount of heat absorbed from the hot reservoir
Step 2: Calculate the amount of heat rejected:
Qc= 0.6×3000 = 1800 J
Therefore, the engine rejects 1800 J of heat to the cold reservoir in each
cycle.
Question 5
Question
A heat engine operates between a hot reservoir at 500◦Cand a cold reservoir
at 20◦C. If the engine has an efficiency of 35
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(C) +
273.15. Hot reservoir temperature: Thot = 500 + 273.15 = 773.15 K
Cold reservoir temperature: Tcold = 20 + 273.15 = 293.15 K
Step 2: Calculate the efficiency of the engine using the Carnot efficiency
formula:
Efficiency = 1 −Tcold
Thot
4
0.35 = 1 −293.15
773.15
Step 3: Calculate the heat input Qin per 1000 kJ: Let Qin be the amount of
heat input from the hot reservoir per 1000 kJ of heat:
Qin =1000
Efficiency
Step 4: Calculate the work output Wout per 1000 kJ:
Wout = Efficiency ×Qin
Step 5: Calculate the minimum possible amount of work Wmin that the
engine can perform per 1000 kJ of heat input:
Wmin =Wout −Qin
Calculate Wmin using the values calculated in the previous steps.
Question 6
Question
A heat engine operates between two reservoirs at temperatures T1and T2(with
T1> T2) with an efficiency of η. Prove that the efficiency of the engine cannot
be 100
Solution
Let Q1be the heat absorbed from the hot reservoir at temperature T1and Q2
be the heat rejected to the cold reservoir at temperature T2.
Step 1: Write the expression for efficiency η. The efficiency of a heat engine
is defined as
η= 1 −Q2
Q1
Step 2: Apply the second law of thermodynamics. According to the sec-
ond law of thermodynamics, the net work output of a heat engine (in absolute
magnitude) cannot exceed the net heat input. Mathematically:
|W|=|Q1−Q2|≤|Q1|
Step 3: Derive an expression for the efficiency in terms of T1and T2. Using
the definition of efficiency and the fact that Q1=|W|+Q2, we have:
η= 1 −Q2
W+Q2
= 1 −Q2
Q1
≤1−Q2
W+Q2
η≤1−Q2
W+Q2
≤1−Q2
Q2
= 1
Step 4: Conclude that the efficiency cannot be 100Since η≤1, the efficiency
of the engine cannot be 100
5
Question 7
Question
A Carnot engine operating between two heat reservoirs absorbs 600 J of heat
from the high-temperature reservoir and exhausts 400 J to the low-temperature
reservoir in each cycle. Calculate the efficiency of this engine. If the low-
temperature reservoir is at a temperature of 300 K, determine the temperature
of the high-temperature reservoir.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine: The efficiency
(η) of a Carnot engine is given by:
η= 1 −TL
TH
where TLis the absolute temperature of the low-temperature reservoir and TH
is the absolute temperature of the high-temperature reservoir.
Step 2: Calculate the efficiency of the engine: Given that the engine absorbs
600 J of heat from the high-temperature reservoir and exhausts 400 J to the
low-temperature reservoir in each cycle, the net work done by the engine in each
cycle is:
W=QH−QC= 600 −400 = 200 J
The efficiency can be calculated as:
η=W
QH
=200
600 =1
3= 0.3333
Step 3: Set up the equation to determine the temperature of the high-
temperature reservoir: Substitute the values into the efficiency formula:
0.3333 = 1 −300
TH
Step 4: Solve for the temperature of the high-temperature reservoir:
300
TH
= 1 −0.3333
300
TH
= 0.6667
TH=300
0.6667 = 450 K
Therefore, the efficiency of the engine is 0.3333 or 33.33
6
Question 8
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the reservoir at tem-
perature Thand exhausts Qcof heat to the reservoir at temperature Tc. Prove
that the efficiency of the Carnot engine is given by η= 1 −Tc
Thusing the second
law of thermodynamics.
Solution
Step 1: Let’s start by using the first law of thermodynamics which states that
for any cyclic process, the net heat absorbed must equal the net work done by
the engine.
According to the first law of thermodynamics:
Qh=W+Qc
Step 2: Next, we’ll use the definition of efficiency for a heat engine. The
efficiency, η, of a heat engine is defined as the ratio of the work done by the
engine to the heat absorbed from the hot reservoir. Mathematically, this is
given by:
η=W
Qh
Step 3: Since the Carnot engine is the most efficient heat engine possible,
its efficiency is equal to the efficiency of a Carnot engine, which is given by:
ηCarnot = 1 −Tc
Th
Step 4: Now, we’ll substitute the values of work and heat into the efficiency
equation. From Step 1, W=Qh−Qc, so:
η=Qh−Qc
Qh
Step 5: We can simplify the expression for efficiency by substituting Qc=
Qh−W. This gives:
η=Qh−(Qh−W)
Qh
Step 6: Further simplifying, we get:
η=W
Qh
= 1 −Qc
Qh
Step 7: Finally, we express Qcand Qhin terms of Tcand Th. From the
second law of thermodynamics, we know that for a reversible process:
Qh
Th
=Qc
Tc
7
Step 8: Solving for Qc, we get:
Qc=Tc
Th
Qh
Step 9: Substituting this expression back into the efficiency equation, we
get:
η= 1 −Tc
Th
Therefore, the efficiency of the Carnot engine is given by η= 1 −Tc
Thusing
the second law of thermodynamics.
Question 9
Question
A Carnot heat engine operates between a high-temperature reservoir at 600 K
and a low-temperature reservoir at 300 K. If the engine absorbs 1000 J of heat
from the high-temperature reservoir in each cycle, calculate the efficiency of the
engine. Is this efficiency greater or less than the efficiency of a Carnot engine
operating between reservoirs at 400 K and 300 K?
Solution
Step 1: Determine the efficiency of the Carnot heat engine operating between
600 K and 300 K. Given temperatures TH= 600 K and TC= 300 K, we can
calculate the efficiency using the formula for Carnot efficiency:
η= 1 −TC
TH
η= 1 −300
600
η= 1 −0.5
η= 0.5
Step 2: Calculate the work done by the engine. Since the efficiency of the
engine is 0.5, the work done by the engine is half the heat absorbed from the
high-temperature reservoir:
W=ηQH
W= 0.5×1000 J
W= 500 J
Step 3: Compare the efficiencies of the two Carnot engines. For the Carnot
engine operating between 400 K and 300 K, the efficiency would be:
η′= 1 −TC
TH
8
η′= 1 −300
400
η′= 1 −0.75
η′= 0.25
Since 0.25 <0.5, the efficiency of the Carnot engine operating between 600 K
and 300 K is greater than the efficiency of the engine operating between 400 K
and 300 K.
Question 10
Question
A heat engine operates between two reservoirs at 400◦C and 50◦C. 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. We are given that TH= 400◦C and TC= 50◦C.
We are also given that the efficiency of the engine is 40
Efficiency = 1 −TC
TH
Step 1: Calculate the efficiency of the engine with the given temperatures.
Substitute TH= 400◦C and TC= 50◦C into the efficiency equation:
Efficiency = 1 −50 + 273
400 + 273 = 1 −323
673 ≈0.5201
So, the efficiency of the engine with the given temperatures is 52.01
Step 2: Calculate the maximum possible efficiency of the engine if the reser-
voir temperatures were reversed. In this case, the new hot reservoir temperature
T′
H= 50◦C and the new cold reservoir temperature T′
C= 400◦C.
The maximum possible efficiency of the engine with the reversed tempera-
tures is given by
Efficiency = 1 −T′
C
T′
H
= 1 −400 + 273
50 + 273 = 1 −673
323 ≈0.5867
Therefore, the maximum possible efficiency of the engine if the reservoir
temperatures were reversed is approximately 58.67
9
Question 11
Question
A Carnot heat engine operates between reservoirs at temperatures of 500 K
and 300 K. If the engine absorbs 600 J of heat from the hot reservoir in each
cycle, what is the efficiency of the engine? Also, explain how the second law of
thermodynamics applies to this situation.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir (300 K) and THis the tem-
perature of the hot reservoir (500 K).
Step 2: Substitute the values into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Step 3: Therefore, the efficiency of the Carnot engine is 40
Step 4: The second law of thermodynamics states that it is impossible for a
heat engine to have an efficiency of 100
Step 5: In this case, even though the Carnot engine is the most efficient heat
engine possible, its efficiency is limited by the temperature difference between
the hot and cold reservoirs as well as the irreversible processes inside the engine.
Step 6: The second law of thermodynamics also implies that no process is
possible in which the sole result would be the transfer of heat from a cooler
to a hotter body without any other change. This principle sets limits on the
efficiency of all heat engines.
Question 12
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(where
Th> Tc). If the heat engine has an efficiency of 25
Solution
Let’s denote the ratio Th
Tcas r, the efficiency of the heat engine as η= 0.25, the
heat input as Qh, and the heat output as Qc. According to the second law of
thermodynamics for a heat engine, the efficiency can be expressed as:
η=W
Qh
= 1 −Qc
Qh
10
where Wis the work output of the engine. We also know that
Qc
Qh
=Tc
Th
Combining these equations, we have:
0.25 = 1 −Tc
Th
Solving for Tc
Thwill give us the ratio we are looking for.
Step 1: Simplify the equation to solve for Tc
Th.
Tc
Th
= 1 −0.25 = 0.75
Step 2: Therefore, the ratio Tc
This 0.75. To find the ratio Th
Tc, we take the
reciprocal:
r=1
0.75 =4
3
Step 3: Thus, the required ratio Th
Tcis 4
3.
Question 13
Question
A certain heat engine takes in 500 J of heat from a high-temperature reservoir
at 400 K and exhausts 300 J of heat to a low-temperature reservoir at 200 K.
Calculate the efficiency of the 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
Given that the engine takes in 500 J of heat and exhausts 300 J of heat, the
useful work output can be calculated as:
Useful Work Output = Heat Input −Heat Output
Useful Work Output = 500 J −300 J = 200 J
Therefore, the efficiency of the engine is:
Efficiency = 200 J
500 J = 0.4 = 40%
11
Step 2: Interpretation of efficiency: An efficiency of 40
Step 3: Analyzing the violation of the second law of thermodynamics: The
second law of thermodynamics states that no heat engine can be more effi-
cient than a Carnot engine working between the same two temperatures. The
theoretical maximum efficiency of a Carnot engine can be calculated using the
formula:
EfficiencyCarnot = 1 −Tlow
Thigh
Given that Tlow = 200 K and Thigh = 400 K, the maximum efficiency of a
Carnot engine operating between these temperatures would be:
EfficiencyCarnot = 1 −200
400 = 0.5 = 50%
Since the actual efficiency of the engine (40
Question 14
Question
A Carnot heat engine operates between a hot reservoir at 500
°
C and a cold
reservoir at 200
°
C. The engine absorbs 5000 J of heat energy from the hot
reservoir in each cycle. Calculate: (a) The efficiency of the engine (b) The
amount of heat energy rejected to the cold reservoir (c) Determine if this engine
violates the second law of thermodynamics.
Solution
(a) To find the efficiency of the engine, we can use the formula for the efficiency
of a Carnot engine:
Efficiency = 1 −Tcold
Thot
Step 1: Convert the temperatures to kelvin.
Thot = 500 + 273 = 773 K
Tcold = 200 + 273 = 473 K
Step 2: Substitute the values into the efficiency formula.
Efficiency = 1 −473
773 = 1 −0.611 = 0.389 = 38.9%
So, the efficiency of the engine is 38.9
(b) The amount of heat energy rejected to the cold reservoir can be found
using the formula:
Heat rejected = Heat absorbed −Work done
12
Since this is a Carnot engine, the work done can be determined using:
Work done = Efficiency ×Heat absorbed
Step 3: Calculate the work done.
Work done = 0.389 ×5000 = 1945 J
Step 4: Calculate the heat rejected.
Heat rejected = 5000 −1945 = 3055 J
So, the amount of heat energy rejected to the cold reservoir is 3055 J.
(c) This engine does not violate the second law of thermodynamics because
the efficiency is less than 100
Question 15
Question
A Carnot 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
during each cycle and expels 2400 J of heat to the cold reservoir during each
cycle, calculate the efficiency of the engine. Also, discuss how the efficiency of
this engine compares to the maximum possible efficiency for any heat engine
operating between the same two reservoirs.
Solution
Step 1: Determine the efficiency of the Carnot engine. Let Qhbe the heat
absorbed from the hot reservoir and Qcbe the heat expelled to the cold reservoir
during each cycle. The efficiency (η) of a Carnot engine is given by the formula:
η= 1 −Tc
Th
Given that Qh= 4000 J and Qc= 2400 J, we know that the efficiency can also
be expressed as:
η= 1 −Qc
Qh
Plugging in the values, we get:
η= 1 −2400
4000 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot engine is 40
13
Step 2: Comparison with maximum possible efficiency. The maximum pos-
sible efficiency for any heat engine operating between the same two reservoirs
can be found using the formula:
Max efficiency = 1 −Tc
Th
Plugging in the given temperatures, we get:
Max efficiency = 1 −Tc
Th
= 1 −2400
4000 = 1 −0.6=0.4
Comparing the efficiency of the Carnot engine with the maximum possi-
ble efficiency, we see that the Carnot engine achieves the maximum possible
efficiency for a heat engine operating between the given two reservoirs.
Question 16
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 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.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency for a Carnot engine:
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 600 K and the cold reservoir
temperature is 300 K, we have:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot engine is 50%.
Step 2: Calculate the amount of heat rejected to the cold reservoir. Since
the engine absorbs 4000 J of heat from the hot reservoir in each cycle, by
conservation of energy, the amount of heat rejected to the cold reservoir must
also be 4000 J.
Hence, the amount of heat rejected to the cold reservoir is 4000 J.
14
Question 17
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, with
Th> Tc. The engine absorbs Qhamount of heat from the hot reservoir and
discards Qcamount of heat to the cold reservoir. Calculate the efficiency of this
heat engine in terms of Th,Tc,Qh, and Qc.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency of a
heat engine is given by the formula:
Efficiency = 1 −Qc
Qh
Step 2: Determine the expression for Qcin terms of Th,Tc, and Qh: Ac-
cording to the second law of thermodynamics, we have:
Qc
Tc
=Qh
Th
Solving for Qc, we get:
Qc=Tc
Th
Qh
Step 3: Substitute the expression for Qcinto the efficiency formula:
Efficiency = 1 −Tc
Th
Therefore, the efficiency of the heat engine in terms of Th,Tc,Qh, and Qc
is 1 −Tc
Th.
Question 18
Question
A heat engine operates between two heat reservoirs at temperatures TH= 500 K
and TC= 300 K. The engine consumes 500 J of heat from the hot reservoir in
each cycle. Calculate the maximum efficiency of the engine according to the
second law of thermodynamics.
Solution
Step 1: Determine the maximum efficiency of the engine using Carnot’s theo-
rem. According to Carnot’s theorem, the maximum efficiency of a heat engine
operating between two reservoirs at temperatures THand TCis given by:
ηmax = 1 −TC
TH
15
Step 2: Substitute the given values into the efficiency formula. Given TH=
500 K and TC= 300 K, we can calculate the maximum efficiency:
ηmax = 1 −300
500 = 1 −0.6=0.4
Step 3: Convert the efficiency to a percentage. To express the efficiency as
a percentage, we multiply by 100:
ηmax = 0.4×100% = 40%
Therefore, the maximum efficiency of the heat engine operating between the
two given heat reservoirs is 40
Question 19
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine has an efficiency of 40%. Determine the maximum possible
efficiency for a heat engine operating between reservoirs at these temperatures.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
η= 1 −Tcold
Thot
Step 2: Substitute the given temperatures into the equation:
η= 1 −300
600
Step 3: Simplify the expression:
η= 1 −1
2=1
2= 50%
Answer: The maximum possible efficiency for a heat engine operating be-
tween reservoirs at 600 K and 300 K is 50%.
Question 20
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir. If the efficiency of the engine is η, prove that the
efficiency can be expressed as η= 1 −Tc
Th.
16
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency (η)
of a heat engine is defined as the ratio of the work done by the engine to the
heat absorbed from the hot reservoir:
η=W
Qh
Step 2: Apply the first law of thermodynamics for the engine: The first law
of thermodynamics states that the net work done by the engine is equal to the
difference between the heat absorbed and the heat exhausted by the engine:
W=Qh−Qc
Step 3: Substitute the expression for work done from Step 2 into the effi-
ciency equation from Step 1:
η=Qh−Qc
Qh
Step 4: Rearrange the expression using algebraic manipulation:
η= 1 −Qc
Qh
Step 5: Apply the definition of efficiency from the Carnot cycle: For a Carnot
engine, the efficiency is given by the formula:
η= 1 −Tc
Th
Step 6: Equate the equations for efficiency from Step 4 and Step 5:
1−Qc
Qh
= 1 −Tc
Th
Step 7: Simplify the equation to obtain the desired expression for efficiency:
η= 1 −Tc
Th
Therefore, the efficiency of the heat engine can be expressed as η= 1 −Tc
Th.
Question 21
Question
A heat engine operates between a hot reservoir at 500K and a cold reservoir
at 300K. If the engine absorbs 2000J of heat from the hot reservoir during
each cycle, what is the maximum efficiency of the engine? Also, determine the
amount of heat rejected to the cold reservoir during each cycle.
17
Solution
Let’s denote the heat absorbed from the hot reservoir as Qh= 2000 J, the
temperature of the hot reservoir as Th= 500 K, and the temperature of the
cold reservoir as Tc= 300 K.
Step 1: Find the efficiency of the engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Tc
Th
Substitute the given values:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Therefore, the maximum efficiency of the engine is 40
Step 2: Find the amount of heat rejected to the cold reservoir. The
amount of heat rejected to the cold reservoir during each cycle can be calculated
using the formula:
Qc=Qh−W
where Wis the work done by the engine during each cycle. Since the engine is
working in a cycle, the work done is equal to the difference between the heat
absorbed and the heat rejected.
W=Qh−Qc
Substitute the given values and solve for the heat rejected:
Qc= 2000 −(2000 ×0.4) = 2000 −800 = 1200 J
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 1200 J.
Question 22
Question
A Carnot engine operates between a hot reservoir at a temperature of 500 K
and a cold reservoir at a temperature of 300 K. If it absorbs 500 J of heat from
the hot reservoir in each cycle, determine:
1. The efficiency of the engine.
2. The amount of heat rejected to the cold reservoir in each cycle.
18
Solution
1. To find the efficiency of the Carnot engine, we use the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs, re-
spectively.
Step 1: Given Thot = 500 K and Tcold = 300 K, we plug these values into
the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4
2. To find the amount of heat rejected to the cold reservoir in each cycle,
we can use the fact that for a Carnot engine, the net work output in each cycle
is equal to the difference in the heat absorbed from the hot reservoir and the
heat rejected to the cold reservoir. Therefore:
Qcold =Tcold
Thot
·Qhot
where Qcold is the heat rejected to the cold reservoir and Qhot is the heat
absorbed from the hot reservoir.
Step 2: Given Qhot = 500 J, we calculate the heat rejected to the cold
reservoir:
Qcold =300
500 ·500 = 300 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
300 J .
Question 23
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold,
where Thot = 500 K and Tcold = 300 K. The engine absorbs 500 J of heat
from the hot reservoir in each cycle and exhausts 300 J to the cold reservoir.
Calculate the efficiency of the engine.
Solution
Step 1: Calculate the efficiency of the engine.
The efficiency of a heat engine is given by the formula:
Efficiency = 1 −Tcold
Thot
19
Given that Thot = 500 K and Tcold = 300 K, we can substitute these values
into the formula:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4
Therefore, the efficiency of the engine is 40
Question 24
Question
A heat engine operates between a hot reservoir at 600◦Cand a cold reservoir
at 20◦C. The engine has an efficiency of 40
Solution
Step 1: Convert the temperatures to Kelvin. Given: Hot reservoir temperature,
Thot = 600◦C= 600 + 273 = 873 K
Cold reservoir temperature, Tcold = 20◦C= 20 + 273 = 293 K
Step 2: Calculate the maximum efficiency using Carnot efficiency formula.
The Carnot efficiency is given by:
Maximum efficiency, ηCarnot = 1 −Tcold
Thot
Substitute the values:
ηCarnot = 1 −293
873
ηCarnot = 1 −293
873
ηCarnot = 1 −293
873
ηCarnot = 1 −0.3360
ηCarnot = 0.6640 or 66.40%
Step 3: Analyze the obtained result. The maximum efficiency for this engine
(66.40
Question 25
Question
A heat engine operates between two reservoirs at temperatures T1= 400 K and
T2= 100 K. If the engine extracts 2000 J of heat from the hot reservoir in each
cycle, calculate the maximum possible efficiency of the engine. Also, discuss
whether this efficiency violates the second law of thermodynamics.
20
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = 1 −T2
T1
Step 2: Substitute the given values into the formula:
Efficiency = 1 −100
400 = 1 −1
4=3
4= 75%
Step 3: Discuss whether this efficiency violates the second law of thermody-
namics. According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Question 26
Question
A heat engine operates with a hot reservoir temperature of 500 K and a cold
reservoir temperature of 300 K. The engine absorbs 2000 J of heat from the hot
reservoir in each cycle and exhausts 1400 J to the cold reservoir.
a) Calculate the efficiency of the heat engine. b) Determine the maximum
possible efficiency of a heat engine operating between these two reservoir tem-
peratures. c) Explain how the second law of thermodynamics relates to the
efficiency of heat engines.
Solution
a) To calculate the efficiency of the heat engine, we can use the formula:
Efficiency = 1 −Qc
Qh
Where Qcis the heat expelled to the cold reservoir and Qhis the heat
absorbed from the hot reservoir.
Step 1: Calculate the efficiency using the given values:
Efficiency = 1 −1400
2000
Efficiency = 1 −0.7 = 0.3
Therefore, the efficiency of the heat engine is 30
b) The maximum possible efficiency of a heat engine operating between two
reservoir temperatures Thand Tcis given by Carnot efficiency:
EfficiencyCarnot = 1 −Tc
Th
21
Step 2: Calculate the Carnot efficiency using the given temperatures:
EfficiencyCarnot = 1 −300
500
EfficiencyCarnot = 1 −0.6=0.4
Therefore, the maximum possible efficiency of a heat engine operating be-
tween 500 K and 300 K is 40
c) The second law of thermodynamics states that it is impossible for any
heat engine to be 100
Question 27
Question
A heat engine operating between a hot reservoir at 600 K and a cold reservoir
at 300 K has an efficiency of 40%. If the engine absorbs 2000 J of energy from
the hot reservoir in each cycle, calculate:
1. The work done by the engine in each cycle.
2. The heat energy rejected to the cold reservoir in each cycle.
Solution
Let’s denote the efficiency of the heat engine as η, the heat absorbed from the
hot reservoir as Qh, the work done by the engine as W, and the heat rejected
to the cold reservoir as Qc. We know that the efficiency is given by:
η=W
Qh
Step 1: Calculate the work done by the engine in each cycle. Given that
the efficiency of the engine is 40% or 0.4, we have:
η= 0.4 = W
Qh
0.4 = W
2000 J
W= 0.4×2000 J
W= 800 J
Therefore, the work done by the engine in each cycle is 800 J.
Step 2: Calculate the heat energy rejected to the cold reservoir in each
cycle. Since the efficiency of the engine is given by:
η=Qh−Qc
Qh
22
We can rearrange this equation to find Qc:
Qc= (1 −η)×Qh
Qc= (1 −0.4) ×2000 J
Qc= 0.6×2000 J
Qc= 1200 J
Therefore, the heat energy rejected to the cold reservoir in each cycle is 1200
J.
Question 28
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. If the engine produces work W, prove that the efficiency
of the engine is given by η= 1 −Tc
Th.
Solution
Step 1: According to the first law of thermodynamics, the net work done by
the engine is equal to the difference between the heat absorbed and the heat
released:
W=Qh−Qc
Step 2: The efficiency of the engine is defined as the ratio of the work done
by the engine to the heat absorbed from the hot reservoir:
η=W
Qh
Step 3: Substituting the expression for Wfrom Step 1 into the efficiency
formula:
η=Qh−Qc
Qh
Step 4: Simplifying the expression:
η= 1 −Qc
Qh
Step 5: We can relate Qcand Qhto the temperatures Tcand Thusing the
Carnot efficiency:
ηC= 1 −Tc
Th
=Qh−Qc
Qh
23
Step 6: If we compare the Carnot efficiency with the efficiency of the engine,
we see that they are equal:
η=ηC= 1 −Tc
Th
Therefore, the efficiency of the engine is given by η= 1 −Tc
Th.
Question 29
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc(Th> Tc). The engine absorbs 600 J of heat from the hot reservoir and
exhausts 400 J to the cold reservoir for each cycle. Find the efficiency of the
engine.
Solution
Let’s denote the efficiency of the Carnot engine as η. We know that the efficiency
of a Carnot engine is given by the formula:
η= 1 −Tc
Th
Step 1: Calculate the efficiency using the given values of heat exchange:
η= 1 −Tc
Th
We are given that the engine absorbs 600 J of heat (Qh= 600 J) and exhausts
400 J (Qc= 400 J) for each cycle. By the first law of thermodynamics, the net
work done by the engine in each cycle is given by:
W=Qh−Qc
This work output is used to express the efficiency in terms of the heat ex-
change:
η= 1 −Qc
Qh
Substitute the given values of heat exchange:
η= 1 −400
600
η= 1 −2
3
η=1
3
24
Step 2: Express the efficiency in percentage:
Efficiency(%) = η×100%
Substitute the value of η:
Efficiency(%) = 1
3×100%
Efficiency(%) = 33.33%
Therefore, the efficiency of the Carnot heat engine is 33.33
Question 30
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and has an
efficiency of η. Prove that the efficiency of any heat engine operating between the
same two temperatures is less than that of a Carnot engine operating between
the same two temperatures.
Solution
Let’s consider a Carnot engine operating between the same two reservoirs at
temperatures Thand Tc.
Step 1: Identify the efficiency of the Carnot engine. The efficiency of a
Carnot engine is given by:
ηCarnot = 1 −Tc
Th
Step 2: Identify the efficiency of the given engine. The efficiency of the
given engine is ηand the heat absorbed from the hot reservoir is Qh. Therefore,
the work output of the engine is given by:
Wgiven =ηQh
and the heat rejected to the cold reservoir is:
Qc=Qh−Wgiven =Qh(1 −η)
Also, we have Qc=Tc∆S, where ∆Sis the change in entropy.
Step 3: Compare the efficiency of the given engine with that of the Carnot
engine. From step 2, we have:
Qh(1 −η) = Tc∆S
25
Since η < 1, we have 1 −η > 0. Thus, ∆S > 0. Now, the heat absorbed
from the hot reservoir for the Carnot engine is Qh. Using the second law of
thermodynamics, we have:
∆SCarnot =Qh
Th
−Qc
Tc
≥0
Substitute the values of Qh,Qc, and ∆SCarnot:
Th
Th
−Tc
Tc
= 1 −Tc
Th
≥0
Therefore, 1 −Tc
Th≥0, which implies that η < ηCarnot. Hence, the efficiency of
any heat engine operating between the same two temperatures is less than that
of a Carnot engine operating between the same two temperatures.
Question 31
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
exhausts Qcof heat to the cold reservoir in each cycle. If the efficiency of the
engine is η, show that η= 1 −Tc
Th.
Solution
Step 1: Recall the efficiency of a Carnot heat 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 denote the net work done by the engine in each cycle as W.
By the first law of thermodynamics, we have
W=Qh−Qc
Step 3: Since the engine is operating between two heat reservoirs, the heat
absorbed from the hot reservoir must equal the work done plus the heat rejected
to the cold reservoir. Therefore, we have
Qh=W+Qc
Step 4: The efficiency of the engine is defined as the ratio of the work done
to the heat input from the hot reservoir. Hence, we can write
η=W
Qh
26
Step 5: Substituting W=Qh−Qcinto the equation for efficiency, we get
η=Qh−Qc
Qh
Step 6: Simplifying the expression, we find
η= 1 −Qc
Qh
Step 7: We know that for a Carnot engine, the efficiency is given by
η= 1 −Tc
Th
Step 8: Equating the two expressions for efficiency, we have
1−Tc
Th
= 1 −Qc
Qh
Step 9: Since Qc
Qh=Tc
Th, we have shown that
η= 1 −Tc
Th
Question 32
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(where
Th> Tc). The heat engine consumes 4000 J of heat from the hot reservoir and
produces 2000 J of work output. Determine the efficiency of the heat engine
and discuss whether this violates the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = Useful work output
Heat input
Step 2: Given that the heat input to the engine is 4000 J and the work output
is 2000 J, we can plug these values into the formula to find the efficiency:
Efficiency = 2000
4000 = 0.5
Step 3: The efficiency of the heat engine is 50
27
Step 4: According to the second law of thermodynamics, the efficiency of a
heat engine is limited by the Carnot efficiency, which is given by:
Carnot efficiency = 1 −Tc
Th
Step 5: Calculate the Carnot efficiency using the given temperatures:
Carnot efficiency = 1 −Tc
Th
= 1 −2000
4000 = 0.5
Step 6: The Carnot efficiency of the heat engine operating between the
temperatures Thand Tcis 50
Step 7: Since the actual efficiency of the heat engine (50
Question 33
Question
A Carnot heat engine operates between two reservoirs at 500 K and 300 K. If the
heat engine absorbs 1000 J of heat energy from the high-temperature reservoir
in each cycle, calculate:
1. The efficiency of the Carnot engine.
2. The heat energy rejected by the engine 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
where TCis the temperature of the low-temperature reservoir and THis the
temperature of the high-temperature reservoir.
Given that TC= 300 K and TH= 500 K, we can substitute these values into
the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot engine is 40%.
Step 2: Calculate the heat energy rejected by the engine to the low-
temperature reservoir in each cycle. We can use the efficiency formula to find
the heat energy rejected:
Efficiency = Work done by engine
Heat energy absorbed from high-temperature reservoir
28
Since the efficiency is 0.4 and the heat energy absorbed from the high-
temperature reservoir is 1000 J, we can rearrange the formula to find the work
done by the engine:
Work done by engine = Efficiency ×Heat energy absorbed
Work done by engine = 0.4×1000 = 400 J
Since the total energy input is 1000 J and the work done by the engine is
400 J, the heat energy rejected to the low-temperature reservoir is:
Heat energy rejected = Total energy input −Work done by engine
Heat energy rejected = 1000 −400 = 600 J
Therefore, the heat energy rejected by the engine to the low-temperature
reservoir in each cycle is 600 J.
Question 34
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tcwhere Th> Tc. If the engine absorbs 600 J of heat energy from the high-
temperature reservoir and exhausts 375 J to the low-temperature reservoir dur-
ing each cycle, what is the efficiency of the engine? Explain your answer in
terms of the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a Carnot heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given values into the formula. The engine absorbs
600 J of heat energy from the high-temperature reservoir (Th) and exhausts 375
J to the low-temperature reservoir (Tc). Therefore, Th= 600 J and Tc= 375 J.
Step 3: Calculate the efficiency using the formula:
Efficiency = 1 −375
600
Step 4: Simplify the expression to find the efficiency:
Efficiency = 1 −375
600 = 1 −5
8=3
8= 0.375
Step 5: The efficiency of the engine is 0.375 or 37.5
29
Question 35
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine is able to extract 4000 J of heat from the hot reservoir
and delivers 2500 J of work per cycle. Calculate the efficiency of the engine and
determine the minimum value of Tcif Th= 500 K.
Solution
Step 1: Calculate the efficiency of the engine. The efficiency of a heat engine is
given by the formula:
Efficiency = Useful work output
Heat input
Given that the engine delivers 2500 J of work and extracts 4000 J of heat, the
efficiency is:
Efficiency = 2500 J
4000 J ×100% = 62.5%
Step 2: Determine the minimum value of Tc. The efficiency of a heat engine
operating between two reservoirs is given by:
Efficiency = 1 −Tc
Th
Given that the efficiency is 62.5
0.625 = 1 −Tc
500
Tc
500 = 0.375
Tc= 0.375 ×500 = 187.5 K
Therefore, the efficiency of the engine is 62.5
30
