PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 4
Liberty University
Question 1
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 6000 J of heat from the hot reservoir and expels
4000 J of heat to the cold reservoir during each complete cycle, calculate: (a)
The efficiency of the engine. (b) The maximum work the engine can perform
per cycle. (c) If the engine is operating at 50
Solution
Step 1: Calculate the efficiency of the engine.
Efficiency, η=Useful work done
Input heat =W
Qh
=Qh−Qc
Qh
where Qhis the heat absorbed from the hot reservoir and Qcis the heat expelled
to the cold reservoir. Given that Qh= 6000 J and Qc= 4000 J, we have:
η=6000 −4000
6000 =2000
6000 =1
3= 33.33%
Step 2: Calculate the maximum work the engine can perform per cycle. The
maximum work is given by:
W=Qh−Qc= 6000 −4000 = 2000 J
Step 3: Determine the temperatures of the hot and cold reservoirs for
50Given that the efficiency, η= 0.5 (50
0.5 = Th−Tc
Th
=⇒Th= 2Tc
Substitute this into the Carnot efficiency formula:
η=Th−Tc
Th
=2Tc−Tc
2Tc
=Tc
2Tc
= 0.5
Solving for Tc:
0.5 = Tc
2Tc
=⇒Tc= 2Tc=⇒Tc=1
2Th
Therefore, the temperatures of the hot and cold reservoirs are in the ratio 2:1.
Question 2
Question
A heat engine operates between a hot reservoir at a temperature of 700 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and delivers 2000 J of work output in each
cycle. 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:
Efficiency = Work Output
Heat Input
Step 2: Substitute the given values into the formula:
Efficiency = 2000 J
4000 J = 0.5
Step 3: The efficiency of the engine is 0.5 or 50%.
Step 4: To determine whether the engine violates the second law of thermo-
dynamics, calculate the efficiency based on the Carnot cycle using the formula:
Maximum Efficiency = 1 −Temperature of Cold Reservoir
Temperature of Hot Reservoir
Step 5: Substitute the given temperatures into the formula:
Maximum Efficiency = 1 −300
700 = 1 −0.4286 = 0.5714
2
Step 6: The maximum efficiency based on the Carnot cycle is 0.5714 or
57.14%.
Step 7: Since the actual efficiency of the engine is 0.5 or 50%, which is less
than the maximum efficiency of 0.5714 or 57.14%, the engine does not violate
the second law of thermodynamics.
Question 3
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th= 500 K and Tc= 300 K. The engine produces 4000 J of heat at Th
in each cycle. Calculate the efficiency of the engine and determine the maximum
amount of work that can be obtained from this engine using the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the Carnot efficiency formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the engine is 40
Step 3: Calculate the maximum work that can be obtained from the engine
using the second law of thermodynamics: The maximum efficiency of a heat
engine operating between two reservoirs at temperatures Thand Tcis given by:
Efficiencymax = 1 −Tc
Th
Step 4: Substitute the given temperatures into the formula:
Efficiencymax = 1 −300
500 = 1 −0.6=0.4
Step 5: The maximum work can be calculated using the formula for effi-
ciency:
Wmax = Efficiencymax ×Qh
Wmax = 0.4×4000 J = 1600 J
Therefore, the maximum amount of work that can be obtained from this
engine is 1600 J.
3
Question 4
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine absorbs 2400 J of heat from the high-temperature reservoir
and rejects 1600 J of heat to the low-temperature reservoir during each cycle.
Calculate the efficiency of the engine.
Solution
1. Recall the efficiency of a heat engine is given by:
η= 1 −QC
QH
where QHis the heat absorbed from the high-temperature reservoir and QCis
the heat rejected to the low-temperature reservoir.
2. We are given that QH= 2400 J and QC= 1600 J. Substitute these values
into the efficiency formula:
η= 1 −1600
2400
3. Simplify the expression:
η= 1 −2
3=1
3
4. Therefore, the efficiency of the heat engine is 1
3or 33.33
Question 5
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. The engine absorbs 4000 J of heat from the hot reservoir and performs
2200 J of work. Calculate the efficiency of the engine. Is this efficiency consistent
with the second law of thermodynamics?
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Heat input
Step 2: First, we need to calculate the heat input. The heat input is the
amount of heat absorbed from the hot reservoir, which is given as 4000 J.
4
Step 3: Next, we calculate the efficiency using the given values:
Efficiency = 2200 J
4000 J
Step 4: Simplify the expression to find the efficiency:
Efficiency = 11
20 = 0.55 or 55%
Step 5: The efficiency of the engine is 55
Step 6: The second law of thermodynamics states that no heat engine can be
more efficient than a Carnot engine operating between the same two reservoirs.
The maximum efficiency of a Carnot engine is given by the Carnot efficiency
formula:
EfficiencyCarnot = 1 −Tc
Th
Step 7: Plugging in the values of the cold and hot reservoir temperatures:
EfficiencyCarnot = 1 −300
800 = 1 −3
8=5
8= 0.625 or 62.5%
Step 8: Since the efficiency of the engine (55
Question 6
Question
A heat engine operates between two reservoirs at temperatures THand TC. If
the engine absorbs 1500 J of heat from the hot reservoir and exhausts 800 J to
the cold reservoir during each cycle, what is the efficiency of the engine in terms
of THand TC?
Solution
Let’s denote the efficiency of the heat engine as η.
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η= 1 −QC
QH
,
where QCis the heat exhausted into the cold reservoir and QHis the heat
absorbed from the hot reservoir during each cycle.
Step 2: Given that the engine absorbs 1500 J from the hot reservoir and
exhausts 800 J to the cold reservoir during each cycle, we have:
QH= 1500 J
QC= 800 J
5
Step 3: Substitute the values of QHand QCinto the efficiency formula:
η= 1 −800
1500
Step 4: Simplify the expression to obtain the efficiency in terms of THand
TC:
η= 1 −8
15 =7
15
Step 5: Therefore, the efficiency of the engine in terms of THand TCis
η=7
15 .
Question 7
Question
A heat engine operates between a hot reservoir at Th= 600 K and a cold
reservoir at Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir
in each cycle and exhausts 1200 J to the cold reservoir in each cycle. Determine
the efficiency of the engine and discuss whether it violates the second law of
thermodynamics.
Solution
Let’s denote the heat absorbed from the hot reservoir as Qh= 2000 J and the
heat exhausted to the cold reservoir as Qc= 1200 J.
Step 1: Calculate the efficiency of the engine using the formula η= 1 −Qc
Qh.
η= 1 −Qc
Qh
= 1 −1200
2000
= 1 −0.6
= 0.4
So, the efficiency of the engine is 40
Step 2: Discuss whether the engine violates the second law of thermody-
namics.
According to the second law of thermodynamics, no heat engine can be 100
Question 8
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 600 J of heat from the hot reservoir and expels
6
300 J to the cold reservoir during each cycle, calculate the efficiency of the
engine and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula to find the effi-
ciency:
Efficiency = 1 −Tc
Th
= 1 −300
600 = 1 −1
2=1
2= 50%
Step 3: Since the efficiency calculated above is less than 100
Question 9
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 maximum amount of
work that can be done in each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine is given by the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs in
Kelvin, respectively.
Given Thot = 600 K and Tcold = 300 K, we can calculate the efficiency:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Calculate the maximum amount of work that can be done in each
cycle. The maximum amount of work that can be done in each cycle by a Carnot
engine is given by the formula:
Maximum Work = Efficiency ×Heat Input
Given that the heat input is 4000 J, the maximum work done in each cycle
is:
Maximum Work = 0.5×4000 = 2000 J
Therefore, the efficiency of the Carnot engine is 50
7
Question 10
Question
A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir
at 300 K. The engine produces 360 J of work per cycle. Calculate the efficiency
of the engine and determine the amount of heat rejected to the cold reservoir
during 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 cold reservoir and This the temperature of
the hot reservoir.
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Step 3: Determine the amount of heat supplied to the engine using the
formula:
Heat input = Work output + Heat rejected
Step 4: Substitute the given work output into the formula:
Heat input = 360 J
Step 5: Calculate the heat rejected to the cold reservoir using the formula:
Heat rejected = Heat input −Work output
Step 6: Substitute the known values into the formula:
Heat rejected = 360 J −360 J = 0 J
Therefore, the efficiency of the engine is 0.4 (or 40
Question 11
Question
A Carnot engine operates between two reservoirs at temperatures T1and T2,
with T1> T2. The engine absorbs heat Q1from the high-temperature reservoir
and exhausts heat Q2to the low-temperature reservoir. If the efficiency of the
engine is η, prove that the efficiency of a Carnot engine is given by the expression
η= 1 −T2
T1
.
8
Solution
Step 1: Recall the efficiency of a Carnot engine is given by
η= 1 −Q2
Q1
.
Step 2: Using the first law of thermodynamics, we know that for a complete
cycle, the net work done by the engine is equal to the difference between the
heat absorbed and the heat released:
W=Q1−Q2.
Step 3: Since the engine is reversible and operates between two reservoirs,
we can express the heat absorbed and heat released in terms of the reservoir
temperatures:
Q1=T1S1, Q2=T2S2,
where S1and S2are the entropy changes associated with each reservoir.
Step 4: The second law of thermodynamics states that the entropy change
of the universe for any reversible process must be zero. Therefore, the sum of
the entropy changes of the reservoirs must be equal:
S1+S2= 0.
Step 5: From Step 3 and Step 4, we have
T1S1+T2S2= 0.
Step 6: Rearranging the above equation, we get
T1S1=−T2S2.
Step 7: Substituting Q1and Q2back into the expression for efficiency, we
have
η= 1 −T2S2
T1S1
= 1 −T2
T1
.
Step 8: Therefore, we have proven that the efficiency of a Carnot engine is
given by η= 1 −T2
T1.
Question 12
Question
A heat engine operates between a hot reservoir at 600◦C and a cold reservoir
at 27◦C. The engine produces 2400 J of work and rejects 4200 J of heat to the
cold reservoir. Determine the efficiency of the engine. Is this engine violating
the second law of thermodynamics?
9
Solution
Step 1: Find the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Calculate the heat input to the engine using the first law of thermo-
dynamics:
Heat input = Useful work output + Heat rejected to cold reservoir
Step 3: Use the second law of thermodynamics to determine if the engine is
violating the law. An engine operating between two reservoirs cannot have an
efficiency greater than the Carnot efficiency:
Efficiency ≤1−Tcold
Thot
where Tcold is the absolute temperature of the cold reservoir and Thot is the
absolute temperature of the hot reservoir.
Question 13
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 in each cycle.
Calculate the maximum theoretical efficiency of this engine.
Solution
To find the maximum theoretical efficiency of the heat engine, we can use the
Carnot efficiency formula:
Efficiency = 1 −Tcold
Thot
where Thot is the temperature of the hot reservoir in Kelvin and Tcold is the
temperature of the cold reservoir in Kelvin.
Step 1: Convert temperatures to Kelvin Given that the hot reservoir
temperature Thot = 500 K and the cold reservoir temperature Tcold = 300 K.
Step 2: Calculate the efficiency Plugging the values into the formula:
Efficiency = 1 −300
500
Efficiency = 1 −0.6 = 0.4
Therefore, the maximum theoretical efficiency of the engine is 40
10
Question 14
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine absorbs 5000 J of heat from the hot reservoir in each cycle.
Calculate the maximum possible efficiency of the engine.
Solution
Step 1: Recall the formula for the efficiency (η) of a heat engine:
η= 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Step 2: Substitute the given values into the formula:
TC= 300 K
TH= 600 K
Step 3: Calculate the efficiency using the formula:
η= 1 −300
600
η= 1 −1
2
η=1
2= 50%
Step 4: Therefore, the maximum possible efficiency of the engine is 50%.
Question 15
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
The engine has an efficiency of 40If the engine absorbs 600 J of heat from the
hot reservoir in each cycle, what is the amount of heat expelled to the cold
reservoir in each cycle?
11
Solution
Let Qin be the heat absorbed from the hot reservoir, Qout be the heat expelled
to the cold reservoir, and ηbe the efficiency of the engine.
Step 1: Recall the formula for the efficiency of a heat engine:
η=Useful work output
Heat input =W
Qin
η=W
Qin
=⇒W=η·Qin
Since the efficiency is given as 40
Step 2: The work done by the engine in each cycle can be calculated using
the efficiency:
W=η·Qin = 0.4×600 J = 240 J
Step 3: The heat expelled to the cold reservoir in each cycle is the difference
between the heat absorbed and the work done:
Qout =Qin −W= 600 J −240 J = 360 J
Step 4: Therefore, the amount of heat expelled to the cold reservoir in each
cycle is 360 J.
Question 16
Question
A Carnot heat engine operates between a hot reservoir at 527◦C and a cold
reservoir at 27◦C. The engine absorbs 6000 J of heat from the hot reservoir in
each cycle. Calculate:
1. The efficiency of the engine.
2. The work done by the engine in each cycle.
3. The amount of heat exhausted to the cold reservoir in each cycle.
Solution
1. To calculate the efficiency of the Carnot engine, we use the formula
Efficiency = 1 −Tcold
Thot
,
where Thot and Tcold are the temperatures of the hot and cold reservoirs in
Kelvin.
Step 1: Convert the temperatures to Kelvin:
Thot = 527 + 273 = 800 K,
12
Tcold = 27 + 273 = 300 K.
Step 2: Calculate the efficiency:
Efficiency = 1 −300
800 = 1 −3
8=5
8= 62.5%.
Therefore, the efficiency of the Carnot engine is 62.5
2. The work done by the engine in each cycle can be calculated using the
formula
Work = Efficiency ×Heat absorbed from hot reservoir.
Step 3: Calculate the work done:
Work = 0.625 ×6000 = 3750 J.
Therefore, the work done by the engine in each cycle is 3750 J.
3. The amount of heat exhausted to the cold reservoir in each cycle is equal
to the difference between the heat absorbed from the hot reservoir and the work
done:
Heat exhausted = Heat absorbed −Work.
Step 4: Calculate the heat exhausted:
Heat exhausted = 6000 −3750 = 2250 J.
Therefore, the amount of heat exhausted to the cold reservoir in each cycle
is 2250 J.
Question 17
Question
A Carnot heat engine operates between a high temperature reservoir at 800 K
and a low temperature reservoir at 300 K. The engine takes in 6,000 J of heat
from the high temperature reservoir in each cycle. Calculate the maximum
efficiency of the engine and the work output per cycle.
Solution
Step 1: Calculate the maximum efficiency of the Carnot engine using the formula
for efficiency:
Efficiency = 1 −TC
TH
where THis the absolute temperature of the high temperature reservoir, and
TCis the absolute temperature of the low temperature reservoir.
Step 2: Convert the temperatures to Kelvin: TH= 800 K and TC= 300 K.
13
Step 3: Substitute the values into the formula for efficiency:
Efficiency = 1 −300
800
Step 4: Calculate the efficiency:
Efficiency = 1 −3
8=5
8= 0.625
Step 5: The maximum efficiency of the Carnot engine is 0.625 or 62.5
Step 6: Calculate the work output per cycle using the formula:
Work output = Efficiency ×Heat input
Step 7: Given that the heat input is 6,000 J, and the efficiency is 0.625,
substitute these values into the formula:
Work output = 0.625 ×6,000
Step 8: Calculate the work output:
Work output = 0.625 ×6,000 = 3,750 J
Step 9: The maximum efficiency of the engine is 62.5
Question 18
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 2000 J of heat from the hot reservoir during each
cycle and delivers 1200 J of work. Determine: (a) The efficiency of the engine.
(b) The amount of heat rejected to the cold reservoir during each cycle.
Solution
Step 1: We can start by calculating the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input
Given that the engine absorbs 2000 J of heat and delivers 1200 J of work, we
have:
Efficiency = 1200 J
2000 J
Step 2: Now we can calculate the efficiency:
Efficiency = 0.6 = 60%
14
Step 3: Next, we can determine the amount of heat rejected to the cold
reservoir during each cycle by using the fact that the heat input is equal to the
sum of the work done and the heat rejected:
Heat input = Useful work output + Heat rejected
Step 4: Since the engine absorbs 2000 J of heat and delivers 1200 J of work,
we can rearrange the equation to solve for the heat rejected:
Heat rejected = Heat input −Useful work output = 2000 J −1200 J
Step 5: Calculating the heat rejected:
Heat rejected = 800 J
Therefore, during each cycle, 800 J of heat is rejected to the cold reservoir.
Question 19
Question
A Carnot engine operates between two reservoirs at temperatures Thot and Tcold.
The engine takes in Qhot of heat from the hot reservoir and expels Qcold to the
cold reservoir. If the efficiency of this engine is η, prove that the efficiency can
also be expressed as η= 1 −Tcold
Thot .
Solution
Step 1: From the definition of efficiency, we have
η= 1 −Qcold
Qhot
.
Step 2: By the first law of thermodynamics, the net work done by the engine
is Wnet =Qhot −Qcold.
Step 3: The efficiency of a Carnot engine is given by η=Wnet
Qhot . Substituting
the expression for Wnet into this formula, we have
η=Qhot −Qcold
Qhot
= 1 −Qcold
Qhot
.
Step 4: Now, we know that for a Carnot engine, the efficiency is given by
η= 1 −Tcold
Thot . We have to show that this is equivalent to the expression we
derived earlier.
Step 5: From the Carnot efficiency formula, we have η= 1 −Tcold
Thot . Multi-
plying both sides by Qhot gives Qhotη=Qhot −Qcold.
Step 6: Dividing both sides of our previous expression η= 1 −Qcold
Qhot by Qhot
yields Qhotη=Qhot −Qcold.
Step 7: Thus, we have shown that η= 1−Tcold
Thot is equivalent to the expression
for efficiency we derived earlier.
15
Question 20
Question
A Carnot heat engine operates between a hot reservoir at 800 K and a cold
reservoir at 400 K. If the engine absorbs 5000 J of heat from the hot reservoir
in each cycle, calculate the efficiency of the engine. Also, discuss how this effi-
ciency compares to the maximum possible efficiency for a heat engine operating
between the same two reservoirs.
Solution
Step 1: Calculate the efficiency of the Carnot heat engine. The efficiency of a
Carnot heat engine is given by the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given that the hot reservoir temperature TH= 800 K and the cold reservoir
temperature TC= 400 K, we can substitute these values into the formula to find
the efficiency:
Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Thus, the efficiency of the Carnot heat engine is 50
Step 2: Compare the efficiency of the engine to the maximum possible effi-
ciency. The maximum possible efficiency of any heat engine operating between
the same two reservoirs is given by the formula:
Max Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
In this case, the maximum efficiency of a heat engine operating between the
400 K and 800 K reservoirs would also be:
Max Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot heat engine is the same as the max-
imum possible efficiency for a heat engine operating between the same two
reservoirs.
Question 21
Question
A heat engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc, where Th> Tc. The engine absorbs Qhjoules of
16
heat from the hot reservoir and exhausts Qcjoules of heat to the cold reservoir.
If the engine does 250 kJ of work per cycle, find the efficiency of the engine in
terms of Qhand Th. Also, explain how this result is related to the second law
of thermodynamics.
Solution
Step 1: We know that the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Input heat energy
Step 2: The useful work output is the work done by the engine per cycle,
which is 250 kJ.
Step 3: The input heat energy is the heat energy absorbed from the hot
reservoir, Qh.
Step 4: Therefore, the efficiency of the engine can be expressed as:
Efficiency = 250 kJ
Qh
Step 5: We also know that for a heat engine operating between two reservoirs,
the efficiency is given by:
Efficiency = 1 −Tc
Th
Step 6: Equating the two expressions for efficiency gives:
250 kJ
Qh
= 1 −Tc
Th
Step 7: Solving for the efficiency in terms of Qhand Th:
Efficiency = 250 kJ
Qh
= 1 −Tc
Th
Step 8: This result demonstrates the relationship between the efficiency of
the engine, the input heat energy, and the temperatures of the reservoirs. It is
related to the second law of thermodynamics, which states that no heat engine
can be 100
Question 22
Question
A heat engine operates between two reservoirs at temperatures 600 K and 300
K. The engine takes in 2000 J of heat from the high-temperature reservoir in
each cycle and delivers 800 J of work. Calculate the efficiency of the engine. Is
this efficiency consistent with the second law of thermodynamics?
17
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Substitute the given values into the formula:
Efficiency = 800 J
2000 J
Step 3: Simplify the expression to find the efficiency:
Efficiency = 4
10 = 0.4 = 40%
Step 4: Discuss whether this efficiency is consistent with the second law
of thermodynamics. The efficiency of a heat engine is given by the Carnot
efficiency:
EfficiencyCarnot = 1 −Tc
Th
Step 5: Substitute the given temperatures into the Carnot efficiency formula:
EfficiencyCarnot = 1 −300
600 = 1 −1
2=1
2= 0.5 = 50%
Step 6: Compare the efficiency of the engine (40
Question 23
Question
A heat engine operates between two reservoirs at temperatures THand TC.
The engine absorbs QHof heat from the hot reservoir and exhausts QCof heat
to the cold reservoir. If the efficiency of this engine is ηand the second law
of thermodynamics states that QC≥TC
THQH, prove that the efficiency of the
engine is given by η≤1−TC
TH.
Solution
Step 1: We know that the efficiency of a heat engine is given by
η= 1 −QC
QH
Step 2: Given QC≥TC
THQH, we can rewrite the efficiency expression as
η≤1−TC
TH
QH
QH
= 1 −TC
TH
18
Step 3: Therefore, we have proved that the efficiency of the engine is given
by η≤1−TC
TH.
Question 24
Question
A Carnot heat engine operates between two reservoirs at temperatures Th= 600
K and Tc= 300 K. The engine absorbs 800 J of heat from the hot reservoir
during each cycle. Calculate the efficiency of the engine and determine the
amount of heat rejected to the cold reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 0.5
Thus, the efficiency of the Carnot engine is 50
Step 3: To determine the amount of heat rejected to the cold reservoir during
each cycle, we use the fact that for a Carnot engine:
Efficiency = work done
heat absorbed
Step 4: Rearranging the formula to solve for work done:
work done = Efficiency ×heat absorbed
Step 5: Substitute the values for efficiency and heat absorbed:
work done = 0.5×800 J = 400 J
Step 6: Since the total energy input is 800 J, and 400 J is converted to work,
the remaining 400 J must be rejected to the cold reservoir.
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 400 J.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir, while doing work Wduring a complete cycle.
19
Prove that the maximum efficiency of the heat engine is given by ηmax =
1−Tc
Th.
Solution
Step 1: We need to determine the efficiency of the heat engine. The efficiency
of a heat engine is given by the ratio of the work done by the engine to the heat
absorbed from the hot reservoir:
Efficiency η=W
Qh
Step 2: Express Qhin terms of Qcand W. Since the engine is operating
in a cycle, the net heat absorbed by the engine must be equal to the net work
done by the engine plus the heat exhausted to the cold reservoir:
Qh=W+Qc
Step 3: Substitute the relationship between Qhand Qcback into the effi-
ciency expression.
η=W
W+Qc
Step 4: Express the heat exhausted to the cold reservoir in terms of the
temperatures. From the second law of thermodynamics, we know that the heat
exhausted to the cold reservoir can be expressed as:
Qc=Tc∆S
where ∆Sis the change in entropy.
Step 5: Express ∆Sin terms of Qhand Tc. Using the relationship between
the change in entropy and heat transfer:
∆S=Qh
Th
−Qc
Tc
Step 6: Substitute the expression for Qcin terms of entropy into the effi-
ciency formula.
η=W
W+TcQh
Th−Qc
Tc
Step 7: Simplify the expression and express the maximum efficiency. After
simplifying and rearranging terms, the efficiency can be expressed as:
η= 1 −Tc
Th
Therefore, the maximum efficiency of the heat engine is given by ηmax =
1−Tc
Th.
20
Question 26
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. If the engine has an efficiency of 40
Solution
Let’s denote the efficiency of the given heat engine as ηgiven = 0.40 and the
temperatures of the hot and cold reservoirs as Thot = 600 K and Tcold = 300 K,
respectively. We want to calculate the maximum theoretical efficiency of a heat
engine operating between these two reservoirs, which is governed by Carnot’s
theorem.
Step 1: Calculate the Carnot efficiency using the formula:
ηCarnot = 1 −Tcold
Thot
.
Step 2: Substitute the given temperatures into the formula:
ηCarnot = 1 −300
600 = 1 −1
2=1
2= 0.50.
Step 3: Compare the given efficiency with the Carnot efficiency: Since the
given efficiency ηgiven = 0.40 is less than the Carnot efficiency ηCarnot = 0.50, we
conclude that the given efficiency is less than the maximum theoretical efficiency
of a heat engine operating between the two given reservoirs according to the
second law of thermodynamics.
Question 27
Question
A Carnot heat engine operates between two reservoirs at temperatures of 500
K and 300 K. If the engine absorbs 400 J of heat from the high-temperature
reservoir in each cycle, what is the efficiency of the engine? Additionally, discuss
how the second law of thermodynamics applies to this scenario.
Solution
Step 1: Calculate the work done by the Carnot heat engine. Given that the
engine absorbs 400 J of heat from the high-temperature reservoir in each cycle,
the work done by the engine is given by the difference between the heat absorbed
from the hot reservoir (Qh) and the heat rejected to the cold reservoir (Qc) in
each cycle. This can be calculated using the formula for the efficiency of a
Carnot engine:
Efficiency = 1 −Tc
Th
21
Step 2: Calculate the efficiency of the Carnot heat engine. First, we need
to calculate the temperatures in Kelvin by adding 273 to the Celsius values
given: - High-temperature reservoir: Th= 500 K - Low-temperature reservoir:
Tc= 300 K
Now we can plug these values into the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot heat engine operating between these
two reservoirs is 40
Step 3: Discuss the second law of thermodynamics. The second law of
thermodynamics states that in any cyclic process, the efficiency of a heat engine
is less than 100
Question 28
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
1. The work done by the engine in each cycle.
2. The amount of heat rejected to the cold reservoir in each cycle.
3. Comment on whether this engine violates the second law of thermody-
namics.
Solution
1. To find the work done by the engine in each cycle, we will use the formula
for efficiency (η) of a heat engine:
η= 1 −Qout
Qin
where ηis the efficiency, Qout is the heat rejected to the cold reservoir, and Qin
is the heat absorbed from the hot reservoir. Given that the efficiency is 40
0.40 = 1 −Qout
2400
Qout = 0.60 ×2400 = 1440 J
The work done by the engine is the difference between the heat absorbed
and the heat rejected:
W=Qin −Qout = 2400 −1440 = 960 J
22
2. The amount of heat rejected to the cold reservoir in each cycle is Qout,
which we found to be 1440 J.
3. The efficiency of the engine is less than 1, which is consistent with the
second law of thermodynamics. The second law states that no heat engine can
be 100
Question 29
Question
A Carnot heat engine operates between reservoirs at temperatures THand TC
with TH> TC. If the efficiency of the engine is η, show that the maximum
possible efficiency of a heat engine operating between the same two reservoirs
is 1 −TC
TH.
Solution
Step 1: Recall the efficiency of a Carnot heat engine: The efficiency of a Carnot
heat engine is given by the formula
η= 1 −TC
TH
Step 2: Consider an arbitrary heat engine operating between the same reser-
voirs. Let’s denote the efficiency of this arbitrary heat engine as η′.
Step 3: Applying the second law of thermodynamics. According to the
second law of thermodynamics, the efficiency of a real heat engine must always
be less than or equal to the efficiency of a Carnot heat engine operating between
the same two reservoirs. Therefore, we have
η′≤η
Step 4: Substitute the expressions for ηand η′. Substitute the expressions
for ηand η′:
η′≤1−TC
TH
Step 5: Rearrange the inequality. Rearrange the inequality to isolate η′:
η′≤1−TC
TH
η′≤1−TC
TH
Step 6: Conclusion. Therefore, the maximum possible efficiency of a heat
engine operating between the same two reservoirs is 1 −TC
TH.
23
Question 30
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tcwith Th> Tc. If the engine absorbs 5000 J of heat from the hot reservoir
and exhausts 3000 J of heat to the cold reservoir in one cycle, calculate: (a) The
efficiency of the engine. (b) The work done by the engine. (c) The maximum
possible efficiency of an engine operating between the same two reservoirs.
Solution
Step 1: Find the efficiency of the engine. The efficiency of a Carnot heat engine
is given by the formula:
Efficiency = 1−Tc
Th×100%
Given: Qh= 5000 J (heat absorbed from hot reservoir) Qc= 3000 J (heat
exhausted to cold reservoir)
Efficiency = 1−Qc
Qh×100%
Efficiency = 1−3000
5000×100%
Efficiency = 0.4×100%
Efficiency = 40%
Therefore, the efficiency of the engine is 40
Step 2: Find the work done by the engine. The work done by the engine in
one cycle is given by the formula:
Work done = Qh−Qc
Given: Qh= 5000 J Qc= 3000 J
Work done = 5000 J −3000 J
Work done = 2000 J
Therefore, the work done by the engine is 2000 J.
Step 3: Find the maximum possible efficiency of an engine operating between
the same two reservoirs. The maximum possible efficiency of any heat engine
operating between two reservoirs at temperatures Thand Tcis given by the
formula:
Maximum efficiency = 1−Tc
Th×100%
24
Given: Th> Tc
Therefore, for the given two reservoirs, the maximum possible efficiency is
Maximum efficiency = 1−Tc
Th×100%
Therefore, the maximum possible efficiency is 1−Tc
Th×100%.
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the heat engine absorbs 5000 J of heat from the reservoir at Thand
rejects 3000 J of heat to the reservoir at Tc, what is the efficiency of the heat
engine? Is this operation consistent with the second law of thermodynamics?
Justify your answer.
Solution
Step 1: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Qc
Qh
where Qcis the heat rejected to the cold reservoir and Qhis the heat absorbed
from the hot reservoir.
Given that Qh= 5000 J and Qc= 3000 J, we can substitute these values
into the efficiency formula:
Efficiency = 1 −3000
5000 = 1 −0.6=0.4
Therefore, the efficiency of the heat engine is 40
Step 2: Determine if the operation is consistent with the second law of
thermodynamics. According to the second law of thermodynamics, the efficiency
of a heat engine is limited by the Carnot efficiency:
Carnot efficiency = 1 −Tc
Th
Given that Th> Tc, the Carnot efficiency will always be less than 1, which
means no heat engine can be 100
Carnot efficiency = 1 −Tc
Th
<1
Since the efficiency of the heat engine we calculated (40
25
Question 32
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine takes in 600 J of heat from the hot reservoir in each cycle and
expels 360 J of heat to the cold reservoir in each cycle. Calculate the efficiency of
the engine and discuss whether this violates the second law of thermodynamics.
Solution
Step 1: We can start by calculating the efficiency of the engine using the formula:
Efficiency = 1 −Qcold
Qhot
Step 2: Given that Qhot = 600 J and Qcold = 360 J, we can plug these values
into the formula to find:
Efficiency = 1 −360
600
Step 3: Calculating this expression gives us:
Efficiency = 1 −360
600 = 1 −0.6=0.4
Step 4: Therefore, the efficiency of the engine is 0.4 or 40
Step 5: According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Question 33
Question
A heat engine operates between a hot reservoir at 650◦Cand a cold reservoir
at 20◦C. The engine has an efficiency of 30%. Calculate the maximum possible
efficiency of this heat engine according to the second law of thermodynamics.
Solution
Step 1: Convert the temperatures to Kelvin by adding 273. Hot reservoir tem-
perature (TH) = 650 + 273 = 923 K
Cold reservoir temperature (TC) = 20 + 273 = 293 K
Step 2: Calculate the maximum possible efficiency using the Carnot effi-
ciency formula:
Efficiencymax = 1 −TC
TH
26
Step 3: Substitute the given temperatures into the formula to find the max-
imum efficiency.
Efficiencymax = 1 −293
923
Efficiencymax = 1 −0.3173
Efficiencymax ≈0.6827 or 68.27%
Therefore, the maximum possible efficiency of the heat engine according to
the second law of thermodynamics is 68.27%.
Question 34
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the hot reservoir and delivers W
of work. If the efficiency of the engine is η, prove that the maximum efficiency
of the engine is given by ηmax = 1 −Tc
Th.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is defined as the ratio of the work output to the heat input:
η=W
Qh
Step 2: Next, consider the second law of thermodynamics: The second law
of thermodynamics states that no cyclic engine can be 100
Step 3: To find the maximum efficiency of the engine, we can analyze the
Carnot cycle which gives the maximum efficiency for a heat engine.
Step 4: In a Carnot cycle, the efficiency ηmax is given by:
ηmax = 1 −Tc
Th
Step 5: Comparing this with the efficiency formula, we see that this expres-
sion represents the maximum possible efficiency of any heat engine operating
between two temperature reservoirs.
Step 6: Therefore, the maximum efficiency of the engine is given by ηmax =
1−Tc
Th.
Question 35
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. If the engine absorbs 4 kJ of heat from the hot reservoir in each cycle,
determine the efficiency of the engine.
27
Step 3: Determine the temperatures of the hot and cold reservoirs for
50Given that the efficiency, η= 0.5 (50
0.5 = Th−Tc
Th
=⇒Th= 2Tc
Substitute this into the Carnot efficiency formula:
η=Th−Tc
Th
=2Tc−Tc
2Tc
=Tc
2Tc
= 0.5
Solving for Tc:
0.5 = Tc
2Tc
=⇒Tc= 2Tc=⇒Tc=1
2Th
Therefore, the temperatures of the hot and cold reservoirs are in the ratio 2:1.
Question 2
Question
A heat engine operates between a hot reservoir at a temperature of 700 K and
a cold reservoir at a temperature of 300 K. The engine absorbs 4000 J of heat
from the hot reservoir in each cycle and delivers 2000 J of work output in each
cycle. 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:
Efficiency = Work Output
Heat Input
Step 2: Substitute the given values into the formula:
Efficiency = 2000 J
4000 J = 0.5
Step 3: The efficiency of the engine is 0.5 or 50%.
Step 4: To determine whether the engine violates the second law of thermo-
dynamics, calculate the efficiency based on the Carnot cycle using the formula:
Maximum Efficiency = 1 −Temperature of Cold Reservoir
Temperature of Hot Reservoir
Step 5: Substitute the given temperatures into the formula:
Maximum Efficiency = 1 −300
700 = 1 −0.4286 = 0.5714
2
Step 6: The maximum efficiency based on the Carnot cycle is 0.5714 or
57.14%.
Step 7: Since the actual efficiency of the engine is 0.5 or 50%, which is less
than the maximum efficiency of 0.5714 or 57.14%, the engine does not violate
the second law of thermodynamics.
Question 3
Question
A Carnot heat engine operates between two reservoirs at temperatures Thand
Tc, where Th= 500 K and Tc= 300 K. The engine produces 4000 J of heat at Th
in each cycle. Calculate the efficiency of the engine and determine the maximum
amount of work that can be obtained from this engine using the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the Carnot efficiency formula:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the engine is 40
Step 3: Calculate the maximum work that can be obtained from the engine
using the second law of thermodynamics: The maximum efficiency of a heat
engine operating between two reservoirs at temperatures Thand Tcis given by:
Efficiencymax = 1 −Tc
Th
Step 4: Substitute the given temperatures into the formula:
Efficiencymax = 1 −300
500 = 1 −0.6=0.4
Step 5: The maximum work can be calculated using the formula for effi-
ciency:
Wmax = Efficiencymax ×Qh
Wmax = 0.4×4000 J = 1600 J
Therefore, the maximum amount of work that can be obtained from this
engine is 1600 J.
3
Question 4
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine absorbs 2400 J of heat from the high-temperature reservoir
and rejects 1600 J of heat to the low-temperature reservoir during each cycle.
Calculate the efficiency of the engine.
Solution
1. Recall the efficiency of a heat engine is given by:
η= 1 −QC
QH
where QHis the heat absorbed from the high-temperature reservoir and QCis
the heat rejected to the low-temperature reservoir.
2. We are given that QH= 2400 J and QC= 1600 J. Substitute these values
into the efficiency formula:
η= 1 −1600
2400
3. Simplify the expression:
η= 1 −2
3=1
3
4. Therefore, the efficiency of the heat engine is 1
3or 33.33
Question 5
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir at
300 K. The engine absorbs 4000 J of heat from the hot reservoir and performs
2200 J of work. Calculate the efficiency of the engine. Is this efficiency consistent
with the second law of thermodynamics?
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Heat input
Step 2: First, we need to calculate the heat input. The heat input is the
amount of heat absorbed from the hot reservoir, which is given as 4000 J.
4
Step 3: Next, we calculate the efficiency using the given values:
Efficiency = 2200 J
4000 J
Step 4: Simplify the expression to find the efficiency:
Efficiency = 11
20 = 0.55 or 55%
Step 5: The efficiency of the engine is 55
Step 6: The second law of thermodynamics states that no heat engine can be
more efficient than a Carnot engine operating between the same two reservoirs.
The maximum efficiency of a Carnot engine is given by the Carnot efficiency
formula:
EfficiencyCarnot = 1 −Tc
Th
Step 7: Plugging in the values of the cold and hot reservoir temperatures:
EfficiencyCarnot = 1 −300
800 = 1 −3
8=5
8= 0.625 or 62.5%
Step 8: Since the efficiency of the engine (55
Question 6
Question
A heat engine operates between two reservoirs at temperatures THand TC. If
the engine absorbs 1500 J of heat from the hot reservoir and exhausts 800 J to
the cold reservoir during each cycle, what is the efficiency of the engine in terms
of THand TC?
Solution
Let’s denote the efficiency of the heat engine as η.
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η= 1 −QC
QH
,
where QCis the heat exhausted into the cold reservoir and QHis the heat
absorbed from the hot reservoir during each cycle.
Step 2: Given that the engine absorbs 1500 J from the hot reservoir and
exhausts 800 J to the cold reservoir during each cycle, we have:
QH= 1500 J
QC= 800 J
5
Step 3: Substitute the values of QHand QCinto the efficiency formula:
η= 1 −800
1500
Step 4: Simplify the expression to obtain the efficiency in terms of THand
TC:
η= 1 −8
15 =7
15
Step 5: Therefore, the efficiency of the engine in terms of THand TCis
η=7
15 .
Question 7
Question
A heat engine operates between a hot reservoir at Th= 600 K and a cold
reservoir at Tc= 300 K. The engine absorbs 2000 J of heat from the hot reservoir
in each cycle and exhausts 1200 J to the cold reservoir in each cycle. Determine
the efficiency of the engine and discuss whether it violates the second law of
thermodynamics.
Solution
Let’s denote the heat absorbed from the hot reservoir as Qh= 2000 J and the
heat exhausted to the cold reservoir as Qc= 1200 J.
Step 1: Calculate the efficiency of the engine using the formula η= 1 −Qc
Qh.
η= 1 −Qc
Qh
= 1 −1200
2000
= 1 −0.6
= 0.4
So, the efficiency of the engine is 40
Step 2: Discuss whether the engine violates the second law of thermody-
namics.
According to the second law of thermodynamics, no heat engine can be 100
Question 8
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the engine absorbs 600 J of heat from the hot reservoir and expels
6
300 J to the cold reservoir during each cycle, calculate the efficiency of the
engine and determine whether it violates the second law of thermodynamics.
Solution
Step 1: Recall the formula for the efficiency of a heat engine:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula to find the effi-
ciency:
Efficiency = 1 −Tc
Th
= 1 −300
600 = 1 −1
2=1
2= 50%
Step 3: Since the efficiency calculated above is less than 100
Question 9
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 maximum amount of
work that can be done in each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine is given by the formula:
Efficiency = 1 −Tcold
Thot
where Tcold and Thot are the temperatures of the cold and hot reservoirs in
Kelvin, respectively.
Given Thot = 600 K and Tcold = 300 K, we can calculate the efficiency:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Calculate the maximum amount of work that can be done in each
cycle. The maximum amount of work that can be done in each cycle by a Carnot
engine is given by the formula:
Maximum Work = Efficiency ×Heat Input
Given that the heat input is 4000 J, the maximum work done in each cycle
is:
Maximum Work = 0.5×4000 = 2000 J
Therefore, the efficiency of the Carnot engine is 50
7
Question 10
Question
A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir
at 300 K. The engine produces 360 J of work per cycle. Calculate the efficiency
of the engine and determine the amount of heat rejected to the cold reservoir
during 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 cold reservoir and This the temperature of
the hot reservoir.
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4
Step 3: Determine the amount of heat supplied to the engine using the
formula:
Heat input = Work output + Heat rejected
Step 4: Substitute the given work output into the formula:
Heat input = 360 J
Step 5: Calculate the heat rejected to the cold reservoir using the formula:
Heat rejected = Heat input −Work output
Step 6: Substitute the known values into the formula:
Heat rejected = 360 J −360 J = 0 J
Therefore, the efficiency of the engine is 0.4 (or 40
Question 11
Question
A Carnot engine operates between two reservoirs at temperatures T1and T2,
with T1> T2. The engine absorbs heat Q1from the high-temperature reservoir
and exhausts heat Q2to the low-temperature reservoir. If the efficiency of the
engine is η, prove that the efficiency of a Carnot engine is given by the expression
η= 1 −T2
T1
.
8
Solution
Step 1: Recall the efficiency of a Carnot engine is given by
η= 1 −Q2
Q1
.
Step 2: Using the first law of thermodynamics, we know that for a complete
cycle, the net work done by the engine is equal to the difference between the
heat absorbed and the heat released:
W=Q1−Q2.
Step 3: Since the engine is reversible and operates between two reservoirs,
we can express the heat absorbed and heat released in terms of the reservoir
temperatures:
Q1=T1S1, Q2=T2S2,
where S1and S2are the entropy changes associated with each reservoir.
Step 4: The second law of thermodynamics states that the entropy change
of the universe for any reversible process must be zero. Therefore, the sum of
the entropy changes of the reservoirs must be equal:
S1+S2= 0.
Step 5: From Step 3 and Step 4, we have
T1S1+T2S2= 0.
Step 6: Rearranging the above equation, we get
T1S1=−T2S2.
Step 7: Substituting Q1and Q2back into the expression for efficiency, we
have
η= 1 −T2S2
T1S1
= 1 −T2
T1
.
Step 8: Therefore, we have proven that the efficiency of a Carnot engine is
given by η= 1 −T2
T1.
Question 12
Question
A heat engine operates between a hot reservoir at 600◦C and a cold reservoir
at 27◦C. The engine produces 2400 J of work and rejects 4200 J of heat to the
cold reservoir. Determine the efficiency of the engine. Is this engine violating
the second law of thermodynamics?
9
Solution
Step 1: Find the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Calculate the heat input to the engine using the first law of thermo-
dynamics:
Heat input = Useful work output + Heat rejected to cold reservoir
Step 3: Use the second law of thermodynamics to determine if the engine is
violating the law. An engine operating between two reservoirs cannot have an
efficiency greater than the Carnot efficiency:
Efficiency ≤1−Tcold
Thot
where Tcold is the absolute temperature of the cold reservoir and Thot is the
absolute temperature of the hot reservoir.
Question 13
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 in each cycle.
Calculate the maximum theoretical efficiency of this engine.
Solution
To find the maximum theoretical efficiency of the heat engine, we can use the
Carnot efficiency formula:
Efficiency = 1 −Tcold
Thot
where Thot is the temperature of the hot reservoir in Kelvin and Tcold is the
temperature of the cold reservoir in Kelvin.
Step 1: Convert temperatures to Kelvin Given that the hot reservoir
temperature Thot = 500 K and the cold reservoir temperature Tcold = 300 K.
Step 2: Calculate the efficiency Plugging the values into the formula:
Efficiency = 1 −300
500
Efficiency = 1 −0.6 = 0.4
Therefore, the maximum theoretical efficiency of the engine is 40
10
Question 14
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. The engine absorbs 5000 J of heat from the hot reservoir in each cycle.
Calculate the maximum possible efficiency of the engine.
Solution
Step 1: Recall the formula for the efficiency (η) of a heat engine:
η= 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Step 2: Substitute the given values into the formula:
TC= 300 K
TH= 600 K
Step 3: Calculate the efficiency using the formula:
η= 1 −300
600
η= 1 −1
2
η=1
2= 50%
Step 4: Therefore, the maximum possible efficiency of the engine is 50%.
Question 15
Question
A heat engine operates between two reservoirs at temperatures Thot and Tcold.
The engine has an efficiency of 40If the engine absorbs 600 J of heat from the
hot reservoir in each cycle, what is the amount of heat expelled to the cold
reservoir in each cycle?
11
Solution
Let Qin be the heat absorbed from the hot reservoir, Qout be the heat expelled
to the cold reservoir, and ηbe the efficiency of the engine.
Step 1: Recall the formula for the efficiency of a heat engine:
η=Useful work output
Heat input =W
Qin
η=W
Qin
=⇒W=η·Qin
Since the efficiency is given as 40
Step 2: The work done by the engine in each cycle can be calculated using
the efficiency:
W=η·Qin = 0.4×600 J = 240 J
Step 3: The heat expelled to the cold reservoir in each cycle is the difference
between the heat absorbed and the work done:
Qout =Qin −W= 600 J −240 J = 360 J
Step 4: Therefore, the amount of heat expelled to the cold reservoir in each
cycle is 360 J.
Question 16
Question
A Carnot heat engine operates between a hot reservoir at 527◦C and a cold
reservoir at 27◦C. The engine absorbs 6000 J of heat from the hot reservoir in
each cycle. Calculate:
1. The efficiency of the engine.
2. The work done by the engine in each cycle.
3. The amount of heat exhausted to the cold reservoir in each cycle.
Solution
1. To calculate the efficiency of the Carnot engine, we use the formula
Efficiency = 1 −Tcold
Thot
,
where Thot and Tcold are the temperatures of the hot and cold reservoirs in
Kelvin.
Step 1: Convert the temperatures to Kelvin:
Thot = 527 + 273 = 800 K,
12
Tcold = 27 + 273 = 300 K.
Step 2: Calculate the efficiency:
Efficiency = 1 −300
800 = 1 −3
8=5
8= 62.5%.
Therefore, the efficiency of the Carnot engine is 62.5
2. The work done by the engine in each cycle can be calculated using the
formula
Work = Efficiency ×Heat absorbed from hot reservoir.
Step 3: Calculate the work done:
Work = 0.625 ×6000 = 3750 J.
Therefore, the work done by the engine in each cycle is 3750 J.
3. The amount of heat exhausted to the cold reservoir in each cycle is equal
to the difference between the heat absorbed from the hot reservoir and the work
done:
Heat exhausted = Heat absorbed −Work.
Step 4: Calculate the heat exhausted:
Heat exhausted = 6000 −3750 = 2250 J.
Therefore, the amount of heat exhausted to the cold reservoir in each cycle
is 2250 J.
Question 17
Question
A Carnot heat engine operates between a high temperature reservoir at 800 K
and a low temperature reservoir at 300 K. The engine takes in 6,000 J of heat
from the high temperature reservoir in each cycle. Calculate the maximum
efficiency of the engine and the work output per cycle.
Solution
Step 1: Calculate the maximum efficiency of the Carnot engine using the formula
for efficiency:
Efficiency = 1 −TC
TH
where THis the absolute temperature of the high temperature reservoir, and
TCis the absolute temperature of the low temperature reservoir.
Step 2: Convert the temperatures to Kelvin: TH= 800 K and TC= 300 K.
13
Step 3: Substitute the values into the formula for efficiency:
Efficiency = 1 −300
800
Step 4: Calculate the efficiency:
Efficiency = 1 −3
8=5
8= 0.625
Step 5: The maximum efficiency of the Carnot engine is 0.625 or 62.5
Step 6: Calculate the work output per cycle using the formula:
Work output = Efficiency ×Heat input
Step 7: Given that the heat input is 6,000 J, and the efficiency is 0.625,
substitute these values into the formula:
Work output = 0.625 ×6,000
Step 8: Calculate the work output:
Work output = 0.625 ×6,000 = 3,750 J
Step 9: The maximum efficiency of the engine is 62.5
Question 18
Question
A heat engine operates between a hot reservoir at 800 K and a cold reservoir
at 300 K. The engine absorbs 2000 J of heat from the hot reservoir during each
cycle and delivers 1200 J of work. Determine: (a) The efficiency of the engine.
(b) The amount of heat rejected to the cold reservoir during each cycle.
Solution
Step 1: We can start by calculating the efficiency of the engine using the formula:
Efficiency = Useful work output
Heat input
Given that the engine absorbs 2000 J of heat and delivers 1200 J of work, we
have:
Efficiency = 1200 J
2000 J
Step 2: Now we can calculate the efficiency:
Efficiency = 0.6 = 60%
14
Step 3: Next, we can determine the amount of heat rejected to the cold
reservoir during each cycle by using the fact that the heat input is equal to the
sum of the work done and the heat rejected:
Heat input = Useful work output + Heat rejected
Step 4: Since the engine absorbs 2000 J of heat and delivers 1200 J of work,
we can rearrange the equation to solve for the heat rejected:
Heat rejected = Heat input −Useful work output = 2000 J −1200 J
Step 5: Calculating the heat rejected:
Heat rejected = 800 J
Therefore, during each cycle, 800 J of heat is rejected to the cold reservoir.
Question 19
Question
A Carnot engine operates between two reservoirs at temperatures Thot and Tcold.
The engine takes in Qhot of heat from the hot reservoir and expels Qcold to the
cold reservoir. If the efficiency of this engine is η, prove that the efficiency can
also be expressed as η= 1 −Tcold
Thot .
Solution
Step 1: From the definition of efficiency, we have
η= 1 −Qcold
Qhot
.
Step 2: By the first law of thermodynamics, the net work done by the engine
is Wnet =Qhot −Qcold.
Step 3: The efficiency of a Carnot engine is given by η=Wnet
Qhot . Substituting
the expression for Wnet into this formula, we have
η=Qhot −Qcold
Qhot
= 1 −Qcold
Qhot
.
Step 4: Now, we know that for a Carnot engine, the efficiency is given by
η= 1 −Tcold
Thot . We have to show that this is equivalent to the expression we
derived earlier.
Step 5: From the Carnot efficiency formula, we have η= 1 −Tcold
Thot . Multi-
plying both sides by Qhot gives Qhotη=Qhot −Qcold.
Step 6: Dividing both sides of our previous expression η= 1 −Qcold
Qhot by Qhot
yields Qhotη=Qhot −Qcold.
Step 7: Thus, we have shown that η= 1−Tcold
Thot is equivalent to the expression
for efficiency we derived earlier.
15
Question 20
Question
A Carnot heat engine operates between a hot reservoir at 800 K and a cold
reservoir at 400 K. If the engine absorbs 5000 J of heat from the hot reservoir
in each cycle, calculate the efficiency of the engine. Also, discuss how this effi-
ciency compares to the maximum possible efficiency for a heat engine operating
between the same two reservoirs.
Solution
Step 1: Calculate the efficiency of the Carnot heat engine. The efficiency of a
Carnot heat engine is given by the formula:
Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given that the hot reservoir temperature TH= 800 K and the cold reservoir
temperature TC= 400 K, we can substitute these values into the formula to find
the efficiency:
Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Thus, the efficiency of the Carnot heat engine is 50
Step 2: Compare the efficiency of the engine to the maximum possible effi-
ciency. The maximum possible efficiency of any heat engine operating between
the same two reservoirs is given by the formula:
Max Efficiency = 1 −TC
TH
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
In this case, the maximum efficiency of a heat engine operating between the
400 K and 800 K reservoirs would also be:
Max Efficiency = 1 −400
800 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot heat engine is the same as the max-
imum possible efficiency for a heat engine operating between the same two
reservoirs.
Question 21
Question
A heat engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc, where Th> Tc. The engine absorbs Qhjoules of
16
heat from the hot reservoir and exhausts Qcjoules of heat to the cold reservoir.
If the engine does 250 kJ of work per cycle, find the efficiency of the engine in
terms of Qhand Th. Also, explain how this result is related to the second law
of thermodynamics.
Solution
Step 1: We know that the efficiency of a heat engine is given by the formula:
Efficiency = Useful work output
Input heat energy
Step 2: The useful work output is the work done by the engine per cycle,
which is 250 kJ.
Step 3: The input heat energy is the heat energy absorbed from the hot
reservoir, Qh.
Step 4: Therefore, the efficiency of the engine can be expressed as:
Efficiency = 250 kJ
Qh
Step 5: We also know that for a heat engine operating between two reservoirs,
the efficiency is given by:
Efficiency = 1 −Tc
Th
Step 6: Equating the two expressions for efficiency gives:
250 kJ
Qh
= 1 −Tc
Th
Step 7: Solving for the efficiency in terms of Qhand Th:
Efficiency = 250 kJ
Qh
= 1 −Tc
Th
Step 8: This result demonstrates the relationship between the efficiency of
the engine, the input heat energy, and the temperatures of the reservoirs. It is
related to the second law of thermodynamics, which states that no heat engine
can be 100
Question 22
Question
A heat engine operates between two reservoirs at temperatures 600 K and 300
K. The engine takes in 2000 J of heat from the high-temperature reservoir in
each cycle and delivers 800 J of work. Calculate the efficiency of the engine. Is
this efficiency consistent with the second law of thermodynamics?
17
Solution
Step 1: Calculate the efficiency of the heat engine using the formula:
Efficiency = Useful work output
Heat input
Step 2: Substitute the given values into the formula:
Efficiency = 800 J
2000 J
Step 3: Simplify the expression to find the efficiency:
Efficiency = 4
10 = 0.4 = 40%
Step 4: Discuss whether this efficiency is consistent with the second law
of thermodynamics. The efficiency of a heat engine is given by the Carnot
efficiency:
EfficiencyCarnot = 1 −Tc
Th
Step 5: Substitute the given temperatures into the Carnot efficiency formula:
EfficiencyCarnot = 1 −300
600 = 1 −1
2=1
2= 0.5 = 50%
Step 6: Compare the efficiency of the engine (40
Question 23
Question
A heat engine operates between two reservoirs at temperatures THand TC.
The engine absorbs QHof heat from the hot reservoir and exhausts QCof heat
to the cold reservoir. If the efficiency of this engine is ηand the second law
of thermodynamics states that QC≥TC
THQH, prove that the efficiency of the
engine is given by η≤1−TC
TH.
Solution
Step 1: We know that the efficiency of a heat engine is given by
η= 1 −QC
QH
Step 2: Given QC≥TC
THQH, we can rewrite the efficiency expression as
η≤1−TC
TH
QH
QH
= 1 −TC
TH
18
Step 3: Therefore, we have proved that the efficiency of the engine is given
by η≤1−TC
TH.
Question 24
Question
A Carnot heat engine operates between two reservoirs at temperatures Th= 600
K and Tc= 300 K. The engine absorbs 800 J of heat from the hot reservoir
during each cycle. Calculate the efficiency of the engine and determine the
amount of heat rejected to the cold reservoir during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula for
efficiency:
Efficiency = 1 −Tc
Th
Step 2: Substitute the given temperatures into the formula:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 0.5
Thus, the efficiency of the Carnot engine is 50
Step 3: To determine the amount of heat rejected to the cold reservoir during
each cycle, we use the fact that for a Carnot engine:
Efficiency = work done
heat absorbed
Step 4: Rearranging the formula to solve for work done:
work done = Efficiency ×heat absorbed
Step 5: Substitute the values for efficiency and heat absorbed:
work done = 0.5×800 J = 400 J
Step 6: Since the total energy input is 800 J, and 400 J is converted to work,
the remaining 400 J must be rejected to the cold reservoir.
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 400 J.
Question 25
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and exhausts heat
Qcto the cold reservoir, while doing work Wduring a complete cycle.
19
Prove that the maximum efficiency of the heat engine is given by ηmax =
1−Tc
Th.
Solution
Step 1: We need to determine the efficiency of the heat engine. The efficiency
of a heat engine is given by the ratio of the work done by the engine to the heat
absorbed from the hot reservoir:
Efficiency η=W
Qh
Step 2: Express Qhin terms of Qcand W. Since the engine is operating
in a cycle, the net heat absorbed by the engine must be equal to the net work
done by the engine plus the heat exhausted to the cold reservoir:
Qh=W+Qc
Step 3: Substitute the relationship between Qhand Qcback into the effi-
ciency expression.
η=W
W+Qc
Step 4: Express the heat exhausted to the cold reservoir in terms of the
temperatures. From the second law of thermodynamics, we know that the heat
exhausted to the cold reservoir can be expressed as:
Qc=Tc∆S
where ∆Sis the change in entropy.
Step 5: Express ∆Sin terms of Qhand Tc. Using the relationship between
the change in entropy and heat transfer:
∆S=Qh
Th
−Qc
Tc
Step 6: Substitute the expression for Qcin terms of entropy into the effi-
ciency formula.
η=W
W+TcQh
Th−Qc
Tc
Step 7: Simplify the expression and express the maximum efficiency. After
simplifying and rearranging terms, the efficiency can be expressed as:
η= 1 −Tc
Th
Therefore, the maximum efficiency of the heat engine is given by ηmax =
1−Tc
Th.
20
Question 26
Question
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at
300 K. If the engine has an efficiency of 40
Solution
Let’s denote the efficiency of the given heat engine as ηgiven = 0.40 and the
temperatures of the hot and cold reservoirs as Thot = 600 K and Tcold = 300 K,
respectively. We want to calculate the maximum theoretical efficiency of a heat
engine operating between these two reservoirs, which is governed by Carnot’s
theorem.
Step 1: Calculate the Carnot efficiency using the formula:
ηCarnot = 1 −Tcold
Thot
.
Step 2: Substitute the given temperatures into the formula:
ηCarnot = 1 −300
600 = 1 −1
2=1
2= 0.50.
Step 3: Compare the given efficiency with the Carnot efficiency: Since the
given efficiency ηgiven = 0.40 is less than the Carnot efficiency ηCarnot = 0.50, we
conclude that the given efficiency is less than the maximum theoretical efficiency
of a heat engine operating between the two given reservoirs according to the
second law of thermodynamics.
Question 27
Question
A Carnot heat engine operates between two reservoirs at temperatures of 500
K and 300 K. If the engine absorbs 400 J of heat from the high-temperature
reservoir in each cycle, what is the efficiency of the engine? Additionally, discuss
how the second law of thermodynamics applies to this scenario.
Solution
Step 1: Calculate the work done by the Carnot heat engine. Given that the
engine absorbs 400 J of heat from the high-temperature reservoir in each cycle,
the work done by the engine is given by the difference between the heat absorbed
from the hot reservoir (Qh) and the heat rejected to the cold reservoir (Qc) in
each cycle. This can be calculated using the formula for the efficiency of a
Carnot engine:
Efficiency = 1 −Tc
Th
21
Step 2: Calculate the efficiency of the Carnot heat engine. First, we need
to calculate the temperatures in Kelvin by adding 273 to the Celsius values
given: - High-temperature reservoir: Th= 500 K - Low-temperature reservoir:
Tc= 300 K
Now we can plug these values into the efficiency formula:
Efficiency = 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the Carnot heat engine operating between these
two reservoirs is 40
Step 3: Discuss the second law of thermodynamics. The second law of
thermodynamics states that in any cyclic process, the efficiency of a heat engine
is less than 100
Question 28
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
1. The work done by the engine in each cycle.
2. The amount of heat rejected to the cold reservoir in each cycle.
3. Comment on whether this engine violates the second law of thermody-
namics.
Solution
1. To find the work done by the engine in each cycle, we will use the formula
for efficiency (η) of a heat engine:
η= 1 −Qout
Qin
where ηis the efficiency, Qout is the heat rejected to the cold reservoir, and Qin
is the heat absorbed from the hot reservoir. Given that the efficiency is 40
0.40 = 1 −Qout
2400
Qout = 0.60 ×2400 = 1440 J
The work done by the engine is the difference between the heat absorbed
and the heat rejected:
W=Qin −Qout = 2400 −1440 = 960 J
22
2. The amount of heat rejected to the cold reservoir in each cycle is Qout,
which we found to be 1440 J.
3. The efficiency of the engine is less than 1, which is consistent with the
second law of thermodynamics. The second law states that no heat engine can
be 100
Question 29
Question
A Carnot heat engine operates between reservoirs at temperatures THand TC
with TH> TC. If the efficiency of the engine is η, show that the maximum
possible efficiency of a heat engine operating between the same two reservoirs
is 1 −TC
TH.
Solution
Step 1: Recall the efficiency of a Carnot heat engine: The efficiency of a Carnot
heat engine is given by the formula
η= 1 −TC
TH
Step 2: Consider an arbitrary heat engine operating between the same reser-
voirs. Let’s denote the efficiency of this arbitrary heat engine as η′.
Step 3: Applying the second law of thermodynamics. According to the
second law of thermodynamics, the efficiency of a real heat engine must always
be less than or equal to the efficiency of a Carnot heat engine operating between
the same two reservoirs. Therefore, we have
η′≤η
Step 4: Substitute the expressions for ηand η′. Substitute the expressions
for ηand η′:
η′≤1−TC
TH
Step 5: Rearrange the inequality. Rearrange the inequality to isolate η′:
η′≤1−TC
TH
η′≤1−TC
TH
Step 6: Conclusion. Therefore, the maximum possible efficiency of a heat
engine operating between the same two reservoirs is 1 −TC
TH.
23
Question 30
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tcwith Th> Tc. If the engine absorbs 5000 J of heat from the hot reservoir
and exhausts 3000 J of heat to the cold reservoir in one cycle, calculate: (a) The
efficiency of the engine. (b) The work done by the engine. (c) The maximum
possible efficiency of an engine operating between the same two reservoirs.
Solution
Step 1: Find the efficiency of the engine. The efficiency of a Carnot heat engine
is given by the formula:
Efficiency = 1−Tc
Th×100%
Given: Qh= 5000 J (heat absorbed from hot reservoir) Qc= 3000 J (heat
exhausted to cold reservoir)
Efficiency = 1−Qc
Qh×100%
Efficiency = 1−3000
5000×100%
Efficiency = 0.4×100%
Efficiency = 40%
Therefore, the efficiency of the engine is 40
Step 2: Find the work done by the engine. The work done by the engine in
one cycle is given by the formula:
Work done = Qh−Qc
Given: Qh= 5000 J Qc= 3000 J
Work done = 5000 J −3000 J
Work done = 2000 J
Therefore, the work done by the engine is 2000 J.
Step 3: Find the maximum possible efficiency of an engine operating between
the same two reservoirs. The maximum possible efficiency of any heat engine
operating between two reservoirs at temperatures Thand Tcis given by the
formula:
Maximum efficiency = 1−Tc
Th×100%
24
Given: Th> Tc
Therefore, for the given two reservoirs, the maximum possible efficiency is
Maximum efficiency = 1−Tc
Th×100%
Therefore, the maximum possible efficiency is 1−Tc
Th×100%.
Question 31
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. If the heat engine absorbs 5000 J of heat from the reservoir at Thand
rejects 3000 J of heat to the reservoir at Tc, what is the efficiency of the heat
engine? Is this operation consistent with the second law of thermodynamics?
Justify your answer.
Solution
Step 1: Calculate the efficiency of the heat engine. The efficiency of a heat
engine is given by the formula:
Efficiency = 1 −Qc
Qh
where Qcis the heat rejected to the cold reservoir and Qhis the heat absorbed
from the hot reservoir.
Given that Qh= 5000 J and Qc= 3000 J, we can substitute these values
into the efficiency formula:
Efficiency = 1 −3000
5000 = 1 −0.6=0.4
Therefore, the efficiency of the heat engine is 40
Step 2: Determine if the operation is consistent with the second law of
thermodynamics. According to the second law of thermodynamics, the efficiency
of a heat engine is limited by the Carnot efficiency:
Carnot efficiency = 1 −Tc
Th
Given that Th> Tc, the Carnot efficiency will always be less than 1, which
means no heat engine can be 100
Carnot efficiency = 1 −Tc
Th
<1
Since the efficiency of the heat engine we calculated (40
25
Question 32
Question
A heat engine operates between a hot reservoir at 500 K and a cold reservoir at
300 K. The engine takes in 600 J of heat from the hot reservoir in each cycle and
expels 360 J of heat to the cold reservoir in each cycle. Calculate the efficiency of
the engine and discuss whether this violates the second law of thermodynamics.
Solution
Step 1: We can start by calculating the efficiency of the engine using the formula:
Efficiency = 1 −Qcold
Qhot
Step 2: Given that Qhot = 600 J and Qcold = 360 J, we can plug these values
into the formula to find:
Efficiency = 1 −360
600
Step 3: Calculating this expression gives us:
Efficiency = 1 −360
600 = 1 −0.6=0.4
Step 4: Therefore, the efficiency of the engine is 0.4 or 40
Step 5: According to the second law of thermodynamics, no heat engine can
have an efficiency of 100
Question 33
Question
A heat engine operates between a hot reservoir at 650◦Cand a cold reservoir
at 20◦C. The engine has an efficiency of 30%. Calculate the maximum possible
efficiency of this heat engine according to the second law of thermodynamics.
Solution
Step 1: Convert the temperatures to Kelvin by adding 273. Hot reservoir tem-
perature (TH) = 650 + 273 = 923 K
Cold reservoir temperature (TC) = 20 + 273 = 293 K
Step 2: Calculate the maximum possible efficiency using the Carnot effi-
ciency formula:
Efficiencymax = 1 −TC
TH
26
Step 3: Substitute the given temperatures into the formula to find the max-
imum efficiency.
Efficiencymax = 1 −293
923
Efficiencymax = 1 −0.3173
Efficiencymax ≈0.6827 or 68.27%
Therefore, the maximum possible efficiency of the heat engine according to
the second law of thermodynamics is 68.27%.
Question 34
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the hot reservoir and delivers W
of work. If the efficiency of the engine is η, prove that the maximum efficiency
of the engine is given by ηmax = 1 −Tc
Th.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is defined as the ratio of the work output to the heat input:
η=W
Qh
Step 2: Next, consider the second law of thermodynamics: The second law
of thermodynamics states that no cyclic engine can be 100
Step 3: To find the maximum efficiency of the engine, we can analyze the
Carnot cycle which gives the maximum efficiency for a heat engine.
Step 4: In a Carnot cycle, the efficiency ηmax is given by:
ηmax = 1 −Tc
Th
Step 5: Comparing this with the efficiency formula, we see that this expres-
sion represents the maximum possible efficiency of any heat engine operating
between two temperature reservoirs.
Step 6: Therefore, the maximum efficiency of the engine is given by ηmax =
1−Tc
Th.
Question 35
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. If the engine absorbs 4 kJ of heat from the hot reservoir in each cycle,
determine the efficiency of the engine.
27
Solution
Step 1: Calculate the heat absorbed by the engine from the hot reservoir in each
cycle. Given that the engine absorbs 4 kJ of heat, we have Qh= 4 kJ.
Step 2: Calculate the efficiency of the Carnot engine. The efficiency of a
Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
where Tcis the temperature of the cold reservoir and This the temperature of
the hot reservoir.
Given Tc= 300 K and Th= 600 K, we can substitute these values into the
formula to find the efficiency:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Therefore, the efficiency of the Carnot engine is 50%.
28
