Heat Engine Refrigerator Multiple Choice Question And Answers

Heat Engine Refrigerator

Question 1. The efficiency of a Carnot engine operating between the temperatures of 100°C and -23°C will be

1. \(\frac{100-23}{273}\)
2. \(\frac{100+23}{373}\)
3. \(\frac{100+23}{100}\)
4. \(\frac{100-23}{100}\)

Answer: 2. \(\frac{100+23}{373}\)

Temperature of source = T₁ = (273 + 100) K

= 373 K,

Temperature of sin k = T₂ = (273 – 23 K)

= 250 K.

The efficiency of the Carnot engine is

⇒ \(\eta=1-\frac{T_2}{T_1}\)

= \(\frac{T_1-T_2}{T_1}\)

= \(\frac{123}{373}\)

= \(\frac{100+23}{373}\)

Question 2. An engine takes heat from a reservoir and converts \(\frac{1}{6}\) of it into work. If the temperature of the sink is reduced by 62 °C, the efficiency of the engine becomes double. The temperatures of the source and sink must be

1. 90°C, 37°C
2. 99°C,37°C
3. 372°C,37°C
4. 206°C, 37°C

Answer: 2. 99°C,37°C

Efficiency = \(\frac{\text { output }}{\text { input }}=\frac{\text { work done }(W)}{\text { heat absorbed }(Q)}\)

⇒ \(1-\frac{T_2}{T_1}=\frac{1}{6}\)…..(1) [∵ W = \(\frac{Q}{6}\)]

When the temperature T₂ of the sink is reduced by 62°C, the new temperature T₂ = T₂– 62, and

⇒ \(\eta^{\prime}=1-\frac{T_2^{\prime}}{T_1}\)

= \(2 \eta=2\left(\frac{1}{6}\right)\)

= \(\frac{1}{3}\)

⇒ \(1-\left(\frac{T_2-62}{T_1}\right)\)

= \(\frac{1}{3}\)…(2)

From(1), \(\frac{T_2}{T_1}=\frac{5}{6}\) and from(2), \(\frac{T_2}{T_1}=\frac{2}{3}+\frac{62}{T_1}\)

Equating \(\frac{T_2}{T_1}\) as mentioned above, \(\frac{5}{6}=\frac{2}{3}+\frac{62}{T_1}\)

T₁ = 372K

= 99°C,

and \(T_2=\frac{5}{6} T_1\)

= 310K

= 37°C.

∴ The required temperatures are 99 °C and 37°C.

Question 3. The temperatures of the source and sink of a heat engine are 127°C and 27°C respectively. A technician claims its efficiency to be 26%.

1. It is impossible.
2. It is possible with high probability.
3. It is possible with low probability.
4. The data is insufficient.

Answer: 1. It is impossible.

Given that T₁ = (273 + 127) K

= 400 K

T₂ = (273 + 27) K

= 300 K.

The efficiency of an ideal engine (Carnot engine),

⇒ \(\eta=1-\frac{T_2}{T_1}\)

= \(1-\frac{300}{400}\)

= \(\frac{1}{4}\)

= 25%

No engine can have an efficiency more than that of a Carnot engine.

Hence, the claim of 26% efficiency is impossible.

Question 4. The efficiency of a Carnot engine is 50% and the temperature of the sink is 500 K. If the temperature of the source is kept constant and the efficiency of the engine is to be raised to 60% then the required temperature of the sink will be

1. 600 K
2. 500 K
3. 400 K
4. 100 K

Answer: 3. 400 K

Efficiency \(\eta=1-\frac{T_2}{T_1}\)

Given that T₂ = 500 K and \(\eta=50 \%=\frac{1}{2} \Rightarrow \frac{1}{2}=1-\frac{500}{T_1}\)

hence T₁ = 1000 K.

For 60% efficiency, let the temperature of the sink be T₂

⇒ \(\frac{60}{100}=1-\frac{T_2^{\prime}}{T_1}=1-\frac{T_2^{\prime}}{1000}\)

∴ T₂ = 400 K

Question 5. An ideal gas heat engine operates as a Carnot cycle between 227°C and 127°C. It absorbs 6 kcal of heat at a higher temperature. The amount of heat (in kcal) converted into work is equal to

1. 1.6
2. 1.2
3. 4.8
4. 3.5

Answer: 2. 1.2

Temperature of source = T₁ = (227 + 273) K

= 500 K.

Temperature ofsink = T₂ = (127 + 273) K

= 400 K.

∴ efficiency \(\eta=1-\frac{T_2}{T_1}\)

= \(1-\frac{400}{500}\)

= \(\frac{1}{5}\)

But \(\eta=\frac{\text { output }(W)}{\text { input }(Q)}\)

= \(\frac{W}{6 \mathrm{kcal}}\)

= \(\frac{1}{5}\)

∴ work W = \(\frac{6kcal}{5}\)

= 1.2 kcal

Question 6. An engine has an efficiency of \(\frac{1}{6}\) When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperature of the source is

1. 120°C
2. 35°C
3. 99°C
4. 65°C

Answer: 3. 99°C

Let T₁ = temperature of the source.

Given that efficiency n = \(\frac{1}{6}\)

⇒ \(\eta=1-\frac{T_2}{T_1} \Rightarrow \frac{1}{6}=1-\frac{T_2}{T_1}\)….(1)

When T₂ is reduced to T₂– 62, efficiency gets doubled. So,

⇒ \(\frac{2}{6}=1-\frac{T_2-62}{T_1}\)…..(2)

Solving (1) and (2), T₁ = 372 K = 99°C.

Question 7. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of the source be increased so as to increase its efficiency by 50% of the original efficiency?

1. 250 K
2. 370 K
3. 270 K
4. 390 K

Answer: 1. 250 K

Efficiency of Carnot engine is \(\eta=1-\frac{T_2}{T_1}\)

Given that \(\eta=40 \%=\frac{4}{10}\), the temperature of sink = T₂ = 300 K.

Temperature of source = T₁.

∴ \(\frac{4}{10}=1-\frac{300}{T_1}\)

⇒ T₁ = 500 K

Let the temperature T₁ of the source be increased by ΔT₁ so that the increased efficiency becomes

⇒ \(\eta^{\prime}=40 \%+50 \% \text { of } \eta=\frac{40}{100}+\frac{50}{100} \times \frac{40}{100}=\frac{60}{100}\)

Hence,

⇒ \(\eta^{\prime}=\frac{60}{100}\)

= \(1-\frac{T_2}{T_1+\Delta T_1}\)

⇒ \(\frac{60}{100}=1-\frac{300}{500+\Delta T_1}\)

⇒ \(\Delta T_1=250 \mathrm{~K}\)

Question 8. Two Carnot engines A and B are operated in series. Engine A receives heat from the source at temperature T₁, and rejects heat to the sink at temperature T. The second engine receives heat at temperature T and rejects a part of it to its sink at temperature T₂. For what value of T are the efficiencies of the two engines equal?

1. \(\frac{T_1+T_2}{2}\)
2. \(\frac{T_1-T_2}{2}\)
3. \(\sqrt{T_1 T_2}\)
4. \(\frac{\sqrt{T_1 T_2}}{2}\)

Answer: 3. \(\sqrt{T_1 T_2}\)

For Carnot engine A, efficiency is \(\eta_{\mathrm{A}}=1-\frac{T}{T_1}\) and for Carnot engine B,

it is \(\eta_{\mathrm{B}}=1-\frac{T_2}{T}\)

Given that \(\eta_{\mathrm{A}}=\eta_{\mathrm{B}}\)

⇒ \(1-\frac{T}{T_1}=1-\frac{T_2}{T}\)

⇒ \(\frac{T}{T_1}=\frac{T_2}{T}\)

Hence, \(T=\sqrt{T_1 T_2}\)

Question 9. The coefficient of performance of a refrigerator is 5. If the temperature inside the freezer is -20°C, the temperature of the surroundings to which it rejects heat is

1. 42 °C
2. 31 °C
3. 21 °C
4. 15°C

Answer: 2. 31 °C

The coefficient of performance (k) of a refrigerator is expressed as

⇒ \(k=\frac{T_2}{T_1-T_2}\)

where T₂ = temperature of the refrigerated space,

and T₁ = temperature of surroundings.

Given that T₂ = -20°C = 253 K.

For k = 5,

⇒ \(5=\frac{253}{T_1-253}\)

⇒ \(T_1=303.6 \mathrm{~K} \approx 31^{\circ} \mathrm{C}\)

Question 10. A Carnot engine having an efficiency of \(\frac{1}{10}\) as a heat engine is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

1. 110 J
2. 90 J
3. 95 J
4. 10 J

Answer: 2. 90 J

When used as a Carnot engine, efficiency \(\eta=1-\frac{T_2}{T_1}\)

Given that n = \(\frac{1}{10}\)

Hence, \(\frac{T_2}{T_1}=\frac{9}{10}\)……(1)

When used as a refrigerator, its coefficient of performance is

⇒ heat extracted from the reservoir (Q₂) / work done on the system

Question 11. A refrigerator works between 4°C and 30°C, It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is

1. 24.65 W
2. 236.5 W
3. 230 W
4. 2.365 W

Answer: 2. 236.5 W

For the refrigerator, T₂ = (4 + 273) K = 277 K.

T₁ = (30 + 273) K

= 303 K.

The coefficient of performance is

⇒ \(\frac{Q_2 \text { (heat absorbed) }}{W(\text { Work done })}=\frac{Q_2}{Q_1-Q_2}\)

= \(\frac{1}{\frac{Q_1}{Q_2}-1}\)

⇒ \(\frac{Q_2}{W}=\frac{1}{\frac{T_1}{T_2}-1}\)

= \(\frac{T_2}{T_1-T_2}\)

Given that Q₂ = 600 cal s^-1 = 600 x 4.2 J s^-1.

∴ required power is,

⇒ \(W=Q_2 \frac{\left(T_1-T_2\right)}{T_2}\)

= \(\frac{\left(600 \times 4.2 \mathrm{~J} \mathrm{~s}^{-1}\right)(26 \mathrm{~K})}{(277 \mathrm{~K})}\)

= 236.5 W.

Question 12. The temperature inside a refrigerator is t₂°C and the room temperature is t₁°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

1. \(\frac{t_1+t_2}{t_1+273} \mathrm{~J}\)
2. \(\frac{t_1+273}{t_1-t_2} \mathrm{~J}\)
3. \(\frac{t_1}{t_1-t_2} \mathrm{~J}\)
4. \(\frac{t_1+273}{t_1-273} \mathrm{~J}\)

Answer: 2. \(\frac{t_1+273}{t_1-t_2} \mathrm{~J}\)

For a refrigerator, the ratio

⇒ \(\frac{Q_1 \text { (heat delivered to the room) }}{Q_2 \text { (heat absorbed) }}=\frac{T_1}{T_2}\)

But Q₁– Q₂ = W, so Q₂ = Q₁– W, where W = electrical energy is consumed.

∴ \(\frac{Q_1}{Q_1-W}=\frac{T_1}{T_2} \Rightarrow \frac{1}{1-\frac{W}{Q_1}}=\frac{T_1}{T_2} \Rightarrow Q_1=\frac{W T_1}{T_1-T_2}\)

Given that T₁ = 273 + t₁, T₂ = 273 + t₂, W = 1 J.

∴ \(Q_1=\frac{(1 \mathrm{~J})\left(273+t_1\right)}{\left(t_1-t_2\right)}=\frac{273+t_1}{t_1-t_2} \mathrm{~J}\)

Question 13. Determine the efficiency of a Carnot engine if, during its adiabatic expansion, the volume is increased to 3 times the initial volume and y = 1.5.

1. \(1-\frac{1}{\sqrt{2}}\)
2. \(1-\frac{1}{\sqrt{3}}\)
3. \(1-\frac{1}{\sqrt{2}}\)
4. \(1-\frac{1}{\sqrt{3}}\)

Answer: 2. \(1-\frac{1}{\sqrt{3}}\)

During the adiabatic expansion,

⇒ \(T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{1}{3}\right)^{1.5-1}\)

= \(\frac{1}{\sqrt{3}}\)

Hence, efficiency is \(\eta=1-\frac{T_2}{T_1}=1-\frac{1}{\sqrt{3}}\)

Question 14. In a refrigerator, the heat absorbed from the source is 800 J and the heat supplied to the sink is 500 J. The coefficient of performance is

1. \(\frac{5}{8}\)
2. \(\frac{8}{5}\)
3. \(\frac{5}{3}\)
4. \(\frac{3}{5}\)

Answer: 3. \(\frac{5}{3}\)

The coefficient of performance of a refrigerator is defined as

⇒ \(k=\frac{\text { heat absorbed }\left(Q_2\right)}{\text { work done on the coolant }(W)}\)

⇒ \(\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}\)

Given that Q₁ = 800J,

Q₂ = 500J,

Hence k = \(\frac{5}{3}\)

Question 15. A Carnot engine works between 27°C and127°C. The heat supplied by the source is 500 J. The heat delivered to the sink is

1. 100 J
2. 667 J
3. 375 J
4. 500 J

Answer: 3. 375 J

In a Carnot engine,

⇒ \(\frac{Q_2}{Q_1}=\frac{T_2}{T_1}\)

= \(\frac{27+273}{127+273}\)

= \(\frac{3}{4}\)

∴ heat delivered to the sink,

Q₂ = \(\frac{3}{4}\)

Q₁ = \(\frac{3}{4}\) x 500J

= 375J.

Question 16. A Carnot engine has an efficiency of \(\frac{1}{6}\). When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are respectively

1. 99°C, 37°C
2. 124°C, 62°C
3. 37°C, 99°C
4. 62°C, 124°C

Answer: 1. 99°C,37°C

The efficiency of the Carnot engine is

⇒ \(\eta=\frac{1}{6}=1-\frac{T_2}{T_1}\)

⇒ \(\frac{T_2}{T_1}=\frac{5}{6}\)

When T₂ is reduced by 62°C, the efficiency becomes

⇒ \(\eta^{\prime}=2 \eta\)

= \(\frac{1}{3}\)

= \(1-\frac{T_2-62}{T_1}\)

⇒ \(\frac{T_2-62}{T_1}=\frac{2}{3}\)

Dividing (2) by (1),

⇒ \(\frac{T_2-62}{T_2}=\frac{2}{3} \times \frac{6}{5}\)

= \(\frac{4}{5}\)

=> 5T₂ – 310 = 4T₂

=> T₂ = 310 K

= (310- 273) °C

= 37°C.

⇒ \(T_1=\frac{6}{5} T_2\)

= \(\frac{6}{5}(310 \mathrm{~K})\)

= 372 K

= (372- 273) °C

= 99°C

Question 17. TwoCarnotenginesAand B areoperatedinseries. EngineAreceives heat at T₁ (= 600 K) and rejects it to a reservoir at temperature T₂. Engine B receives heat rejected by Engine A and in turn, rejects heat to a heat reservoir at T₃ (= 400 K). If the work outputs by the two engines are equal, the temperature T₂ is equal to

1. 600 K
2. 500 K
3. 400 K
4. 300 K

Answer: 2. 500 K

The diagram shows the working of Carnot engines A and B in series.

For A: Q₁ = Q₂ + W₁,

So, W₁ = Q₁– Q₂.

For B: Q₂ = Q₃ + W₂,

So W₂ = Q₂-Q₃.

Given that work, the output is the same for both A and B, hence W₁ = W₂

Q₁ – Q₂ = Q₂-Q₃

=> Q₁ + Q₃ = 2Q₂

⇒ \(\frac{Q_1}{Q_2}+\frac{Q_3}{Q_2}=2\)

⇒ \(\frac{T_1}{T_2}+\frac{T_3}{T_2}=2\)

Substituting, \(\frac{600 \mathrm{~K}}{T_2}+\frac{400 \mathrm{~K}}{T_2}=2\)

⇒ T₂ = 500 K.