PHYS 305 - INTRODUCTION TO MODERN PHYSICS Heat engines, efficiency, and the second law of thermodynamics Question Bank Set 5 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Heat engines,

efficiency, and the second law of

thermodynamics

Question Bank - Set 5

Liberty University

Question 1

Question

A heat engine operates between a hot reservoir at a temperature of 600 K and

a cold reservoir at a temperature of 300 K. The engine absorbs 5,000 J of heat

from the hot reservoir in each cycle and exhausts 2,500 J to the cold reservoir

in each cycle. Calculate the efficiency of this heat engine.

Solution

Step 1: Calculate the net work done by the engine. Given that the engine

absorbs Qhot = 5,000 J of heat and exhausts Qcold = 2,500 J, the net work

done by the engine is:

W=Qhot −Qcold = 5,000 J −2,500 J = 2,500 J

Step 2: Calculate the efficiency of the engine. The efficiency of a heat engine

is given by the formula:

Efficiency = Net work done

Absorbed heat =W

Qhot

Substitute the values of Wand Qhot into the formula:

Efficiency = 2,500 J

5,000 J = 0.5

Therefore, the efficiency of this heat engine is 50

Question 2

Question

A heat engine operates between a hot reservoir at 500 K and a cold reservoir

at 300 K. The engine absorbs 2000 J of heat from the hot reservoir during

each cycle and expels 1200 J of heat to the cold reservoir during each cycle.

Calculate the efficiency of the engine. Is this engine violating the second law of

thermodynamics?

Solution

Step 1: Recall that the efficiency of a heat engine is given by the formula:

Efficiency = 1 −Heat expelled

Heat absorbed

Step 2: Substitute the values given in the question:

Efficiency = 1 −1200

2000

Step 3: Perform the calculation:

Efficiency = 1 −0.6 = 0.4

Step 4: Therefore, the efficiency of the engine is 40

Step 5: The engine is not violating the second law of thermodynamics be-

cause the efficiency is less than 100

Question 3

Question

A Carnot heat engine operates between two heat reservoirs at temperatures Th

and Tcwith Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir

and exhausts 2500 J to the cold reservoir per cycle. Calculate the efficiency of

the engine and determine the amount of work done per cycle.

Solution

Step 1: Calculate the efficiency of the Carnot heat engine using the formula:

Efficiency = 1 −Tc

Step 2: Substitute the given temperatures Thand Tc:

Efficiency = 1 −Tc

= 1 −2500

4000 = 1 −0.625 = 0.375

Step 3: The efficiency of the engine is 0.375 or 37.5

Step 4: Calculate the amount of work done per cycle using the formula:

Work done = Heat absorbed −Heat rejected

Step 5: Substitute the given values of heat absorbed and heat rejected:

Work done = 4000 J −2500 J = 1500 J

Step 6: The amount of work done per cycle by the engine is 1500 J.

Question 4

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 in one cycle, determine: (a) The efficiency of

the engine (b) The maximum work output of the engine if it operates between

reservoirs at temperatures 500 K and 300 K.

Solution

(a) To determine the efficiency of the engine, we first need to find the amount

of work done by the engine in one cycle.

Step 1: Find the net work done by the engine in one cycle.

The net work done by the engine in one cycle can be calculated using the

First Law of Thermodynamics:

Net Work = Heat Input −Heat Output

Given that the engine absorbs 6000 J of heat from the hot reservoir and

expels 4000 J of heat to the cold reservoir, the net work done by the engine is:

Net Work = 6000 J −4000 J = 2000 J

Step 2: Find the efficiency of the engine.

The efficiency of the engine is given by the formula:

Efficiency = Net Work

Heat Input

Substitute the known values:

Efficiency = 2000 J

6000 J =1

3= 33.3%

Therefore, the efficiency of the engine is 33.3

(b) To find the maximum work output of the engine, we can use Carnot’s

theorem which states that the maximum efficiency of a heat engine is given by:

Efficiencymax = 1 −Tc

where Tcand Thare the temperatures of the cold and hot reservoirs, respec-

tively.

Step 3: Calculate the maximum work output.

Given that the temperatures of the cold and hot reservoirs are 300 K and

500 K respectively, substitute the values into the formula:

Efficiencymax = 1 −300

500 = 1 −3

5=2

5= 0.4 = 40%

The maximum work output is equal to the efficiency multiplied by the heat

input:

Maximum Work Output = Efficiencymax ×Heat Input

Maximum Work Output = 0.4×6000 J = 2400 J

Therefore, the maximum work output of the engine is 2400 J.

Question 5

Question

A heat engine operates between a hot reservoir at 127◦C and a cold reservoir

at 27◦C. The engine produces 2000 J of work while absorbing 5000 J of heat

from the hot reservoir. Determine the efficiency of the engine. Is this engine

operating in violation of the second law of thermodynamics?

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = Useful work output

Heat input

Step 2: Calculate the heat input to the engine: Since the engine absorbs

5000 J of heat from the hot reservoir, the heat input is 5000 J.

Step 3: Calculate the useful work output of the engine, which is given as

2000 J.

Step 4: Substitute the values into the efficiency formula:

Efficiency = 2000

5000 = 0.4 = 40%

Step 5: No, this engine is not operating in violation of the second law of

thermodynamics. The efficiency is less than 100

Question 6

Question

A heat engine operates between two reservoirs at temperatures Th= 600 K

and Tc= 300 K. The engine absorbs 1000 J of heat from the hot reservoir in

each cycle and expels 600 J to the cold reservoir. Calculate the efficiency of the

engine and discuss if it violates the second law of thermodynamics.

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = 1 −Tc

where Tcis the temperature of the cold reservoir and This the temperature of

the hot reservoir.

Step 1: Calculate the efficiency of the engine.

Efficiency = 1 −300

600 = 1 −0.5=0.5

Step 2: Discuss whether the engine violates the second law of thermodynam-

ics. According to the second law of thermodynamics, the maximum efficiency

of a heat engine operating between two reservoirs at temperatures Thand Tcis

given by:

Maximum Efficiency = 1 −Tc

In this case, the maximum efficiency would be:

Max Efficiency = 1 −300

600 = 1 −0.5=0.5

Since the actual efficiency of the engine is equal to the maximum efficiency, the

engine does not violate the second law of thermodynamics.

Question 7

Question

A heat engine operates between a hot reservoir at 600 K and a cold reservoir at

300 K. The engine absorbs 1000 J of heat from the hot reservoir in each cycle.

Determine: a) The efficiency of this heat engine. b) The maximum amount of

work that can be done by this engine in each cycle. c) Verify the second law of

thermodynamics for this engine.

Solution

a) The efficiency of a heat engine is given by the formula:

Efficiency = 1 −Tc

where Tc= temperature of the cold reservoir = 300 K, Th= temperature of the

hot reservoir = 600 K.

Step 1: Substitute the values into the formula:

Efficiency = 1 −300

600 = 1 −1

2=1

2= 50%

b) The maximum amount of work that can be done by the engine in each

cycle is given by the formula:

Work = Heat absorbed ×Efficiency

Given that the engine absorbs 1000 J of heat in each cycle, and the efficiency is

50%, we have:

Work = 1000 ×1

2= 500 J

c) The second law of thermodynamics states that no heat engine can be 100%

efficient, meaning that some heat must always be lost to the surroundings. If

we assume that the efficiency of the engine is 100%, then all the heat absorbed

would be converted into work, and the heat rejected to the cold reservoir would

be 0 J. However, in reality, some heat must be rejected to the cold reservoir.

In this case, the engine is only 50% efficient, so it can only convert half of

the absorbed heat into work. The remaining 500 J of heat must be rejected to

the cold reservoir. This verifies the second law of thermodynamics.

Question 8

Question

A Carnot engine operates between two reservoirs at temperatures THand TC,

where TH> TC. If the engine absorbs 4000 J of heat from the hot reservoir and

has an efficiency of 50

1. The work done by the engine.

2. The heat rejected to the cold reservoir.

Solution

We can start by recalling the formula for the efficiency of a Carnot engine:

Efficiency = 1 −TC

Step 1: Find the work done by the engine. Given that the efficiency

is 50

0.5=1−TC

Solving for TC, we get:

TC= 0.5TH

The work done by the engine can be calculated using:

Work = Efficiency ×Input Heat

Substitute the values and solve for the work:

Work = 0.5×4000

Work = 2000 J

Therefore, the work done by the engine is 2000 J.

Step 2: Find the heat rejected to the cold reservoir. The heat

rejected to the cold reservoir is the difference between the heat absorbed from

the hot reservoir and the work done by the engine:

Heat Rejected = Heat Absorbed −Work

Heat Rejected = 4000 −2000

Heat Rejected = 2000 J

Thus, the heat rejected to the cold reservoir is 2000 J.

Question 9

Question

A heat engine operates between two reservoirs at temperatures Th= 600 K and

Tc= 300 K. The engine takes in 500 J of heat from the hot reservoir and expels

300 J to the cold reservoir in each cycle. Determine:

1. The efficiency of the engine.

2. The maximum possible efficiency of a heat engine operating between these

two temperatures.

Solution

1. To find the efficiency of the engine, we can use the formula:

Efficiency = Useful work done

Input heat

We are given that the engine takes in 500 J of heat from the hot reservoir

and expels 300 J to the cold reservoir in each cycle. Therefore, the useful work

done is the difference between these two values, which is 500 J - 300 J = 200 J.

So, the efficiency of the engine is:

Efficiency = 200 J

500 J = 0.4 = 40%

2. The maximum possible efficiency of a heat engine operating between two

reservoirs at temperatures Thand Tcis given by the Carnot efficiency:

EfficiencyCarnot = 1 −Tc

Substitute the given temperatures into the formula:

EfficiencyCarnot = 1 −300 K

600 K = 1 −0.5 = 0.5 = 50%

Therefore, the maximum possible efficiency of a heat engine operating be-

tween these two temperatures is 50

Question 10

Question

A heat engine operates between two reservoirs at temperatures Thot and Tcold.

The engine absorbs 6000 J of heat from the hot reservoir and produces 2000 J

of work. The cold reservoir absorbs the remaining heat rejected by the engine.

Calculate the efficiency of the engine and determine the minimum amount of

heat rejected to the cold reservoir in this process.

Solution

Step 1: Calculate the efficiency of the engine using the formula for efficiency:

Efficiency = Useful work output

Heat input

Given that the engine absorbs 6000 J of heat and produces 2000 J of work,

we have

Efficiency = 2000

6000 =1

3= 33.33%

Step 2: Determine the heat rejected to the cold reservoir using the first law

of thermodynamics, which states that the net heat input to a system equals

the net work done by the system plus the net increase in internal energy of the

system.

Since the engine absorbs 6000 J of heat and produces 2000 J of work, the

net heat input is 6000 J - 2000 J = 4000 J.

Therefore, the heat rejected to the cold reservoir is 4000 J.

Question 11

Question

A heat engine operates between two reservoirs at temperatures Thand Tc, where

Th> Tc. If the heat engine has an efficiency of 40(a) The expression for the

efficiency of this engine in terms of these temperatures. (b) The maximum

efficiency possible for this engine. (c) If the engine actually operates at its

maximum efficiency, what is the relationship between the heat absorbed at

the high temperature reservoir and the heat rejected at the low temperature

reservoir?

Solution

(a) Let us denote the efficiency of the engine as η. The efficiency of a heat

engine can be defined as the ratio of work output to the heat input:

η=W

where Wis the work output and Qhis the heat absorbed from the high-

temperature reservoir.

We also know that the efficiency of a Carnot engine (the most efficient type

of heat engine) is given by η= 1 −Tc

Th.

Setting η=W

Qh= 1 −Tc

Th= 0.40, we can solve for W/Qhin terms of Tcand

Th.

η= 1 −Tc

= 0.40

(b) The maximum efficiency for any heat engine operating between two

reservoirs at temperatures Thand Tcis given by the Carnot efficiency:

ηCarnot = 1 −Tc

Here, This the temperature of the hot reservoir and Tcis the temperature of

the cold reservoir. Since Th> Tc, the maximum efficiency occurs when Tc= 0.

Therefore, ηmax = 1 −0 = 1.

Thus, the maximum efficiency possible for this engine is 1 or 100

ship between the heat absorbed at the high temperature reservoir Qhand the

heat rejected at the low temperature reservoir Qccan be determined using the

equation for efficiency:

η=W

= 1 −Tc

Since the engine is operating at its maximum efficiency, we have η= 1. Substi-

tuting this into the equation above gives:

1 = W

⇒W=Qh

This relationship tells us that when the engine operates at maximum efficiency,

the work output Wis equal to the heat absorbed Qh.

Thus, W=Qh.

Question 12

Question

A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir

at 300 K. The engine takes in 1200 J of heat from the hot reservoir in each cycle.

Calculate the efficiency of the engine.

Solution

Step 1: Calculate the efficiency of a Carnot engine using the formula:

Efficiency = 1 −Tc

where This the absolute temperature of the hot reservoir and Tcis the absolute

temperature of the cold reservoir.

Step 2: Convert the temperatures to absolute scale: Th= 800 K and Tc=

300 K.

Step 3: Substitute the values into the efficiency formula: Efficiency = 1−300

800

Step 4: Simplify the expression: Efficiency = 1 −3

8=5

Step 5: Therefore, the efficiency of the Carnot engine is 5

8or 62.5%.

Question 13

Question

A heat engine operating between two reservoirs produces 2000 J of work while

absorbing 4000 J of heat from the high-temperature reservoir. Calculate the

efficiency of the heat engine and determine the amount of heat rejected to the

low-temperature reservoir.

Solution

Let QHbe the heat absorbed from the high-temperature reservoir, QLbe the

heat rejected to the low-temperature reservoir, and Wbe the work done by the

engine.

Step 1: Recall the efficiency of a heat engine is given by the formula

Efficiency (ε) = Work done (W)

Heat absorbed (QH)

Given W= 2000 J and QH= 4000 J, we can plug these values in to find the

efficiency.

ε=2000

4000 = 0.5

Therefore, the efficiency of the heat engine is 0.5 or 50%.

Step 2: To find the heat rejected to the low-temperature reservoir, we use

the fact that the net work done by the engine is the difference between the

heat absorbed from the high-temperature reservoir and the heat rejected to the

low-temperature reservoir:

W=QH−QL

Rearranging the equation, we have

QL=QH−W= 4000 −2000 = 2000 J

Hence, the amount of heat rejected to the low-temperature reservoir is 2000

Question 14

Question

A heat engine operates between two reservoirs at temperatures Thand Tc, where

Th> Tc. The engine absorbs heat Qhfrom the hot reservoir and exhausts heat

Qcto the cold reservoir. If the efficiency of the engine is denoted by η(where

η=W

Qhand Wis the work done by the engine), show that the efficiency of the

engine is given by η= 1 −Tc

Th.

Solution

Step 1: The work done by the engine is given by the difference between the heat

absorbed and the heat exhausted:

W=Qh−Qc

Step 2: The efficiency of the engine is defined as the ratio of the work done

by the engine to the heat absorbed from the hot reservoir:

η=W

Step 3: Substituting W=Qh−Qcinto the equation for efficiency:

η=Qh−Qc

Step 4: Rearranging the terms:

η= 1 −Qc

Step 5: Using the First Law of Thermodynamics (Qh=W+Qc) and rear-

ranging for Qc:

Qc=Qh−W

Step 6: Substituting the expression for Qcinto the efficiency equation:

η= 1 −Qh−W

Step 7: Since W=Qh−Qc, we can write W=Qh−(Qh−W):

W=Qh−Qh+W

W=W

Step 8: Substituting back into the equation for efficiency:

η= 1 −Qh−(Qh−Qc)

Step 9: Simplifying further:

η= 1 −Qc

Step 10: Finally, using the definition of thermal efficiency (η= 1 −Qc

Qh) and

Qc=Tc∆S, where ∆Sis the change in entropy of the cold reservoir, we have:

η= 1 −Tc∆S

Step 11: Recall that for a reversible process, ∆S=Q

T. Therefore, ∆S=Qc

Tc.

Substituting this into the efficiency equation:

η= 1 −Tc·Qc

η= 1 −Qc

Step 12: Since Qh=Qc+W, we have Qc=Qh−W.

Therefore, η= 1 −Qh−W

Qhand simplifying further gives:

η= 1 −Tc

Hence, the efficiency of the engine is given by η= 1 −Tc

Th.

Question 15

Question

A Carnot engine operates between a high-temperature reservoir at 500 K and a

low-temperature reservoir at 300 K. If the engine absorbs 500 J of heat from the

high-temperature reservoir in each cycle, calculate the efficiency of the engine.

Solution

Step 1: Determine the efficiency of the Carnot engine using the formula for

efficiency:

Efficiency = 1 −TC

where THis the absolute temperature of the high-temperature reservoir and TC

is the absolute temperature of the low-temperature reservoir.

Given that TH= 500 K and TC= 300 K, we can calculate the efficiency:

Efficiency = 1 −300

500 = 1 −0.6=0.4

Therefore, the efficiency of the Carnot engine is 0.4 or 40%.

Step 2: Calculate the work done by the engine in each cycle using the for-

mula:

Work done = Efficiency ×Heat absorbed from high-temperature reservoir

Given that the engine absorbs 500 J of heat, and the efficiency is 0.4, the

work done can be calculated as:

Work done = 0.4×500 = 200 J

Therefore, the work done by the Carnot engine in each cycle is 200 J.

Question 16

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 expels Qc

of heat to the cold reservoir. If the efficiency of the engine is η, prove that the

efficiency ηis given by

η= 1 −Tc

Solution

Step 1: The efficiency, η, of a heat engine is defined as the ratio of the work

done by the engine to the heat energy absorbed from the hot reservoir:

η=Work output

Heat input from hot reservoir.

Step 2: By the first law of thermodynamics, the work output by the engine is

equal to the difference between the heat energy absorbed from the hot reservoir,

Qh, and the heat energy expelled to the cold reservoir, Qc:

Work output = Qh−Qc.

Step 3: Therefore, the efficiency can be written as

η=Qh−Qc

Step 4: Using the definition of efficiency, we know that

η= 1 −Qc

Step 5: From the second law of thermodynamics, we have that for a heat

engine operating between two reservoirs at temperatures Thand Tc:

−Qh

≤0.

Step 6: Rearranging terms, we get

≤Tc

Step 7: Substituting this inequality back into our expression for efficiency,

we get

η= 1 −Qc

≥1−Tc

Step 8: Therefore, we have proven that the efficiency ηof a heat engine

operating between two reservoirs at temperatures Thand Tcis given by

η= 1 −Tc

Question 17

Question

A heat engine operates between two reservoirs at temperatures Thand Tc, where

Th> Tc. The engine absorbs Qhof heat from the reservoir at temperature Th

and exhausts Qcof heat to the reservoir at temperature Tc. Given that the

efficiency of this heat engine is η, express ηin terms of Qhand Qc.

Solution

Step 1: Recall that the efficiency of a heat engine is defined as the ratio of the

work done by the engine to the heat input:

η=Work output

Heat input

Step 2: The work output of the engine can be expressed as the difference

between the heat input and the heat rejected:

Work output = Qh−Qc

Step 3: Substitute the expression for work output into the efficiency equa-

tion:

η=Qh−Qc

Step 4: Simplify the expression by dividing out Qh:

η= 1 −Qc

Therefore, the efficiency of the heat engine in terms of Qhand Qcis η=

1−Qc

Qh.

Question 18

Question

A Carnot heat engine operates between two reservoirs at temperatures THand

TC(where TH> TC). If the engine absorbs 5000 J of heat from the hot reservoir

and has an efficiency of 40

1. The net work done by the engine.

2. The temperature of the cold reservoir if the temperature of the hot reser-

voir is 500 K.

Solution

1. The efficiency of a Carnot heat engine is given by the formula:

Efficiency = 1 −TC

Given that the efficiency is 40

0.40 = 1 −TC

500

Solving for TC, we get:

TC= 300 K

The net work done by the engine can be calculated using the formula for effi-

ciency:

Efficiency = Net work done

Heat absorbed from hot reservoir

Plugging in the values, we have:

0.40 = Net work done

5000

Solving for the net work done by the engine:

Net work done = 0.40 ×5000 = 2000 J

2. The temperature of the cold reservoir is already determined to be 300 K

from the efficiency calculation.

Question 19

Question

A heat engine operates between a hot reservoir at 600 K and a cold reservoir

at 300 K. If the engine produces 4000 J of work in each cycle, determine the

efficiency of the engine. Also, discuss whether this engine violates the second

law of thermodynamics.

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = 1 −Tc

where Tcis the temperature of the cold reservoir and This the temperature of

the hot reservoir.

Step 1: Plug in the given values:

Efficiency = 1 −300

600 = 1 −1

2=1

2= 0.5

Step 2: Verify the efficiency obtained in Step 1.

Based on the second law of thermodynamics, no heat engine can be more

efficient than a Carnot engine operating between the same two reservoirs. The

efficiency of a Carnot engine is given by:

EfficiencyCarnot = 1 −Tc

Step 2: Calculate the efficiency of the Carnot engine using the given tem-

peratures:

EfficiencyCarnot = 1 −300

600 =1

2= 0.5

Since the efficiency of the actual engine is equal to the efficiency of the

Carnot engine, it does not violate the second law of thermodynamics.

Question 20

Question

A heat engine operates between a cold reservoir at Tc= 300 K and a hot reservoir

at Th= 800 K. The engine absorbs 1200 J of heat from the hot reservoir and

performs 800 J of work during each cycle. Calculate the efficiency of the engine.

Solution

Step 1: Calculate the efficiency of the engine using the formula for efficiency:

Efficiency = 1 −Tc

Step 2: Substitute the given values Tc= 300 K and Th= 800 K into the

formula:

Efficiency = 1 −300

800

Step 3: Simplify the expression:

Efficiency = 1 −3

8=5

Step 4: Finally, convert the efficiency to a percentage:

Efficiency = 5

8×100% = 62.5%

Therefore, the efficiency of the engine is 62.5

Question 21

Question

A heat engine operates between two reservoirs at temperatures Thand Tc(with

Th> Tc). The heat engine absorbs energy Qhfrom the hot reservoir and

exhausts energy Qcto the cold reservoir. If the efficiency of this engine is η, show

that the efficiency of the engine is limited by the second law of thermodynamics,

which states that η≤1−Tc

Th.

Solution

Let’s start by expressing the efficiency of the heat engine in terms of Qhand

Qc:

η=Useful work done

Input heat energy

Step 1: The useful work done by the engine can be expressed as:

Useful work done = Qh−Qc

Step 2: Now we know that for any heat engine, we have:

Input heat energy = Qh

Step 3: Substitute the expressions for useful work done and input heat

energy into the efficiency formula:

η=Qh−Qc

Step 4: Simplify the expression:

η= 1 −Qc

Step 5: From the first law of thermodynamics, we know that for any heat

engine:

Qh=Qc+ Useful work done

Step 6: Rearranging the equation, we get:

Qc=Qh−Useful work done

Step 7: Substitute this expression for Qcback into the efficiency formula:

η= 1 −Qh−Useful work done

Step 8: Simplify further:

η= 1 −Qh−(Qh−Qc)

Step 9:

η= 1 −Qc

Step 10: Recall that Qc=TcScand Qh=ThSh, where Scand Share the

entropies of the cold and hot reservoirs, respectively.

Step 11: Substitute Qcand Qhin terms of temperatures:

η= 1 −TcSc

ThSh

Step 12: Since entropy is always positive, we know that Sc≥0 and Sh≥0.

Hence, Sc/Sh≤1.

Step 13: Therefore, we can conclude:

η≤1−Tc

This inequality proves that the efficiency of the engine is limited by the

second law of thermodynamics.

Question 22

Question

A Carnot engine operates between two heat reservoirs at temperatures 420 K

and 220 K. The engine absorbs 2000 J of heat from the high-temperature reser-

voir in each cycle. Calculate the efficiency of the engine.

Solution

Step 1: Calculate the efficiency of the Carnot engine using the formula for

efficiency:

Efficiency = 1 −Tc

where Tcis the absolute temperature of the cold reservoir and This the absolute

temperature of the hot reservoir.

Step 2: Convert the temperatures from Celsius to Kelvin by adding 273 to

each temperature.

Tc= 220 K, Th= 420 K

Step 3: Substitute the values into the formula to find the efficiency of the

engine.

Efficiency = 1 −220

420 = 1 −11

21 =10

21 ≈0.476

Step 4: The efficiency of the Carnot engine is approximately 0.476 or 47.6

Question 23

Question

A heat engine operates between a source temperature of 500 K and a sink tem-

perature of 300 K. If the engine absorbs 6000 J of heat energy from the source

in each cycle, calculate the efficiency of the engine. Is this engine violating the

second law of thermodynamics?

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = 1 −Heat rejected

Heat absorbed

Step 2: Calculate the heat rejected by the engine using the formula:

Heat rejected = Heat absorbed −Work done

Step 3: Calculate the work done by the engine using the formula:

Work done = Heat absorbed −Heat rejected

Step 4: Calculate the heat rejected by the engine using the formula:

Heat rejected = Qout =Qin ×Tout

Tin

Step 5: Calculate the efficiency of the engine:

Efficiency = 1 −Heat rejected

Heat absorbed

Step 6: Determine if the engine is violating the second law of thermodynam-

ics by checking if the efficiency is less than the Carnot efficiency:

Efficiency ≤Carnot efficiency = 1 −Tsink

Tsource

Question 24

Question

A heat engine operates between two reservoirs at temperatures THand TC. The

engine absorbs QHof heat from the hot reservoir and rejects QCof heat to the

cold reservoir. If the efficiency of the engine is η, prove that the efficiency of

a reversible engine operating between the same reservoirs is greater than the

efficiency of the actual engine.

Solution

Given: - Heat absorbed from the hot reservoir, QH- Heat rejected to the cold

reservoir, QC- Efficiency of the actual engine, η- Temperatures of the hot and

cold reservoirs, THand TC

Efficiency of the actual engine:

η= 1 −QC

Let Q′

Hbe the heat absorbed by the reversible engine operating between the

same reservoirs at temperatures THand TC, and Q′

Cbe the heat rejected by

the reversible engine.

Step 1: Calculate the efficiency of the reversible engine The efficiency η′of

a reversible engine is given by Carnot’s principle:

η′= 1 −TC

Step 2: Prove that η′> η We know that η′= 1 −TC

TH>1−QC

QH=η.

Therefore, the efficiency of a reversible engine operating between the same reser-

voirs is greater than the efficiency of the actual engine.

Question 25

Question

A heat engine operates between a hot reservoir at 500 K and a cold reservoir at

300 K. If the engine absorbs 800 J of heat from the hot reservoir in each cycle,

find:

1. The efficiency of the engine.

2. The maximum possible work output of the engine.

Assume the engine operates in a Carnot cycle.

Solution

1. To find the efficiency of the engine, we can use the formula:

Efficiency = 1 −Tcold

Thot

where Thot is the temperature of the hot reservoir (500 K) and Tcold is the

temperature of the cold reservoir (300 K).

Step 1: Calculate the efficiency.

Efficiency = 1 −300

500 = 1 −0.6 = 0.4 = 40%

2. The maximum possible work output of the engine can be calculated using

the formula for efficiency of a heat engine:

Efficiency = Wout

Qin

where Wout is the work output and Qin is the heat absorbed from the hot

reservoir.

Step 2: Rearrange the formula to find the work output.

Wout = Efficiency ×Qin

Step 3: Substitute the values to calculate the maximum work output.

Wout = 0.4×800 J = 320 J

Therefore, the efficiency of the engine is 40

Question 26

Question

A heat engine operates between two reservoirs at temperatures T1and T2, where

T1> T2. The engine absorbs Q1of heat from the reservoir at temperature T1

and exhausts Q2of heat to the reservoir at temperature T2. Prove that the

efficiency of the engine is given by

η= 1 −T2

= 1 −Q2

Solution

Step 1: The efficiency of a heat engine is defined as the ratio of the useful work

output to the heat input. Mathematically, efficiency is defined as

η=W

where Wis the work done by the engine and Q1is the heat absorbed from the

high-temperature reservoir.

Step 2: The work done by the engine can be determined using the first law

of thermodynamics, which states that the net work output of the engine is equal

to the difference between the heat absorbed from the hot reservoir and the heat

rejected to the cold reservoir, i.e.,

W=Q1−Q2

Step 3: Using the definition of efficiency in Step 1 and the expression for

work done in Step 2, we have

η=Q1−Q2

= 1 −Q2

Step 4: To express the efficiency in terms of temperatures, we can use the

fact that for a reversible heat engine, the heat input and output are related to

the temperatures of the reservoirs by

=Q2

Step 5: Rearranging the above equation, we get

Q2=T2

Step 6: Substituting equation (5) into equation (3), we get

η= 1 −T2

Therefore, the efficiency of the engine is given by η= 1 −T2

T1.

Question 27

Question

A heat engine operates between two reservoirs at temperatures Thand Tc(with

Th> Tc). The engine absorbs Qhamount of heat from the hot reservoir and

produces Wamount of work. Calculate the efficiency of the engine in terms of

Qhand Tc.

Solution

Step 1: Recall the efficiency of a heat engine is given by the formula:

Efficiency = Useful work output

Heat input =W

Step 2: The heat input to the engine is Qhand the heat rejected to the cold

reservoir is Qc. By the first law of thermodynamics, we have:

Qh=W+Qc

Step 3: Since the engine is operating in a cycle, we can apply the second

law of thermodynamics which states that the efficiency of any heat engine is

bounded by the Carnot efficiency:

Efficiency ≤1−Tc

Step 4: Substituting Qh=W+Qcinto the efficiency formula, we get:

Efficiency = W

W+Qc

Step 5: From the second law of thermodynamics, we know that Qc=Qh−W.

Substitute Qc=Qh−Winto the efficiency formula:

Efficiency = W

W+Qh−W

Step 6: Simplifying the expression gives the efficiency in terms of Qhand

Tc:

Efficiency = W

= 1 −Tc

So, the efficiency of the engine in terms of Qhand Tcis 1 −Tc

Th.

Question 28

Question

A heat engine operates between two reservoirs at temperatures Th= 600 K

and Tc= 300 K. The engine extracts 2000 J of heat from the high-temperature

reservoir in each cycle and delivers 1000 J of work to the surroundings in each

cycle.

Calculate the efficiency of the heat engine and determine if it violates the

second law of thermodynamics.

Solution

Step 1: Calculate the efficiency of the heat engine using the formula for efficiency,

η=W

Qh.

Efficiency = W

Efficiency = 1000 J

2000 J

Efficiency = 0.5

Step 2: The efficiency of the engine is found to be 0.5. The efficiency of

a heat engine is given by η= 1 −Tc

Th, where This the temperature of the hot

reservoir and Tcis the temperature of the cold reservoir. Substitute the given

temperatures into the formula to find the maximum efficiency of this engine.

Step 3: Calculate the theoretical maximum efficiency using the formula η=

1−Tc

Th.

Efficiency = 1 −300 K

600 K

Efficiency = 1 −0.5

Efficiency = 0.5

Step 4: Comparing the calculated efficiency (0.5) with the theoretical max-

imum efficiency (0.5), we find that the efficiency of the engine is equal to the

maximum theoretical efficiency. Therefore, the engine does not violate the sec-

ond law of thermodynamics.

Question 29

Question

A Carnot engine operates between two heat reservoirs at temperatures of 500

K and 300 K. The engine absorbs 2500 J of heat from the high-temperature

reservoir in each cycle. Calculate the efficiency of the engine and determine the

amount of heat rejected to the low-temperature reservoir in each cycle.

Solution

Step 1: Calculate the efficiency of the Carnot engine using the formula:

Efficiency = 1 −Tlow

Thigh

where Tlow and Thigh are the temperatures of the low and high heat reservoirs,

respectively.

Given Tlow = 300 K and Thigh = 500 K, we have:

Efficiency = 1 −300

500 = 1 −0.6 = 0.4 = 40%

Step 2: Calculate the amount of heat rejected to the low-temperature reser-

voir in each cycle using the formula:

Heat rejected = Efficiency ×Heat absorbed from high-temperature reservoir

Given that the engine absorbs 2500 J of heat from the high-temperature

reservoir in each cycle, we substitute the efficiency and heat absorbed into the

formula:

Heat rejected = 0.4×2500 J = 1000 J

Therefore, the efficiency of the engine is 40

Question 30

Question

A heat engine operates between a hot reservoir at a temperature of 600 K and a

cold reservoir at a temperature of 300 K. If the engine produces 5000 J of work

output per cycle, determine the efficiency of the engine. Also, discuss how the

efficiency of the engine relates to the second law of thermodynamics.

Solution

Step 1: Calculate the efficiency of the heat engine using the formula for effi-

ciency:

Efficiency = 1 −Tc

where Efficiency is the ratio of work output to heat input Tcis the absolute

temperature of the cold reservoir This the absolute temperature of the hot

reservoir

Step 2: Substitute the given values into the formula: Given, Th= 600Kand

Tc= 300K

Efficiency = 1 −300

600 = 1 −0.5=0.5

So, the efficiency of the heat engine is 50

Step 3: Discussion on the second law of thermodynamics: The second law

of thermodynamics states that heat will naturally flow from a hot reservoir to a

cold reservoir without the need for any work. This law implies that not all the

energy input to a heat engine can be converted into work output. Some of the

energy must be dissipated as waste heat to the cold reservoir. This is captured

by the efficiency formula, where a fraction of the heat input is not converted

into work but instead rejected to the cold reservoir.

In this case, the heat engine’s efficiency of 50

Question 31

Question

A heat engine operates between two reservoirs at temperatures Th= 600 K and

Tc= 300 K. The engine absorbs 600 J of heat from the hot reservoir during

each cycle and exhausts 400 J of heat to the cold reservoir during each cycle.

Calculate the efficiency of this heat engine.

Solution

Step 1: Calculate the efficiency of the heat engine using the formula:

Efficiency = 1 −Tc

Step 2: Substitute the given temperatures into the formula:

Efficiency = 1 −300

600

Step 3: Perform the calculation to find the efficiency:

Efficiency = 1 −1

2=1

2= 0.5

Step 4: The efficiency of the heat engine is 0.5 or 50

Question 32

Question

A heat engine operates between two reservoirs at temperatures Thand Tc(with

Th> Tc). If the efficiency of the engine is 30

Solution

Let’s denote the efficiency of the given heat engine as η, the maximum possible

efficiency as ηmax,Qhas the heat input from the hot reservoir, and Qcas the

heat output to the cold reservoir. According to the Carnot efficiency:

η= 1 −Tc

Given that η= 0.30, we have:

0.30 = 1 −Tc

= 0.70 ⇒Tc= 0.70Th

The efficiency of a Carnot engine operating between the same two tempera-

tures is:

ηmax = 1 −Tc

= 1 −0.70 = 0.30 = 30%

Therefore, the maximum possible efficiency of a heat engine operating be-

tween the same two reservoirs is 30

Question 33

Question

A heat engine operates between a high-temperature reservoir at 500◦Cand a

low-temperature reservoir at 25◦C. The engine absorbs 2000 J of heat from

the high-temperature reservoir in each cycle. Calculate the maximum possible

efficiency of the engine. Is this efficiency feasible based on the second law of

thermodynamics?

Solution

Step 1: Convert the temperatures to Kelvin: Given that Thot = 500◦C=

500 + 273 = 773 K and Tcold = 25◦C= 25 + 273 = 298 K.

Step 2: Calculate the maximum efficiency using Carnot’s efficiency formula:

The maximum efficiency of a heat engine operating between two temperatures

Thot and Tcold is given by the formula:

Efficiency = 1 −Tcold

Thot

Substitute the temperatures to find the maximum efficiency:

Efficiency = 1 −298

773

Efficiency ≈0.6144 or 61.44%

Step 3: Analyze the feasibility based on the second law of thermodynamics:

According to the second law of thermodynamics, no heat engine can have an

efficiency greater than that of a Carnot engine operating between the same two

temperatures. The actual efficiency of a real heat engine will always be less than

the Carnot efficiency. Therefore, the efficiency calculated in this case (61.44

Question 34

Question

A heat engine operates between two reservoirs at temperatures T1= 600 K and

T2= 300 K. The engine absorbs 600 J of heat from the hot reservoir for every

300 J it exhausts to the cold reservoir. Determine the efficiency of the engine.

Solution

Let us denote the heat absorbed from the hot reservoir as Qhand the heat

exhausted to the cold reservoir as Qc. The efficiency of the engine is given by

the formula:

Efficiency = 1 −Qc

Step 1: Determine the ratio of Qcto Qh. Given that the engine absorbs 600

J of heat from the hot reservoir for every 300 J it exhausts to the cold reservoir,

we have: Qc

=300

600 =1

Step 2: Calculate the efficiency of the engine. Using the formula for effi-

ciency:

Efficiency = 1 −Qc

= 1 −1

2=1

Therefore, the efficiency of the engine is 1

2or 50

Question 35

Question

A heat engine operates between two heat reservoirs at temperatures Thot and

Tcold. If the engine absorbs 5000 J of heat from the hot reservoir per cycle and

expels 3000 J of heat to the cold reservoir per cycle, calculate the efficiency

of the engine in terms of the reservoir temperatures Thot and Tcold. Using the

calculated efficiency, determine if the engine violates the second law of thermo-

dynamics.

Solution

Step 1: Recall the definition of efficiency for a heat engine:

Efficiency = Work output

Heat input

Step 2: Let’s first calculate the work output of the engine. According to the

first law of thermodynamics (conservation of energy), the work output is the

difference between the heat input and the heat output:

Work output = Heat input −Heat output

Step 3: Given that the engine absorbs 5000 J of heat from the hot reservoir

and expels 3000 J of heat to the cold reservoir, we can substitute these values

into the equation:

Work output = 5000 J −3000 J = 2000 J

Step 4: Now, we can plug the values of work output and heat input into the

efficiency formula:

Efficiency = 2000 J

5000 J =2

5=40

100 = 40%

Step 5: The efficiency of the engine in terms of the reservoir temperatures

Thot and Tcold is 40

Step 6: According to the second law of thermodynamics, no heat engine can

have an efficiency of 100

Question 2

Question

A heat engine operates between a hot reservoir at 500 K and a cold reservoir

at 300 K. The engine absorbs 2000 J of heat from the hot reservoir during

each cycle and expels 1200 J of heat to the cold reservoir during each cycle.

Calculate the efficiency of the engine. Is this engine violating the second law of

thermodynamics?

Solution

Step 1: Recall that the efficiency of a heat engine is given by the formula:

Efficiency = 1 −Heat expelled

Heat absorbed

Step 2: Substitute the values given in the question:

Efficiency = 1 −1200

2000

Step 3: Perform the calculation:

Efficiency = 1 −0.6 = 0.4

Step 4: Therefore, the efficiency of the engine is 40

Step 5: The engine is not violating the second law of thermodynamics be-

cause the efficiency is less than 100

Question 3

Question

A Carnot heat engine operates between two heat reservoirs at temperatures Th

and Tcwith Th> Tc. The engine absorbs 4000 J of heat from the hot reservoir

and exhausts 2500 J to the cold reservoir per cycle. Calculate the efficiency of

the engine and determine the amount of work done per cycle.

Solution

Step 1: Calculate the efficiency of the Carnot heat engine using the formula:

Efficiency = 1 −Tc

Step 2: Substitute the given temperatures Thand Tc:

Efficiency = 1 −Tc

= 1 −2500

4000 = 1 −0.625 = 0.375

Step 3: The efficiency of the engine is 0.375 or 37.5

Step 4: Calculate the amount of work done per cycle using the formula:

Work done = Heat absorbed −Heat rejected

Step 5: Substitute the given values of heat absorbed and heat rejected:

Work done = 4000 J −2500 J = 1500 J

Step 6: The amount of work done per cycle by the engine is 1500 J.

Question 4

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 in one cycle, determine: (a) The efficiency of

the engine (b) The maximum work output of the engine if it operates between

reservoirs at temperatures 500 K and 300 K.

Solution

(a) To determine the efficiency of the engine, we first need to find the amount

of work done by the engine in one cycle.

Step 1: Find the net work done by the engine in one cycle.

The net work done by the engine in one cycle can be calculated using the

First Law of Thermodynamics:

Net Work = Heat Input −Heat Output

Given that the engine absorbs 6000 J of heat from the hot reservoir and

expels 4000 J of heat to the cold reservoir, the net work done by the engine is:

Net Work = 6000 J −4000 J = 2000 J

Step 2: Find the efficiency of the engine.

The efficiency of the engine is given by the formula:

Efficiency = Net Work

Heat Input

Substitute the known values:

Efficiency = 2000 J

6000 J =1

3= 33.3%

Therefore, the efficiency of the engine is 33.3

(b) To find the maximum work output of the engine, we can use Carnot’s

theorem which states that the maximum efficiency of a heat engine is given by:

Efficiencymax = 1 −Tc

where Tcand Thare the temperatures of the cold and hot reservoirs, respec-

tively.

Step 3: Calculate the maximum work output.

Given that the temperatures of the cold and hot reservoirs are 300 K and

500 K respectively, substitute the values into the formula:

Efficiencymax = 1 −300

500 = 1 −3

5=2

5= 0.4 = 40%

The maximum work output is equal to the efficiency multiplied by the heat

input:

Maximum Work Output = Efficiencymax ×Heat Input

Maximum Work Output = 0.4×6000 J = 2400 J

Therefore, the maximum work output of the engine is 2400 J.

Question 5

Question

A heat engine operates between a hot reservoir at 127◦C and a cold reservoir

at 27◦C. The engine produces 2000 J of work while absorbing 5000 J of heat

from the hot reservoir. Determine the efficiency of the engine. Is this engine

operating in violation of the second law of thermodynamics?

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = Useful work output

Heat input

Step 2: Calculate the heat input to the engine: Since the engine absorbs

5000 J of heat from the hot reservoir, the heat input is 5000 J.

Step 3: Calculate the useful work output of the engine, which is given as

2000 J.

Step 4: Substitute the values into the efficiency formula:

Efficiency = 2000

5000 = 0.4 = 40%

Step 5: No, this engine is not operating in violation of the second law of

thermodynamics. The efficiency is less than 100

Question 6

Question

A heat engine operates between two reservoirs at temperatures Th= 600 K

and Tc= 300 K. The engine absorbs 1000 J of heat from the hot reservoir in

each cycle and expels 600 J to the cold reservoir. Calculate the efficiency of the

engine and discuss if it violates the second law of thermodynamics.

Solution

Step 1: Calculate the efficiency of the engine using the formula:

Efficiency = 1 −Tc

where Tcis the temperature of the cold reservoir and This the temperature of

the hot reservoir.

Step 1: Calculate the efficiency of the engine.

Efficiency = 1 −300

600 = 1 −0.5=0.5

Step 2: Discuss whether the engine violates the second law of thermodynam-

ics. According to the second law of thermodynamics, the maximum efficiency

of a heat engine operating between two reservoirs at temperatures Thand Tcis

given by:

Maximum Efficiency = 1 −Tc

In this case, the maximum efficiency would be:

Max Efficiency = 1 −300

600 = 1 −0.5=0.5

Since the actual efficiency of the engine is equal to the maximum efficiency, the

engine does not violate the second law of thermodynamics.

Question 7

Question

A heat engine operates between a hot reservoir at 600 K and a cold reservoir at

300 K. The engine absorbs 1000 J of heat from the hot reservoir in each cycle.

Determine: a) The efficiency of this heat engine. b) The maximum amount of

work that can be done by this engine in each cycle. c) Verify the second law of

thermodynamics for this engine.