PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Heat engines,
efficiency, and the second law of
thermodynamics
Question Bank - Set 3
Liberty University
Question 1
Question
A heat engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs heat Qhfrom the high-temperature
reservoir and rejects heat Qcto the low-temperature reservoir. If the efficiency
of the engine is η, show that the maximum efficiency of the engine is given by
the Carnot efficiency, ηCarnot = 1 −Tc
Th.
Solution
Step 1: Recall the definition of efficiency for a heat engine,
η=Useful work output
Energy input =W
Qh
.
Step 2: The maximum work output of the engine is achieved by a reversible
process (Carnot cycle), for which the efficiency is given by
ηCarnot = 1 −Tc
Th
,
where Thand Tcare the absolute temperatures of the hot and cold reservoirs,
respectively.
Step 3: For any heat engine operating between the same two reservoirs, the
efficiency will always be less than or equal to the Carnot efficiency. This is a
consequence of the second law of thermodynamics, which states that no heat
engine can be more efficient than a reversible engine operating between the same
two temperature reservoirs.
Therefore, the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by the Carnot efficiency formula.
Question 2
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and rejects heat
Qcto the cold reservoir. Calculate the efficiency of the engine in terms of Qh
and Qc.
Solution
Let us denote the efficiency of the engine as η.
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η= 1 −Qc
Qh
Step 2: Since the engine absorbs heat Qhfrom the hot reservoir and rejects
heat Qcto the cold reservoir, we have:
Qh=W+Qc
where Wis the work done by the engine.
Step 3: We can rewrite the efficiency formula in terms of W:
η= 1 −Qc
W+Qc
Step 4: From the first law of thermodynamics, the work done by the engine
is given by:
W=Qh−Qc
Step 5: Substituting the expression for Winto the efficiency formula, we
get:
η= 1 −Qc
Qh−Qc+Qc
η= 1 −Qc
Qh
Step 6: Therefore, the efficiency of the engine in terms of Qhand Qcis:
η= 1 −Qc
Qh
2
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwith
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and releases heat
Qcto the cold reservoir. If Wis the work done by the engine, show that the
efficiency of the engine is given by η= 1 −Tc
Th.
Solution
Step 1: We first need to recall the definition of the efficiency of a heat engine,
which is given by
η=Work output
Heat input =W
Qh
.
Step 2: We know that according to the first law of thermodynamics, the net
work done by the engine is equal to the difference in the heat absorbed and the
heat released, i.e., W=Qh−Qc.
Step 3: Substituting W=Qh−Qcinto the efficiency formula, we get
η=Qh−Qc
Qh
.
Step 4: Rearranging the terms, we have
η= 1 −Qc
Qh
.
Step 5: Now, from the second law of thermodynamics, the efficiency of a
heat engine is limited by the Carnot efficiency, given by ηCarnot = 1 −Tc
Th.
Step 6: Comparing η= 1 −Qc
Qhwith ηCarnot = 1 −Tc
Th, we see that Qc
Qh=Tc
Th.
Step 7: Substituting Qc
Qh=Tc
Thback into the expression for efficiency, we find
η= 1 −Tc
Th
.
Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
with Th> Tc. The engine absorbs heat energy Qhfrom the reservoir at temper-
ature Thand expels heat energy Qcto the reservoir at temperature Tc. Prove
that the efficiency of the Carnot engine, η, is given by η= 1 −Tc
Th.
3
Solution
Step 1: Recall the efficiency of a heat engine is defined as the ratio of the work
output to the heat input. For the Carnot engine, the efficiency is given by
η=W
Qh
Step 2: With reference to the second law of thermodynamics, during the
Carnot cycle, the ratio of the heat absorbed to the high-temperature reservoir
to the work done by the engine is equal to the ratio of the heat expelled to the
low-temperature reservoir over the work done by the engine:
Qh
Th
=Qc
Tc
Step 3: Rearranging the above equation, we get
Qc
Qh
=Tc
Th
Step 4: Now, we know that the work done by the engine is equal to the
difference between the heat absorbed and the heat expelled:
W=Qh−Qc
Step 5: Substituting the values of Qcand Qhfrom step 3 into the equation
above, we get
W=Qh−Tc
ThQh
Step 6: Simplifying the above expression, we find
W=Qh1−Tc
Th
Step 7: Finally, substituting the value of Winto the efficiency equation from
step 1, we obtain
η=W
Qh
=
Qh1−Tc
Th
Qh
= 1 −Tc
Th
Therefore, the efficiency of a Carnot engine is given by η= 1 −Tc
Th.
Question 5
Question
A Carnot heat engine operates between two reservoirs at temperatures of 500
K and 300 K. If the engine absorbs 4000 J of heat from the high-temperature
reservoir in each cycle, calculate the following: (a) The efficiency of the Carnot
engine. (b) The heat rejected to the low-temperature reservoir in each cycle.
(c) The net work output in each cycle.
4
Solution
(a) Let’s denote the high-temperature reservoir as TH= 500 K and the low-
temperature reservoir as TC= 300 K. The efficiency of a Carnot engine is given
by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4 = 40%
(b) The heat rejected to the low-temperature reservoir in each cycle can be
determined using the formula:
Heat rejected = QC= Efficiency ×QH
Step 2: Calculate the heat rejected to the low-temperature reservoir.
QC= 0.4×4000 J = 1600 J
(c) The net work output in each cycle is given by the difference between the
heat absorbed from the high-temperature reservoir and the heat rejected to the
low-temperature reservoir:
Net work output = Wout =QH−QC
Step 3: Calculate the net work output in each cycle.
Wout = 4000 J −1600 J = 2400 J
Therefore, the answers to the questions are: (a) The efficiency of the Carnot
engine is 40(b) The heat rejected to the low-temperature reservoir in each cycle
is 1600 J. (c) The net work output in each cycle is 2400 J.
Question 6
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and a
cold reservoir at a temperature of 300 K. If the engine produces 1200 J of work
per cycle, calculate the efficiency of the engine. Is this efficiency physically
possible according to the second law of thermodynamics?
5
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for efficiency
(η):
η= 1 −Tc
Th
where η= efficiency of the engine, Tc= temperature of the cold reservoir, Th
= temperature of the hot reservoir.
Given that Th= 500 K and Tc= 300 K, we can plug these values into the
formula to find the efficiency.
Step 2: Calculate the efficiency:
η= 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the engine is 0.4 (or 40
Step 3: Determine whether this efficiency is physically possible according to
the second law of thermodynamics. According to the second law of thermody-
namics, no heat engine can have an efficiency greater than the Carnot efficiency,
which is given by
ηCarnot = 1 −Tc
Th
where ηCarnot is the maximum possible efficiency of any heat engine operating
between the two temperatures.
Step 4: Calculate the Carnot efficiency:
ηCarnot = 1 −300
500 = 1 −0.6=0.4
Step 5: Compare the efficiency of the engine with the Carnot efficiency. The
efficiency of the engine is equal to the Carnot efficiency, which means the engine
is operating at the maximum possible efficiency allowed by the second law of
thermodynamics. Thus, the efficiency of 40
Question 7
Question
A heat engine operates between two reservoirs with temperatures THand TC,
where TH> TC. The engine absorbs QHjoules of heat from the hot reservoir
and expels QCjoules of heat to the cold reservoir during each cycle. If the
efficiency of the engine is η, prove that the efficiency of the engine cannot be
100%.
6
Solution
Let’s assume that the efficiency of the engine is 100% or η= 1. This means
that all the heat absorbed from the hot reservoir is converted into work.
Step 1: Calculate the heat expelled to the cold reservoir. Using the effi-
ciency formula for a heat engine:
η=Useful work done
Heat absorbed = 1
This implies that the useful work done is equal to the heat absorbed from the
hot reservoir:
Useful work done = QH
Since η= 1, the work done is equal to the heat absorbed from the hot reservoir.
Step 2: Apply the conservation of energy. According to the first law of
thermodynamics, we have:
QH= Useful work done + QC
Substitute the values:
QH=QH+QC
QC= 0
This implies that the engine expels no heat to the cold reservoir, which con-
tradicts the given information. Therefore, our assumption that the efficiency is
100% must be incorrect, and the efficiency of the engine cannot be 100%.
Question 8
Question
A Carnot engine operates between two heat reservoirs at temperatures T1and
T2(T1> T2). The engine absorbs heat Q1from the reservoir at temperature
T1,Q1= 4000 J, and exhausts heat Q2to the reservoir at temperature T2. Find
the efficiency of the engine and the amount of heat exhausted to the reservoir
at temperature T2,Q2, in terms of Q1.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine is given by the formula:
Efficiency = 1 −T2
T1
Given T1> T2, we have T1=T2+ ∆T, where ∆T=T1−T2. So, the efficiency
of the Carnot engine is:
Efficiency = 1 −T2
T2+ ∆T= 1 −T2
T1
7
Step 2: Substitute values and calculate. Given Q1= 4000 J, the efficiency
is:
Efficiency = 1 −T2
T1
= 1 −T2
T2+ ∆T
Efficiency = 1 −T2
T2+ (T1−T2)= 1 −T2
T1
Efficiency = 1 −T2
T1
= 1 −T2
T1
= 1 −T2
T1
Step 3: Calculate the heat exhausted to the reservoir at temperature T2,Q2.
Using the first law of thermodynamics, we know that the net work done by the
engine is:
Wnet =Q1−Q2
For a Carnot engine, the net work done is:
Wnet =Q11−T2
T1
And the heat exhausted to the reservoir at temperature T2is:
Q2=Q1−Wnet
Step 4: Substitute values and calculate Q2. Given Q1= 4000 J, we have:
Q2=Q1−Wnet =Q11−T2
T1
Q2= 4000 1−T2
T1
Therefore, the efficiency of the engine is 1 −T2
T1and the heat exhausted to
the reservoir at temperature T2,Q2, is 4000 1−T2
T1.
Question 9
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, with Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
rejects Qcof heat to the cold reservoir. Find an expression for the efficiency of
the Carnot engine in terms of Th,Tc,Qh, and Qc.
8
Solution
Step 1: Recall that the efficiency of any heat engine is defined as the ratio of
the work output to the heat input. For a Carnot engine, the efficiency can be
expressed as
η=W
Qh
.
Step 2: The work output of a Carnot engine is given by the difference in the
heat absorbed from the hot reservoir and the heat rejected to the cold reservoir,
i.e., W=Qh−Qc.
Step 3: Substitute the expression for work output into the efficiency formula
to yield
η=Qh−Qc
Qh
.
Step 4: Recognize that the heat rejected to the cold reservoir can be written
in terms of the heat absorbed from the hot reservoir and the efficiency of the
engine. We have Qc=Qh−W=Qh−ηQh.
Step 5: Substitute the expression for Qcinto the efficiency formula, resulting
in
η=Qh−(Qh−ηQh)
Qh
=Qh−Qh+ηQh
Qh
=ηQh
Qh
.
Step 6: Simplify the expression to find the final formula for the efficiency of
a Carnot engine:
η=η.
Therefore, the efficiency of a Carnot engine is solely dependent on the effi-
ciency itself and is given by η= 1.
Question 10
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the high-temperature reservoir and
rejects heat Qcto the low-temperature reservoir. If the efficiency of the engine
is η, show that the maximum possible efficiency of the engine is given by:
ηmax = 1 −Tc
Th
You may assume the second law of thermodynamics that states that no heat
engine can have an efficiency greater than the efficiency of a Carnot engine
operating between the same two reservoirs.
9
Solution
Step 1: Recall the definition of efficiency of a heat engine: The efficiency of a
heat engine is given by:
η=Useful work output
Input heat
Step 2: Calculate the useful work output: The useful work output Wout is given
by the difference between the heat absorbed Qhand the heat rejected Qc:
Wout =Qh−Qc
Step 3: Substitute the formula for efficiency and the expression for useful work
output into the efficiency equation:
η=Wout
Qh
=Qh−Qc
Qh
Step 4: Rearrange the equation to express efficiency in terms of Th,Tc, and the
unknown efficiency η:
η= 1 −Qc
Qh
= 1 −Tc
Th
Step 5: Therefore, the maximum possible efficiency of the engine is:
ηmax = 1 −Tc
Th
This shows that the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis 1 −Tc
Th, where Th> Tc.
Question 11
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the high-temperature reservoir
and performs Wof work during each complete cycle. Determine the efficiency
of the engine in terms of Qh,W, and the temperatures Thand Tc. Also, explain
how this efficiency is related to the second law of thermodynamics.
Solution
Step 1: To determine the efficiency of the engine, we first need to calculate the
heat rejected to the low-temperature reservoir. Since the engine absorbs Qhof
heat and performs Wof work, the heat rejected Qcis given by the first law of
thermodynamics:
Qh=W+Qc
10
Step 2: The efficiency ηof the engine is defined as the ratio of the net work
output to the heat input:
η=W
Qh
Step 3: Substituting the expression for Qcinto the efficiency equation, we
get:
η=W
W+Qc
Step 4: Rearranging the equation further, we have:
η=1
1 + Qc
W
Step 5: Using the definition of efficiency, Qc=Tc/Th∗Qh, we can rewrite
the efficiency in terms of temperatures Thand Tcas:
η=1
1 + Tc
Th×Qh
W
Step 6: The efficiency of a heat engine is always less than one (expressed as
a percentage) due to the second law of thermodynamics, which states that no
heat engine can be 100
Question 12
Question
A Carnot heat engine operates between a reservoir at 500 K and a reservoir at
300 K. If the engine takes in 1500 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 situation.
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 cold reservoir is at 300 K and the hot reservoir is at 500 K,
we can substitute these values into the formula to find the efficiency:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
11
Step 2: Discuss the second law of thermodynamics. The second law of ther-
modynamics states that heat will naturally flow from a hot reservoir to a cold
one, but it will not flow in the opposite direction without external work being
done. This implies that not all of the heat energy from the high-temperature
reservoir can be converted into work by the engine. The efficiency of the engine
(40
In the context of this Carnot heat engine operating between 500 K and 300 K,
the second law of thermodynamics dictates that some heat energy will inevitably
be lost to the low-temperature reservoir, resulting in a limited efficiency despite
the idealized Carnot cycle.
Question 13
Question
A heat engine operates between two reservoirs at 800 K and 300 K. It has an
efficiency of 40%. Determine the maximum possible efficiency of a heat engine
operating between these two reservoirs.
Solution
Let’s denote the temperature of the hot reservoir as TH= 800 K and the
temperature of the cold reservoir as TC= 300 K. The efficiency of a heat engine
is given by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the current efficiency of the heat engine. Plugging in the
given temperatures into the efficiency formula, we have:
Efficiencycurrent = 1 −300
800 = 1 −3
8=5
8= 0.625 = 62.5%
Step 2: Determine the maximum possible efficiency. The maximum effi-
ciency of a heat engine operating between two temperatures THand TCis given
by Carnot’s efficiency formula:
Efficiencymax = 1 −TC
TH
Plugging in the given temperatures TH= 800 K and TC= 300 K, we have:
Efficiencymax = 1 −300
800 = 1 −3
8=5
8= 0.625 = 62.5%
Therefore, the maximum possible efficiency of a heat engine operating be-
tween these two reservoirs is 62.5%.
12
Question 14
Question
A heat engine operates between two reservoirs at temperatures T1and T2(T1>
T2). The engine has an efficiency of 40
Solution
Let Qhbe the heat extracted from the hot reservoir, Qcbe the heat rejected
to the cold reservoir, and Wbe the work done by the engine. Recall that the
efficiency of the engine is given by the formula:
Efficiency = Work done by the engine
Heat extracted from the hot reservoir
Step 1: Express the efficiency in terms of Qh,Qc, and W.
Efficiency = W
Qh
Given that the efficiency is 40
W
Qh
= 0.40
W= 0.40 ×Qh
Step 2: Apply the first law of thermodynamics. The first law of thermody-
namics states:
Heat in = Work done + Heat out
Qh=W+Qc
Step 3: Substitute the expressions for Wand Qhin terms of Qcinto the
equation from step 2.
500 = 0.40 ×500 + Qc
Qc= 500 −200
Qc= 300 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
300 J.
Question 15
Question
A heat engine operates between two reservoirs at temperatures THand TC
(TH> TC). If the engine absorbs 2000 J of heat from the hot reservoir and
performs 1200 J of work, determine the efficiency of the engine in terms of TH
and TC.
13
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input:
Efficiency = Work output
Heat input
Step 2: In this case, the work output is given as 1200 J and the heat input
is 2000 J. Therefore, the efficiency can be calculated as follows:
Efficiency = 1200 J
2000 J
Step 3: Simplify the expression to find the efficiency:
Efficiency = 3
5= 0.6
Step 4: The efficiency can also be expressed in terms of THand TCusing
the Carnot efficiency formula:
Efficiency = 1 −TC
TH
Step 5: Substitute the given temperatures to express the efficiency in terms
of THand TC:
Efficiency = 1 −TC
TH
= 1 −2000
TH
Step 6: Therefore, the efficiency of the engine in terms of THand TCis
1−2000
TH.
Question 16
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine receives Q1of heat from the reservoir at temperature T1
and exhausts Wof work to do the job. Calculate the efficiency of the engine in
terms of Q1and T1.
Solution
Step 1: Recall that the efficiency of a heat engine, denoted by η, is given by the
ratio of the work done by the engine to the heat input from the hot reservoir.
It can be expressed mathematically as:
η=Work done by the engine
Heat input from the hot reservoir
14
Step 2: The work done by the engine can be calculated using the first law
of thermodynamics, which states that in any thermodynamic process the total
energy of an isolated system remains constant. Therefore, the work done by the
engine is given by:
W=Q1−Q2
Step 3: To find Q2, we use the fact that in a cyclic process, the net heat
absorbed by the working substance equals zero. Therefore,
Q1−Q2= 0
Q2=Q1
Step 4: Now, we substitute Q2=Q1back into the equation for W:
W=Q1−Q1= 0
Step 5: Substituting W= 0 into the efficiency equation, we have:
η=0
Q1
= 0
Step 6: Therefore, the efficiency of the heat engine operating between two
reservoirs at temperatures T1and T2is η= 0.
Question 17
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhamount of heat from the hot
reservoir and rejects Qcamount of heat to the cold reservoir during each cycle.
If Qh= 800 J, Qc= 450 J, and the efficiency of the engine is 60
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −Tc
Th
Step 2: Given that the efficiency is 60
0.60 = 1 −Tc
Th
Step 3: Rearrange the equation to solve for Tc:
Tc= (1 −0.60) Th= 0.40Th
15
Step 4: We are also given the heat absorbed and rejected during each cycle:
Qh= 800 J, Qc= 450 J
Step 5: Since the engine is reversible, the heat transfer ratio is equal to the
temperature ratio: Qh
Qc
=Th
Tc
Step 6: Substitute the given values and Tc= 0.40Thinto the equation above:
800
450 =Th
0.40Th
Step 7: Solve for Th:
800
450 =2= 1
0.40 =Th
0.40Th
Step 8: From the above equation, we find that Th= 800 K.
Step 9: Finally, calculate Tcusing Tc= 0.40Th:
Tc= 0.40 ×800 = 320 K
Therefore, the temperatures of the hot and cold reservoirs are Th= 800 K
and Tc= 320 K, respectively.
Question 18
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc, where Th> Tc. If the efficiency of the engine is η, prove that the
efficiency can be expressed as η= 1 −Tc
Th.
Solution
Step 1: Recall the Carnot efficiency formula, which is given by:
η= 1 −Tc
Th
Step 2: The Carnot efficiency can also be expressed as the ratio of work
output to heat input:
η= 1 −Qc
Qh
Step 3: According to the second law of thermodynamics, for a reversible
heat engine, the ratio of heat added to a system at a lower temperature Tcto
16
the heat rejected to a system at a higher temperature This equal to the ratio
of the absolute temperatures: Qc
Qh
=Tc
Th
Step 4: Substituting this relation into the Carnot efficiency formula gives:
η= 1 −Tc
Th
Step 5: Therefore, the efficiency of a Carnot heat engine operating between
two heat reservoirs at temperatures Thand Tccan be expressed as η= 1 −Tc
Th.
Question 19
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 and exhausts 3000 J of heat to the cold reservoir in each cycle. Determine
the efficiency of the engine and discuss whether this violates the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Substitute the given values into the formula:
Efficiency = 1 −3000
5000
Step 3: Perform the calculation:
Efficiency = 1 −0.6 = 0.4
Step 4: 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 greater than
1−Tcold
Thot
where Tcold is the temperature of the cold reservoir and Thot is the temperature
of the hot reservoir.
17
Step 6: Calculate the maximum efficiency using the given temperatures:
1−300
600 = 0.5
Step 7: The maximum possible efficiency for this engine, based on the given
temperatures, is 0.5 or 50
Step 8: Since the actual efficiency of the engine (40
Question 20
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine absorbs Q1amount of heat from the reservoir at temper-
ature T1and exhausts Q2amount of heat to the reservoir at temperature T2.
Prove that the efficiency of the engine is given by
η= 1 −T2
T1
.
Solution
To prove that the efficiency of the engine is given by η= 1 −T2
T1, we can use the
definition of efficiency for a heat engine:
η=Useful work output
Energy input .
Step 1: Express useful work output in terms of heat absorbed
and heat rejected. The work output of the engine is given by the difference
between the heat absorbed (Q1) and the heat rejected (Q2):
Useful work output = Q1−Q2.
Step 2: Express the energy input in terms of the heat absorbed.
The energy input to the engine is the heat absorbed from the reservoir at tem-
perature T1:
Energy input = Q1.
Step 3: Calculate the efficiency of the engine. Substitute the expres-
sions for useful work output and energy input into the efficiency formula:
η=Q1−Q2
Q1
.
Step 4: Use the definition of efficiency to simplify the expression.
Recall the first law of thermodynamics: Q1= Useful work output + Q2. Sub-
stitute this into the efficiency formula:
η=Q1−(Q1−Q2)
Q1
=Q2
Q1
.
18
Step 5: Use Carnot’s theorem to relate heat and temperature. Ac-
cording to Carnot’s theorem, the maximum efficiency of a heat engine operating
between two temperatures T1and T2is 1 −T2
T1.
Step 6: Compare the obtained efficiency with the maximum effi-
ciency. We have η=Q2
Q1= 1 −T2
T1, which matches the maximum efficiency.
Therefore, we have proved that the efficiency of the engine is η= 1 −T2
T1.
Question 21
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
where Th> Tc. The engine absorbs Qhof heat energy from the high-temperature
reservoir and releases Qcof heat energy to the low-temperature reservoir during
each cycle. Determine the efficiency of the Carnot engine in terms of Th,Tc,
Qh, and Qc.
Solution
Step 1: Recall the efficiency of a Carnot engine, which is given by the formula:
η= 1 −Tc
Th
Step 2: We know that the efficiency can also be expressed in terms of the
heat energies absorbed and released:
η=W
Qh
= 1 −Qc
Qh
Step 3: Equate the two expressions for efficiency:
1−Tc
Th
= 1 −Qc
Qh
Step 4: Simplify the equation:
Qc
Qh
=Tc
Th
Step 5: Therefore, the efficiency of the Carnot engine in terms of Th,Tc,Qh,
and Qcis:
η= 1 −Qc
Qh
= 1 −Tc
Th
19
Question 22
Question
A Carnot engine operates between two heat reservoirs with temperatures TH
and TC, where TH> TC. The engine absorbs 600 J of heat from the reservoir
at THand exhausts 400 J to the reservoir at TC. Determine the efficiency of
this Carnot engine.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −TC
TH
Step 2: Calculate the efficiency using the given temperatures THand TC:
Efficiency = 1 −TC
TH
= 1 −400
600 = 1 −2
3=1
3
Step 3: Therefore, the efficiency of the Carnot engine is 1
3or 33.33
Question 23
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine absorbs QHamount of heat from the hot reservoir
and rejects QCamount of heat to the cold reservoir. If the efficiency of the engine
is η, show that the efficiency of the engine can be expressed as η= 1 −TC
TH.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input. Mathematically, it is given by
η=Wout
Qin
where Wout is the work output and Qin is the heat input.
Step 2: By the first law of thermodynamics, the work output of the engine
is given by
Wout =Qin −Qout
where Qout is the heat rejected to the cold reservoir.
Step 3: The efficiency of the engine can then be rewritten as
η=Qin −Qout
Qin
= 1 −Qout
Qin
20
Step 4: Since the engine absorbs Qin =QHfrom the hot reservoir and
rejects Qout =QCto the cold reservoir, we have
η= 1 −QC
QH
Step 5: Using the definition of efficiency in terms of temperature, we know
that the efficiency of a Carnot engine is given by
η= 1 −TC
TH
Step 6: Comparing the expressions for efficiency, we see that for any heat en-
gine operating between two reservoirs at temperatures THand TC, the efficiency
can be expressed as η= 1 −TC
TH.
Question 24
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and
a cold reservoir at a temperature of 300 K. If the engine has an efficiency of
40%, what is the maximum amount of work that can be extracted from 1000 J
of heat input from the hot reservoir?
Solution
Step 1: Calculate the efficiency of the heat 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: Given that the efficiency is 40% and TC= 300 K and TH= 500 K,
we can solve for the efficiency:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Step 3: The efficiency of the heat engine is 40%. This means that 40% of
the heat input can be converted to work. Therefore, the work output is 40% of
the heat input.
Step 4: Calculate the work output:
Work output = 0.4×1000 J = 400 J
Step 5: The maximum amount of work that can be extracted from 1000 J of
heat input from the hot reservoir is 400 J .
21
Question 25
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 300 K. The engine absorbs 3000 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 during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot heat 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 1: Given, Tc= 300 K and Th= 500 K. Plugging these values into the
formula, we get
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Step 2: Calculate the amount of heat rejected to the cold reservoir during
each cycle using the formula
Heat rejected = Absorbed heat ×Efficiency
Step 2: Given, absorbed heat is 3000 J. Plugging in the values of absorbed
heat and efficiency, we get
Heat rejected = 3000 J ×0.4 = 1200 J
Therefore, the efficiency of the Carnot heat engine is 40
Question 26
Question
A heat engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc. The engine absorbs heat Qhfrom the hot reservoir,
performs work Wduring the cycle, and rejects heat Qcto the cold reservoir.
Prove that the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by ηmax = 1 −Tc
Th.
22
Solution
Step 1: Apply the first law of thermodynamics to the heat 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 rejected:
W=Qh−Qc
Step 2: Write the efficiency of the engine: The efficiency of a heat engine is
defined as the ratio of the work done by the engine to the heat absorbed from
the hot reservoir:
η=W
Qh
Step 3: Substitute the expression for work from Step 1 into the efficiency
expression from Step 2:
η=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: Apply the second law of thermodynamics: The second law of ther-
modynamics states that the efficiency of a heat engine operating between two
reservoirs is always less than the Carnot efficiency, which is the maximum pos-
sible efficiency for a heat engine operating between those two reservoirs:
η≤ηCarnot = 1 −Tc
Th
Step 5: Prove that ηmax = 1 −Tc
Th: Since the efficiency of the engine must be
less than or equal to the Carnot efficiency, then the maximum efficiency ηmax is
equal to the Carnot efficiency:
ηmax = 1 −Tc
Th
Therefore, the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by ηmax = 1 −Tc
Th.
Question 27
Question
A Carnot engine operates between two heat reservoirs at temperatures TH= 600
K and TL= 300 K. The engine absorbs 2000 J of heat from the hot reservoir in
each cycle. Calculate the efficiency of the engine and determine the amount of
heat rejected to the cold reservoir in each cycle.
23
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −TL
TH
Step 1: Given: TH= 600 K, TL= 300 K.
The efficiency of the Carnot engine is:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
QC= Efficiency ×QH
Step 2: Given: QH= 2000 J.
Substitute the values into the formula:
QC=1
2×2000 = 1000 J
Therefore, the efficiency of the Carnot engine is 50
Question 28
Question
A heat engine operates with a thermal efficiency of 40
1. The work output of the engine.
2. The heat expelled to the cold reservoir.
Solution
Let’s denote the following variables:
Qh= heat energy received from the hot reservoir (given as 800 J)
Qc= heat expelled to the cold reservoir
W= work output of the engine
η= thermal efficiency (given as 40
24
Step 1: Calculate the work output of the engine. The thermal efficiency of
an engine is given by the formula:
η=W
Qh
Substitute the values given:
0.4 = W
800
Solve for W:
W= 0.4×800 = 320 J
Therefore, the work output of the engine is 320 J.
Step 2: Calculate the heat expelled to the cold reservoir. The first law of
thermodynamics states that the net energy output must equal the difference
between the energy input and the energy lost to the environment. In this case:
Qh=W+Qc
Substitute the known values:
800 = 320 + Qc
Solve for Qc:
Qc= 800 −320 = 480 J
Therefore, the heat expelled to the cold reservoir is 480 J.
Question 29
Question
A heat engine operating between two heat reservoirs absorbs 800 J of heat
from the high-temperature reservoir at 400 K and expels 300 J of heat to the
low-temperature reservoir. Calculate the efficiency of the engine. Is this result
consistent with the second law of thermodynamics?
Solution
Step 1: We can calculate the efficiency of the engine using the formula for
efficiency:
Efficiency (%) = 1−Heat output
Heat input ×100%
Step 2: Given that the engine absorbs 800 J of heat from the high-temperature
reservoir (Qh= 800 J at Th= 400 K) and expels 300 J to the low-temperature
reservoir, we can find the heat input and heat output:
Qh= 800 J, Qc= 300 J
25
Step 3: Now, substitute the values into the formula for efficiency:
Efficiency (%) = 1−Qc
Qh×100%
Efficiency (%) = 1−300
800×100%
Efficiency (%) = (1 −0.375) ×100%
Efficiency (%) = 0.625 ×100%
Efficiency (%) = 62.5%
Step 4: The efficiency of the engine is 62.5%. This result is consistent with
the second law of thermodynamics, which states that no heat engine can be
100% efficient. Some heat must be dissipated to the surroundings in each cycle,
leading to a decrease in efficiency.
Question 30
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs 1000 J of heat from the hot reservoir and delivers
250 J of work in each cycle. Calculate the efficiency of the engine and discuss
how this result relates to the second law of thermodynamics.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
Efficiency = Work Output
Heat Input .
Step 2: Given that the engine absorbs 1000 J of heat and delivers 250 J of
work, we can substitute these values into the efficiency formula:
Efficiency = 250 J
1000 J.
Step 3: Simplifying the expression gives
Efficiency = 1
4= 0.25 = 25%.
Step 4: The efficiency of the engine is 25
Step 5: This result is in accordance with the second law of thermodynamics,
which states that no heat engine can have an efficiency of 100%. Some heat
must always be rejected to a cooler reservoir, and there will always be some
energy losses due to inefficiencies in the engine.
Step 6: Therefore, the efficiency of the engine being less than 100% confirms
the second law of thermodynamics, which is a fundamental principle governing
the behavior of heat engines.
26
Question 31
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 1500 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:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Convert the temperatures from degrees Celsius to kelvin:
Th= 600 K, Tc= 300 K
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −300
600 = 1 −0.5=0.5
Therefore, the efficiency of the Carnot engine is 50
Step 4: Calculate the amount of heat rejected to the cold reservoir during
each cycle using the formula:
Qc= Efficiency ×Qh
where Qcis the heat rejected to the cold reservoir, Efficiency is the efficiency of
the engine, and Qhis the heat absorbed from the hot reservoir.
Step 5: Substitute the values into the formula:
Qc= 0.5×1500 J = 750 J
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 750 J.
Question 32
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 5000 J of heat from the reservoir at
temperature Th, and 3000 J of heat is rejected to the reservoir at temperature
Tc. Determine the efficiency of this Carnot engine in terms of Thand Tc.
27
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Calculate the efficiency using the given information that the engine
absorbs 5000 J of heat (Qh) and rejects 3000 J of heat (Qc):
Efficiency = 1 −Qc
Qh
= 1 −3000 J
5000 J
= 1 −3
5
=2
5
Step 3: Substitute the temperatures Thand Tcback into the formula for
efficiency:
Efficiency = 1 −Tc
Th
So, the efficiency of the Carnot engine in terms of Thand Tcis 1 −Tc
Th
.
Question 33
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine absorbs QHheat from the hot reservoir and
rejects QCheat to the cold reservoir. If the engine’s efficiency is η, prove that
the efficiency is given by the expression η= 1 −TC
TH.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is given by:
η= 1 −QC
QH
Step 2: Use the First Law of Thermodynamics: The First Law of Thermo-
dynamics states that for a heat engine, the net work done by the engine is equal
to the difference between the heat input QHand the heat output QC:
Wnet =QH−QC
28
Step 3: Write the expression for efficiency in terms of the net work done:
Substitute Wnet =QH−QCinto the expression for efficiency:
η= 1 −QC
QH
= 1 −QC
Wnet +QC
Step 4: Use the definition of efficiency in terms of temperature: The expres-
sion for efficiency in terms of temperatures is given by:
η= 1 −TC
TH
Step 5: Compare the two expressions for efficiency: From Step 3 and Step
4, we have:
1−QC
Wnet +QC
= 1 −TC
TH
Step 6: Simplify the expression:
1−QC
QH
= 1 −TC
TH
QC
QH
=TC
TH
Step 7: Rearrange the expression to obtain the desired result:
η= 1 −TC
TH
Therefore, we have verified that the efficiency of the heat engine is given by
η= 1 −TC
TH.
Question 34
Question
A Carnot heat engine operates between a reservoir at 800 K and a reservoir at
400 K. The engine absorbs 5000 J of heat from the high-temperature reservoir
in each cycle. Calculate the efficiency of the heat engine and determine the min-
imum amount of heat that must be exhausted to the low-temperature reservoir
in each cycle.
Solution
Step 1: Calculate the efficiency of the heat engine using the Carnot efficiency
formula:
Efficiency = 1 −TC
TH
29
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given that TH= 800 K and 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%
Therefore, the efficiency of the heat engine is 50%.
Step 2: Use the efficiency of the engine to find the heat exhausted to the
low-temperature reservoir in each cycle.
The heat absorbed from the high-temperature reservoir is 5000 J in each
cycle.
Since the efficiency is 50%, the engine converts half of the absorbed heat
into work and the remaining half is exhausted to the low-temperature reservoir.
Therefore, the minimum amount of heat exhausted to the low-temperature
reservoir is 5000 J ×0.5 = 2500 J in each cycle.
Question 35
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc
where Th= 600 K and Tc= 300 K. The engine absorbs 2500 J of heat from the
high-temperature reservoir in each cycle. Calculate the efficiency of the engine
and the heat rejected to the low-temperature reservoir in each cycle.
Solution
Step 1: Calculate the efficiency of the engine using the formula for Carnot
efficiency, which is given by
η= 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Substitute the given values Tc= 300 K and Th= 600 K into the
formula to calculate the efficiency.
η= 1 −300
600 = 1 −1
2=1
2= 0.5
Step 3: Therefore, the efficiency of the engine is 50%.
Step 4: Calculate the heat rejected to the low-temperature reservoir in each
cycle using the formula
Qc=ηQh
where Qcis the heat rejected to the low-temperature reservoir, ηis the efficiency
of the engine, and Qhis the heat absorbed from the high-temperature reservoir.
30
engine can be more efficient than a reversible engine operating between the same
two temperature reservoirs.
Therefore, the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by the Carnot efficiency formula.
Question 2
Question
A heat engine operates between two reservoirs at temperatures Thand Tc(with
Th> Tc). The engine absorbs heat Qhfrom the hot reservoir and rejects heat
Qcto the cold reservoir. Calculate the efficiency of the engine in terms of Qh
and Qc.
Solution
Let us denote the efficiency of the engine as η.
Step 1: Recall that the efficiency of a heat engine is given by the formula:
η= 1 −Qc
Qh
Step 2: Since the engine absorbs heat Qhfrom the hot reservoir and rejects
heat Qcto the cold reservoir, we have:
Qh=W+Qc
where Wis the work done by the engine.
Step 3: We can rewrite the efficiency formula in terms of W:
η= 1 −Qc
W+Qc
Step 4: From the first law of thermodynamics, the work done by the engine
is given by:
W=Qh−Qc
Step 5: Substituting the expression for Winto the efficiency formula, we
get:
η= 1 −Qc
Qh−Qc+Qc
η= 1 −Qc
Qh
Step 6: Therefore, the efficiency of the engine in terms of Qhand Qcis:
η= 1 −Qc
Qh
2
Question 3
Question
A heat engine operates between two reservoirs at temperatures Thand Tcwith
Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and releases heat
Qcto the cold reservoir. If Wis the work done by the engine, show that the
efficiency of the engine is given by η= 1 −Tc
Th.
Solution
Step 1: We first need to recall the definition of the efficiency of a heat engine,
which is given by
η=Work output
Heat input =W
Qh
.
Step 2: We know that according to the first law of thermodynamics, the net
work done by the engine is equal to the difference in the heat absorbed and the
heat released, i.e., W=Qh−Qc.
Step 3: Substituting W=Qh−Qcinto the efficiency formula, we get
η=Qh−Qc
Qh
.
Step 4: Rearranging the terms, we have
η= 1 −Qc
Qh
.
Step 5: Now, from the second law of thermodynamics, the efficiency of a
heat engine is limited by the Carnot efficiency, given by ηCarnot = 1 −Tc
Th.
Step 6: Comparing η= 1 −Qc
Qhwith ηCarnot = 1 −Tc
Th, we see that Qc
Qh=Tc
Th.
Step 7: Substituting Qc
Qh=Tc
Thback into the expression for efficiency, we find
η= 1 −Tc
Th
.
Therefore, the efficiency of the heat engine is given by η= 1 −Tc
Th.
Question 4
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
with Th> Tc. The engine absorbs heat energy Qhfrom the reservoir at temper-
ature Thand expels heat energy Qcto the reservoir at temperature Tc. Prove
that the efficiency of the Carnot engine, η, is given by η= 1 −Tc
Th.
3
Solution
Step 1: Recall the efficiency of a heat engine is defined as the ratio of the work
output to the heat input. For the Carnot engine, the efficiency is given by
η=W
Qh
Step 2: With reference to the second law of thermodynamics, during the
Carnot cycle, the ratio of the heat absorbed to the high-temperature reservoir
to the work done by the engine is equal to the ratio of the heat expelled to the
low-temperature reservoir over the work done by the engine:
Qh
Th
=Qc
Tc
Step 3: Rearranging the above equation, we get
Qc
Qh
=Tc
Th
Step 4: Now, we know that the work done by the engine is equal to the
difference between the heat absorbed and the heat expelled:
W=Qh−Qc
Step 5: Substituting the values of Qcand Qhfrom step 3 into the equation
above, we get
W=Qh−Tc
ThQh
Step 6: Simplifying the above expression, we find
W=Qh1−Tc
Th
Step 7: Finally, substituting the value of Winto the efficiency equation from
step 1, we obtain
η=W
Qh
=
Qh1−Tc
Th
Qh
= 1 −Tc
Th
Therefore, the efficiency of a Carnot engine is given by η= 1 −Tc
Th.
Question 5
Question
A Carnot heat engine operates between two reservoirs at temperatures of 500
K and 300 K. If the engine absorbs 4000 J of heat from the high-temperature
reservoir in each cycle, calculate the following: (a) The efficiency of the Carnot
engine. (b) The heat rejected to the low-temperature reservoir in each cycle.
(c) The net work output in each cycle.
4
Solution
(a) Let’s denote the high-temperature reservoir as TH= 500 K and the low-
temperature reservoir as TC= 300 K. The efficiency of a Carnot engine is given
by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the efficiency of the Carnot engine.
Efficiency = 1 −300
500 = 1 −3
5=2
5= 0.4 = 40%
(b) The heat rejected to the low-temperature reservoir in each cycle can be
determined using the formula:
Heat rejected = QC= Efficiency ×QH
Step 2: Calculate the heat rejected to the low-temperature reservoir.
QC= 0.4×4000 J = 1600 J
(c) The net work output in each cycle is given by the difference between the
heat absorbed from the high-temperature reservoir and the heat rejected to the
low-temperature reservoir:
Net work output = Wout =QH−QC
Step 3: Calculate the net work output in each cycle.
Wout = 4000 J −1600 J = 2400 J
Therefore, the answers to the questions are: (a) The efficiency of the Carnot
engine is 40(b) The heat rejected to the low-temperature reservoir in each cycle
is 1600 J. (c) The net work output in each cycle is 2400 J.
Question 6
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and a
cold reservoir at a temperature of 300 K. If the engine produces 1200 J of work
per cycle, calculate the efficiency of the engine. Is this efficiency physically
possible according to the second law of thermodynamics?
5
Solution
Step 1: Calculate the efficiency of the heat engine using the formula for efficiency
(η):
η= 1 −Tc
Th
where η= efficiency of the engine, Tc= temperature of the cold reservoir, Th
= temperature of the hot reservoir.
Given that Th= 500 K and Tc= 300 K, we can plug these values into the
formula to find the efficiency.
Step 2: Calculate the efficiency:
η= 1 −300
500 = 1 −0.6=0.4
Therefore, the efficiency of the engine is 0.4 (or 40
Step 3: Determine whether this efficiency is physically possible according to
the second law of thermodynamics. According to the second law of thermody-
namics, no heat engine can have an efficiency greater than the Carnot efficiency,
which is given by
ηCarnot = 1 −Tc
Th
where ηCarnot is the maximum possible efficiency of any heat engine operating
between the two temperatures.
Step 4: Calculate the Carnot efficiency:
ηCarnot = 1 −300
500 = 1 −0.6=0.4
Step 5: Compare the efficiency of the engine with the Carnot efficiency. The
efficiency of the engine is equal to the Carnot efficiency, which means the engine
is operating at the maximum possible efficiency allowed by the second law of
thermodynamics. Thus, the efficiency of 40
Question 7
Question
A heat engine operates between two reservoirs with temperatures THand TC,
where TH> TC. The engine absorbs QHjoules of heat from the hot reservoir
and expels QCjoules of heat to the cold reservoir during each cycle. If the
efficiency of the engine is η, prove that the efficiency of the engine cannot be
100%.
6
Solution
Let’s assume that the efficiency of the engine is 100% or η= 1. This means
that all the heat absorbed from the hot reservoir is converted into work.
Step 1: Calculate the heat expelled to the cold reservoir. Using the effi-
ciency formula for a heat engine:
η=Useful work done
Heat absorbed = 1
This implies that the useful work done is equal to the heat absorbed from the
hot reservoir:
Useful work done = QH
Since η= 1, the work done is equal to the heat absorbed from the hot reservoir.
Step 2: Apply the conservation of energy. According to the first law of
thermodynamics, we have:
QH= Useful work done + QC
Substitute the values:
QH=QH+QC
QC= 0
This implies that the engine expels no heat to the cold reservoir, which con-
tradicts the given information. Therefore, our assumption that the efficiency is
100% must be incorrect, and the efficiency of the engine cannot be 100%.
Question 8
Question
A Carnot engine operates between two heat reservoirs at temperatures T1and
T2(T1> T2). The engine absorbs heat Q1from the reservoir at temperature
T1,Q1= 4000 J, and exhausts heat Q2to the reservoir at temperature T2. Find
the efficiency of the engine and the amount of heat exhausted to the reservoir
at temperature T2,Q2, in terms of Q1.
Solution
Step 1: Calculate the efficiency of the Carnot engine. The efficiency of a Carnot
engine is given by the formula:
Efficiency = 1 −T2
T1
Given T1> T2, we have T1=T2+ ∆T, where ∆T=T1−T2. So, the efficiency
of the Carnot engine is:
Efficiency = 1 −T2
T2+ ∆T= 1 −T2
T1
7
Step 2: Substitute values and calculate. Given Q1= 4000 J, the efficiency
is:
Efficiency = 1 −T2
T1
= 1 −T2
T2+ ∆T
Efficiency = 1 −T2
T2+ (T1−T2)= 1 −T2
T1
Efficiency = 1 −T2
T1
= 1 −T2
T1
= 1 −T2
T1
Step 3: Calculate the heat exhausted to the reservoir at temperature T2,Q2.
Using the first law of thermodynamics, we know that the net work done by the
engine is:
Wnet =Q1−Q2
For a Carnot engine, the net work done is:
Wnet =Q11−T2
T1
And the heat exhausted to the reservoir at temperature T2is:
Q2=Q1−Wnet
Step 4: Substitute values and calculate Q2. Given Q1= 4000 J, we have:
Q2=Q1−Wnet =Q11−T2
T1
Q2= 4000 1−T2
T1
Therefore, the efficiency of the engine is 1 −T2
T1and the heat exhausted to
the reservoir at temperature T2,Q2, is 4000 1−T2
T1.
Question 9
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, with Th> Tc. The engine absorbs Qhof heat from the hot reservoir and
rejects Qcof heat to the cold reservoir. Find an expression for the efficiency of
the Carnot engine in terms of Th,Tc,Qh, and Qc.
8
Solution
Step 1: Recall that the efficiency of any heat engine is defined as the ratio of
the work output to the heat input. For a Carnot engine, the efficiency can be
expressed as
η=W
Qh
.
Step 2: The work output of a Carnot engine is given by the difference in the
heat absorbed from the hot reservoir and the heat rejected to the cold reservoir,
i.e., W=Qh−Qc.
Step 3: Substitute the expression for work output into the efficiency formula
to yield
η=Qh−Qc
Qh
.
Step 4: Recognize that the heat rejected to the cold reservoir can be written
in terms of the heat absorbed from the hot reservoir and the efficiency of the
engine. We have Qc=Qh−W=Qh−ηQh.
Step 5: Substitute the expression for Qcinto the efficiency formula, resulting
in
η=Qh−(Qh−ηQh)
Qh
=Qh−Qh+ηQh
Qh
=ηQh
Qh
.
Step 6: Simplify the expression to find the final formula for the efficiency of
a Carnot engine:
η=η.
Therefore, the efficiency of a Carnot engine is solely dependent on the effi-
ciency itself and is given by η= 1.
Question 10
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs heat Qhfrom the high-temperature reservoir and
rejects heat Qcto the low-temperature reservoir. If the efficiency of the engine
is η, show that the maximum possible efficiency of the engine is given by:
ηmax = 1 −Tc
Th
You may assume the second law of thermodynamics that states that no heat
engine can have an efficiency greater than the efficiency of a Carnot engine
operating between the same two reservoirs.
9
Solution
Step 1: Recall the definition of efficiency of a heat engine: The efficiency of a
heat engine is given by:
η=Useful work output
Input heat
Step 2: Calculate the useful work output: The useful work output Wout is given
by the difference between the heat absorbed Qhand the heat rejected Qc:
Wout =Qh−Qc
Step 3: Substitute the formula for efficiency and the expression for useful work
output into the efficiency equation:
η=Wout
Qh
=Qh−Qc
Qh
Step 4: Rearrange the equation to express efficiency in terms of Th,Tc, and the
unknown efficiency η:
η= 1 −Qc
Qh
= 1 −Tc
Th
Step 5: Therefore, the maximum possible efficiency of the engine is:
ηmax = 1 −Tc
Th
This shows that the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis 1 −Tc
Th, where Th> Tc.
Question 11
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs Qhof heat from the high-temperature reservoir
and performs Wof work during each complete cycle. Determine the efficiency
of the engine in terms of Qh,W, and the temperatures Thand Tc. Also, explain
how this efficiency is related to the second law of thermodynamics.
Solution
Step 1: To determine the efficiency of the engine, we first need to calculate the
heat rejected to the low-temperature reservoir. Since the engine absorbs Qhof
heat and performs Wof work, the heat rejected Qcis given by the first law of
thermodynamics:
Qh=W+Qc
10
Step 2: The efficiency ηof the engine is defined as the ratio of the net work
output to the heat input:
η=W
Qh
Step 3: Substituting the expression for Qcinto the efficiency equation, we
get:
η=W
W+Qc
Step 4: Rearranging the equation further, we have:
η=1
1 + Qc
W
Step 5: Using the definition of efficiency, Qc=Tc/Th∗Qh, we can rewrite
the efficiency in terms of temperatures Thand Tcas:
η=1
1 + Tc
Th×Qh
W
Step 6: The efficiency of a heat engine is always less than one (expressed as
a percentage) due to the second law of thermodynamics, which states that no
heat engine can be 100
Question 12
Question
A Carnot heat engine operates between a reservoir at 500 K and a reservoir at
300 K. If the engine takes in 1500 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 situation.
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 cold reservoir is at 300 K and the hot reservoir is at 500 K,
we can substitute these values into the formula to find the efficiency:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
11
Step 2: Discuss the second law of thermodynamics. The second law of ther-
modynamics states that heat will naturally flow from a hot reservoir to a cold
one, but it will not flow in the opposite direction without external work being
done. This implies that not all of the heat energy from the high-temperature
reservoir can be converted into work by the engine. The efficiency of the engine
(40
In the context of this Carnot heat engine operating between 500 K and 300 K,
the second law of thermodynamics dictates that some heat energy will inevitably
be lost to the low-temperature reservoir, resulting in a limited efficiency despite
the idealized Carnot cycle.
Question 13
Question
A heat engine operates between two reservoirs at 800 K and 300 K. It has an
efficiency of 40%. Determine the maximum possible efficiency of a heat engine
operating between these two reservoirs.
Solution
Let’s denote the temperature of the hot reservoir as TH= 800 K and the
temperature of the cold reservoir as TC= 300 K. The efficiency of a heat engine
is given by the formula:
Efficiency = 1 −TC
TH
Step 1: Calculate the current efficiency of the heat engine. Plugging in the
given temperatures into the efficiency formula, we have:
Efficiencycurrent = 1 −300
800 = 1 −3
8=5
8= 0.625 = 62.5%
Step 2: Determine the maximum possible efficiency. The maximum effi-
ciency of a heat engine operating between two temperatures THand TCis given
by Carnot’s efficiency formula:
Efficiencymax = 1 −TC
TH
Plugging in the given temperatures TH= 800 K and TC= 300 K, we have:
Efficiencymax = 1 −300
800 = 1 −3
8=5
8= 0.625 = 62.5%
Therefore, the maximum possible efficiency of a heat engine operating be-
tween these two reservoirs is 62.5%.
12
Question 14
Question
A heat engine operates between two reservoirs at temperatures T1and T2(T1>
T2). The engine has an efficiency of 40
Solution
Let Qhbe the heat extracted from the hot reservoir, Qcbe the heat rejected
to the cold reservoir, and Wbe the work done by the engine. Recall that the
efficiency of the engine is given by the formula:
Efficiency = Work done by the engine
Heat extracted from the hot reservoir
Step 1: Express the efficiency in terms of Qh,Qc, and W.
Efficiency = W
Qh
Given that the efficiency is 40
W
Qh
= 0.40
W= 0.40 ×Qh
Step 2: Apply the first law of thermodynamics. The first law of thermody-
namics states:
Heat in = Work done + Heat out
Qh=W+Qc
Step 3: Substitute the expressions for Wand Qhin terms of Qcinto the
equation from step 2.
500 = 0.40 ×500 + Qc
Qc= 500 −200
Qc= 300 J
Therefore, the amount of heat rejected to the cold reservoir in each cycle is
300 J.
Question 15
Question
A heat engine operates between two reservoirs at temperatures THand TC
(TH> TC). If the engine absorbs 2000 J of heat from the hot reservoir and
performs 1200 J of work, determine the efficiency of the engine in terms of TH
and TC.
13
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input:
Efficiency = Work output
Heat input
Step 2: In this case, the work output is given as 1200 J and the heat input
is 2000 J. Therefore, the efficiency can be calculated as follows:
Efficiency = 1200 J
2000 J
Step 3: Simplify the expression to find the efficiency:
Efficiency = 3
5= 0.6
Step 4: The efficiency can also be expressed in terms of THand TCusing
the Carnot efficiency formula:
Efficiency = 1 −TC
TH
Step 5: Substitute the given temperatures to express the efficiency in terms
of THand TC:
Efficiency = 1 −TC
TH
= 1 −2000
TH
Step 6: Therefore, the efficiency of the engine in terms of THand TCis
1−2000
TH.
Question 16
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine receives Q1of heat from the reservoir at temperature T1
and exhausts Wof work to do the job. Calculate the efficiency of the engine in
terms of Q1and T1.
Solution
Step 1: Recall that the efficiency of a heat engine, denoted by η, is given by the
ratio of the work done by the engine to the heat input from the hot reservoir.
It can be expressed mathematically as:
η=Work done by the engine
Heat input from the hot reservoir
14
Step 2: The work done by the engine can be calculated using the first law
of thermodynamics, which states that in any thermodynamic process the total
energy of an isolated system remains constant. Therefore, the work done by the
engine is given by:
W=Q1−Q2
Step 3: To find Q2, we use the fact that in a cyclic process, the net heat
absorbed by the working substance equals zero. Therefore,
Q1−Q2= 0
Q2=Q1
Step 4: Now, we substitute Q2=Q1back into the equation for W:
W=Q1−Q1= 0
Step 5: Substituting W= 0 into the efficiency equation, we have:
η=0
Q1
= 0
Step 6: Therefore, the efficiency of the heat engine operating between two
reservoirs at temperatures T1and T2is η= 0.
Question 17
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs Qhamount of heat from the hot
reservoir and rejects Qcamount of heat to the cold reservoir during each cycle.
If Qh= 800 J, Qc= 450 J, and the efficiency of the engine is 60
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −Tc
Th
Step 2: Given that the efficiency is 60
0.60 = 1 −Tc
Th
Step 3: Rearrange the equation to solve for Tc:
Tc= (1 −0.60) Th= 0.40Th
15
Step 4: We are also given the heat absorbed and rejected during each cycle:
Qh= 800 J, Qc= 450 J
Step 5: Since the engine is reversible, the heat transfer ratio is equal to the
temperature ratio: Qh
Qc
=Th
Tc
Step 6: Substitute the given values and Tc= 0.40Thinto the equation above:
800
450 =Th
0.40Th
Step 7: Solve for Th:
800
450 =2= 1
0.40 =Th
0.40Th
Step 8: From the above equation, we find that Th= 800 K.
Step 9: Finally, calculate Tcusing Tc= 0.40Th:
Tc= 0.40 ×800 = 320 K
Therefore, the temperatures of the hot and cold reservoirs are Th= 800 K
and Tc= 320 K, respectively.
Question 18
Question
A Carnot heat engine operates between two heat reservoirs at temperatures Th
and Tc, where Th> Tc. If the efficiency of the engine is η, prove that the
efficiency can be expressed as η= 1 −Tc
Th.
Solution
Step 1: Recall the Carnot efficiency formula, which is given by:
η= 1 −Tc
Th
Step 2: The Carnot efficiency can also be expressed as the ratio of work
output to heat input:
η= 1 −Qc
Qh
Step 3: According to the second law of thermodynamics, for a reversible
heat engine, the ratio of heat added to a system at a lower temperature Tcto
16
the heat rejected to a system at a higher temperature This equal to the ratio
of the absolute temperatures: Qc
Qh
=Tc
Th
Step 4: Substituting this relation into the Carnot efficiency formula gives:
η= 1 −Tc
Th
Step 5: Therefore, the efficiency of a Carnot heat engine operating between
two heat reservoirs at temperatures Thand Tccan be expressed as η= 1 −Tc
Th.
Question 19
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 and exhausts 3000 J of heat to the cold reservoir in each cycle. Determine
the efficiency of the engine and discuss whether this violates the second law of
thermodynamics.
Solution
Step 1: Calculate the efficiency of the engine using the formula:
Efficiency = 1 −Heat output
Heat input
Step 2: Substitute the given values into the formula:
Efficiency = 1 −3000
5000
Step 3: Perform the calculation:
Efficiency = 1 −0.6 = 0.4
Step 4: 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 greater than
1−Tcold
Thot
where Tcold is the temperature of the cold reservoir and Thot is the temperature
of the hot reservoir.
17
Step 6: Calculate the maximum efficiency using the given temperatures:
1−300
600 = 0.5
Step 7: The maximum possible efficiency for this engine, based on the given
temperatures, is 0.5 or 50
Step 8: Since the actual efficiency of the engine (40
Question 20
Question
A heat engine operates between two reservoirs at temperatures T1and T2, where
T1> T2. The engine absorbs Q1amount of heat from the reservoir at temper-
ature T1and exhausts Q2amount of heat to the reservoir at temperature T2.
Prove that the efficiency of the engine is given by
η= 1 −T2
T1
.
Solution
To prove that the efficiency of the engine is given by η= 1 −T2
T1, we can use the
definition of efficiency for a heat engine:
η=Useful work output
Energy input .
Step 1: Express useful work output in terms of heat absorbed
and heat rejected. The work output of the engine is given by the difference
between the heat absorbed (Q1) and the heat rejected (Q2):
Useful work output = Q1−Q2.
Step 2: Express the energy input in terms of the heat absorbed.
The energy input to the engine is the heat absorbed from the reservoir at tem-
perature T1:
Energy input = Q1.
Step 3: Calculate the efficiency of the engine. Substitute the expres-
sions for useful work output and energy input into the efficiency formula:
η=Q1−Q2
Q1
.
Step 4: Use the definition of efficiency to simplify the expression.
Recall the first law of thermodynamics: Q1= Useful work output + Q2. Sub-
stitute this into the efficiency formula:
η=Q1−(Q1−Q2)
Q1
=Q2
Q1
.
18
Step 5: Use Carnot’s theorem to relate heat and temperature. Ac-
cording to Carnot’s theorem, the maximum efficiency of a heat engine operating
between two temperatures T1and T2is 1 −T2
T1.
Step 6: Compare the obtained efficiency with the maximum effi-
ciency. We have η=Q2
Q1= 1 −T2
T1, which matches the maximum efficiency.
Therefore, we have proved that the efficiency of the engine is η= 1 −T2
T1.
Question 21
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc,
where Th> Tc. The engine absorbs Qhof heat energy from the high-temperature
reservoir and releases Qcof heat energy to the low-temperature reservoir during
each cycle. Determine the efficiency of the Carnot engine in terms of Th,Tc,
Qh, and Qc.
Solution
Step 1: Recall the efficiency of a Carnot engine, which is given by the formula:
η= 1 −Tc
Th
Step 2: We know that the efficiency can also be expressed in terms of the
heat energies absorbed and released:
η=W
Qh
= 1 −Qc
Qh
Step 3: Equate the two expressions for efficiency:
1−Tc
Th
= 1 −Qc
Qh
Step 4: Simplify the equation:
Qc
Qh
=Tc
Th
Step 5: Therefore, the efficiency of the Carnot engine in terms of Th,Tc,Qh,
and Qcis:
η= 1 −Qc
Qh
= 1 −Tc
Th
19
Question 22
Question
A Carnot engine operates between two heat reservoirs with temperatures TH
and TC, where TH> TC. The engine absorbs 600 J of heat from the reservoir
at THand exhausts 400 J to the reservoir at TC. Determine the efficiency of
this Carnot engine.
Solution
Step 1: Recall the formula for the efficiency of a Carnot engine:
Efficiency = 1 −TC
TH
Step 2: Calculate the efficiency using the given temperatures THand TC:
Efficiency = 1 −TC
TH
= 1 −400
600 = 1 −2
3=1
3
Step 3: Therefore, the efficiency of the Carnot engine is 1
3or 33.33
Question 23
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine absorbs QHamount of heat from the hot reservoir
and rejects QCamount of heat to the cold reservoir. If the efficiency of the engine
is η, show that the efficiency of the engine can be expressed as η= 1 −TC
TH.
Solution
Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the
work output to the heat input. Mathematically, it is given by
η=Wout
Qin
where Wout is the work output and Qin is the heat input.
Step 2: By the first law of thermodynamics, the work output of the engine
is given by
Wout =Qin −Qout
where Qout is the heat rejected to the cold reservoir.
Step 3: The efficiency of the engine can then be rewritten as
η=Qin −Qout
Qin
= 1 −Qout
Qin
20
Step 4: Since the engine absorbs Qin =QHfrom the hot reservoir and
rejects Qout =QCto the cold reservoir, we have
η= 1 −QC
QH
Step 5: Using the definition of efficiency in terms of temperature, we know
that the efficiency of a Carnot engine is given by
η= 1 −TC
TH
Step 6: Comparing the expressions for efficiency, we see that for any heat en-
gine operating between two reservoirs at temperatures THand TC, the efficiency
can be expressed as η= 1 −TC
TH.
Question 24
Question
A heat engine operates between a hot reservoir at a temperature of 500 K and
a cold reservoir at a temperature of 300 K. If the engine has an efficiency of
40%, what is the maximum amount of work that can be extracted from 1000 J
of heat input from the hot reservoir?
Solution
Step 1: Calculate the efficiency of the heat 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: Given that the efficiency is 40% and TC= 300 K and TH= 500 K,
we can solve for the efficiency:
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Step 3: The efficiency of the heat engine is 40%. This means that 40% of
the heat input can be converted to work. Therefore, the work output is 40% of
the heat input.
Step 4: Calculate the work output:
Work output = 0.4×1000 J = 400 J
Step 5: The maximum amount of work that can be extracted from 1000 J of
heat input from the hot reservoir is 400 J .
21
Question 25
Question
A Carnot heat engine operates between a hot reservoir at 500 K and a cold
reservoir at 300 K. The engine absorbs 3000 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 during each cycle.
Solution
Step 1: Calculate the efficiency of the Carnot heat 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 1: Given, Tc= 300 K and Th= 500 K. Plugging these values into the
formula, we get
Efficiency = 1 −300
500 = 1 −0.6 = 0.4 = 40%
Step 2: Calculate the amount of heat rejected to the cold reservoir during
each cycle using the formula
Heat rejected = Absorbed heat ×Efficiency
Step 2: Given, absorbed heat is 3000 J. Plugging in the values of absorbed
heat and efficiency, we get
Heat rejected = 3000 J ×0.4 = 1200 J
Therefore, the efficiency of the Carnot heat engine is 40
Question 26
Question
A heat engine operates between a hot reservoir at temperature Thand a cold
reservoir at temperature Tc. The engine absorbs heat Qhfrom the hot reservoir,
performs work Wduring the cycle, and rejects heat Qcto the cold reservoir.
Prove that the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by ηmax = 1 −Tc
Th.
22
Solution
Step 1: Apply the first law of thermodynamics to the heat 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 rejected:
W=Qh−Qc
Step 2: Write the efficiency of the engine: The efficiency of a heat engine is
defined as the ratio of the work done by the engine to the heat absorbed from
the hot reservoir:
η=W
Qh
Step 3: Substitute the expression for work from Step 1 into the efficiency
expression from Step 2:
η=Qh−Qc
Qh
= 1 −Qc
Qh
Step 4: Apply the second law of thermodynamics: The second law of ther-
modynamics states that the efficiency of a heat engine operating between two
reservoirs is always less than the Carnot efficiency, which is the maximum pos-
sible efficiency for a heat engine operating between those two reservoirs:
η≤ηCarnot = 1 −Tc
Th
Step 5: Prove that ηmax = 1 −Tc
Th: Since the efficiency of the engine must be
less than or equal to the Carnot efficiency, then the maximum efficiency ηmax is
equal to the Carnot efficiency:
ηmax = 1 −Tc
Th
Therefore, the maximum efficiency of a heat engine operating between two
reservoirs at temperatures Thand Tcis given by ηmax = 1 −Tc
Th.
Question 27
Question
A Carnot engine operates between two heat reservoirs at temperatures TH= 600
K and TL= 300 K. The engine absorbs 2000 J of heat from the hot reservoir in
each cycle. Calculate the efficiency of the engine and determine the amount of
heat rejected to the cold reservoir in each cycle.
23
Solution
Step 1: Calculate the efficiency of the Carnot engine using the formula:
Efficiency = 1 −TL
TH
Step 1: Given: TH= 600 K, TL= 300 K.
The efficiency of the Carnot engine is:
Efficiency = 1 −300
600 = 1 −1
2=1
2= 50%
Step 2: Calculate the amount of heat rejected to the cold reservoir in each
cycle using the formula:
QC= Efficiency ×QH
Step 2: Given: QH= 2000 J.
Substitute the values into the formula:
QC=1
2×2000 = 1000 J
Therefore, the efficiency of the Carnot engine is 50
Question 28
Question
A heat engine operates with a thermal efficiency of 40
1. The work output of the engine.
2. The heat expelled to the cold reservoir.
Solution
Let’s denote the following variables:
Qh= heat energy received from the hot reservoir (given as 800 J)
Qc= heat expelled to the cold reservoir
W= work output of the engine
η= thermal efficiency (given as 40
24
Step 1: Calculate the work output of the engine. The thermal efficiency of
an engine is given by the formula:
η=W
Qh
Substitute the values given:
0.4 = W
800
Solve for W:
W= 0.4×800 = 320 J
Therefore, the work output of the engine is 320 J.
Step 2: Calculate the heat expelled to the cold reservoir. The first law of
thermodynamics states that the net energy output must equal the difference
between the energy input and the energy lost to the environment. In this case:
Qh=W+Qc
Substitute the known values:
800 = 320 + Qc
Solve for Qc:
Qc= 800 −320 = 480 J
Therefore, the heat expelled to the cold reservoir is 480 J.
Question 29
Question
A heat engine operating between two heat reservoirs absorbs 800 J of heat
from the high-temperature reservoir at 400 K and expels 300 J of heat to the
low-temperature reservoir. Calculate the efficiency of the engine. Is this result
consistent with the second law of thermodynamics?
Solution
Step 1: We can calculate the efficiency of the engine using the formula for
efficiency:
Efficiency (%) = 1−Heat output
Heat input ×100%
Step 2: Given that the engine absorbs 800 J of heat from the high-temperature
reservoir (Qh= 800 J at Th= 400 K) and expels 300 J to the low-temperature
reservoir, we can find the heat input and heat output:
Qh= 800 J, Qc= 300 J
25
Step 3: Now, substitute the values into the formula for efficiency:
Efficiency (%) = 1−Qc
Qh×100%
Efficiency (%) = 1−300
800×100%
Efficiency (%) = (1 −0.375) ×100%
Efficiency (%) = 0.625 ×100%
Efficiency (%) = 62.5%
Step 4: The efficiency of the engine is 62.5%. This result is consistent with
the second law of thermodynamics, which states that no heat engine can be
100% efficient. Some heat must be dissipated to the surroundings in each cycle,
leading to a decrease in efficiency.
Question 30
Question
A heat engine operates between two reservoirs at temperatures Thand Tc, where
Th> Tc. The engine absorbs 1000 J of heat from the hot reservoir and delivers
250 J of work in each cycle. Calculate the efficiency of the engine and discuss
how this result relates to the second law of thermodynamics.
Solution
Step 1: Recall that the efficiency of a heat engine is given by the formula
Efficiency = Work Output
Heat Input .
Step 2: Given that the engine absorbs 1000 J of heat and delivers 250 J of
work, we can substitute these values into the efficiency formula:
Efficiency = 250 J
1000 J.
Step 3: Simplifying the expression gives
Efficiency = 1
4= 0.25 = 25%.
Step 4: The efficiency of the engine is 25
Step 5: This result is in accordance with the second law of thermodynamics,
which states that no heat engine can have an efficiency of 100%. Some heat
must always be rejected to a cooler reservoir, and there will always be some
energy losses due to inefficiencies in the engine.
Step 6: Therefore, the efficiency of the engine being less than 100% confirms
the second law of thermodynamics, which is a fundamental principle governing
the behavior of heat engines.
26
Question 31
Question
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir
at 300 K. The engine absorbs 1500 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:
Efficiency = 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Convert the temperatures from degrees Celsius to kelvin:
Th= 600 K, Tc= 300 K
Step 3: Substitute the values into the efficiency formula:
Efficiency = 1 −300
600 = 1 −0.5=0.5
Therefore, the efficiency of the Carnot engine is 50
Step 4: Calculate the amount of heat rejected to the cold reservoir during
each cycle using the formula:
Qc= Efficiency ×Qh
where Qcis the heat rejected to the cold reservoir, Efficiency is the efficiency of
the engine, and Qhis the heat absorbed from the hot reservoir.
Step 5: Substitute the values into the formula:
Qc= 0.5×1500 J = 750 J
Therefore, the amount of heat rejected to the cold reservoir during each cycle
is 750 J.
Question 32
Question
A Carnot engine operates between two heat reservoirs at temperatures Thand
Tc, where Th> Tc. The engine absorbs 5000 J of heat from the reservoir at
temperature Th, and 3000 J of heat is rejected to the reservoir at temperature
Tc. Determine the efficiency of this Carnot engine in terms of Thand Tc.
27
Solution
Step 1: Recall the efficiency of a Carnot engine is given by the formula:
Efficiency = 1 −Tc
Th
Step 2: Calculate the efficiency using the given information that the engine
absorbs 5000 J of heat (Qh) and rejects 3000 J of heat (Qc):
Efficiency = 1 −Qc
Qh
= 1 −3000 J
5000 J
= 1 −3
5
=2
5
Step 3: Substitute the temperatures Thand Tcback into the formula for
efficiency:
Efficiency = 1 −Tc
Th
So, the efficiency of the Carnot engine in terms of Thand Tcis 1 −Tc
Th
.
Question 33
Question
A heat engine operates between two reservoirs at temperatures THand TC,
where TH> TC. The engine absorbs QHheat from the hot reservoir and
rejects QCheat to the cold reservoir. If the engine’s efficiency is η, prove that
the efficiency is given by the expression η= 1 −TC
TH.
Solution
Step 1: Recall the definition of efficiency for a heat engine: The efficiency ηof
a heat engine is given by:
η= 1 −QC
QH
Step 2: Use the First Law of Thermodynamics: The First Law of Thermo-
dynamics states that for a heat engine, the net work done by the engine is equal
to the difference between the heat input QHand the heat output QC:
Wnet =QH−QC
28
Step 3: Write the expression for efficiency in terms of the net work done:
Substitute Wnet =QH−QCinto the expression for efficiency:
η= 1 −QC
QH
= 1 −QC
Wnet +QC
Step 4: Use the definition of efficiency in terms of temperature: The expres-
sion for efficiency in terms of temperatures is given by:
η= 1 −TC
TH
Step 5: Compare the two expressions for efficiency: From Step 3 and Step
4, we have:
1−QC
Wnet +QC
= 1 −TC
TH
Step 6: Simplify the expression:
1−QC
QH
= 1 −TC
TH
QC
QH
=TC
TH
Step 7: Rearrange the expression to obtain the desired result:
η= 1 −TC
TH
Therefore, we have verified that the efficiency of the heat engine is given by
η= 1 −TC
TH.
Question 34
Question
A Carnot heat engine operates between a reservoir at 800 K and a reservoir at
400 K. The engine absorbs 5000 J of heat from the high-temperature reservoir
in each cycle. Calculate the efficiency of the heat engine and determine the min-
imum amount of heat that must be exhausted to the low-temperature reservoir
in each cycle.
Solution
Step 1: Calculate the efficiency of the heat engine using the Carnot efficiency
formula:
Efficiency = 1 −TC
TH
29
where TCis the temperature of the cold reservoir and THis the temperature of
the hot reservoir.
Given that TH= 800 K and 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%
Therefore, the efficiency of the heat engine is 50%.
Step 2: Use the efficiency of the engine to find the heat exhausted to the
low-temperature reservoir in each cycle.
The heat absorbed from the high-temperature reservoir is 5000 J in each
cycle.
Since the efficiency is 50%, the engine converts half of the absorbed heat
into work and the remaining half is exhausted to the low-temperature reservoir.
Therefore, the minimum amount of heat exhausted to the low-temperature
reservoir is 5000 J ×0.5 = 2500 J in each cycle.
Question 35
Question
A Carnot engine operates between two reservoirs at temperatures Thand Tc
where Th= 600 K and Tc= 300 K. The engine absorbs 2500 J of heat from the
high-temperature reservoir in each cycle. Calculate the efficiency of the engine
and the heat rejected to the low-temperature reservoir in each cycle.
Solution
Step 1: Calculate the efficiency of the engine using the formula for Carnot
efficiency, which is given by
η= 1 −Tc
Th
where Tcis the absolute temperature of the cold reservoir and This the absolute
temperature of the hot reservoir.
Step 2: Substitute the given values Tc= 300 K and Th= 600 K into the
formula to calculate the efficiency.
η= 1 −300
600 = 1 −1
2=1
2= 0.5
Step 3: Therefore, the efficiency of the engine is 50%.
Step 4: Calculate the heat rejected to the low-temperature reservoir in each
cycle using the formula
Qc=ηQh
where Qcis the heat rejected to the low-temperature reservoir, ηis the efficiency
of the engine, and Qhis the heat absorbed from the high-temperature reservoir.
30
Step 5: Substitute the given values η= 0.5 and Qh= 2500 J into the formula
to calculate the heat rejected to the low-temperature reservoir.
Qc= 0.5×2500 = 1250 J
Step 6: Therefore, the heat rejected to the low-temperature reservoir in each
cycle is 1250 J.
31
