## Heat Engine Refrigerator

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

- \(\frac{100-23}{273}\)
- \(\frac{100+23}{373}\)
- \(\frac{100+23}{100}\)
- \(\frac{100-23}{100}\)

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

Temperature of source = T_{1} = (273 + 100) K

= 373 K,

Temperature of sin k = T_{2} = (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**

- 90°C, 37°C
- 99°C,37°C
- 372°C,37°C
- 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 T_{2} of the sink is reduced by 62°C, the new temperature T_{2} = T_{2}– 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), \(\frac{T_2}{T_1}=\frac{2}{3}+\frac{62}{T_1}\)

**difference between heat engine heat pump and refrigerator**

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

T_{1} = 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%.**

- It is impossible.
- It is possible with high probability.
- It is possible with low probability.
- The data is insufficient.

**Answer:** 1. It is impossible.

Given that T_{1} = (273 + 127) K

= 400 K

T_{2} = (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**

- 600 K
- 500 K
- 400 K
- 100 K

**Answer:** 3. 400 K

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

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

hence T_{1} = 1000 K.

For 60% efficiency, let the temperature of the sink be T_{2}

**difference between heat engine heat pump and refrigerator**

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

∴ T_{2} = 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**

- 1.6
- 1.2
- 4.8
- 3.5

**Answer:** 2. 1.2

Temperature of source = T_{1} = (227 + 273) K

= 500 K.

Temperature ofsink = T_{2} = (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**

- 120°C
- 35°C
- 99°C
- 65°C

**Answer:** 3. 99°C

Let T_{1} = 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 T_{2} is reduced to T_{2}– 62, efficiency gets doubled. So,

**difference between heat engine heat pump and refrigerator**

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

Solving (1) and (2), T_{1} = 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?**

- 250 K
- 370 K
- 270 K
- 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 = T_{2} = 300 K.

Temperature of source = T_{1}.

∴ \(\frac{4}{10}=1-\frac{300}{T_1}\)

⇒ T_{1} = 500 K

Let the temperature T_{1} of the source be increased by ΔT_{1} 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 T _{1}, 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 T_{2}. For what value of T are the efficiencies of the two engines equal?**

- \(\frac{T_1+T_2}{2}\)
- \(\frac{T_1-T_2}{2}\)
- \(\sqrt{T_1 T_2}\)
- \(\frac{\sqrt{T_1 T_2}}{2}\)

**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**

- 42 °C
- 31 °C
- 21 °C
- 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 T_{2} = temperature of the refrigerated space,

and T_{1} = temperature of surroundings.

Given that T_{2} = -20°C = 253 K.

**difference between heat engine heat pump and refrigerator**

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**

- 110 J
- 90 J
- 95 J
- 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 (Q_{2}) / 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**

- 24.65 W
- 236.5 W
- 230 W
- 2.365 W

**Answer:** 2. 236.5 W

For the refrigerator, T_{2} = (4 + 273) K = 277 K.

T_{1} = (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 Q_{2} = 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 t _{2}°C and the room temperature is t_{1}°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be**

- \(\frac{t_1+t_2}{t_1+273} \mathrm{~J}\)
- \(\frac{t_1+273}{t_1-t_2} \mathrm{~J}\)
- \(\frac{t_1}{t_1-t_2} \mathrm{~J}\)
- \(\frac{t_1+273}{t_1-273} \mathrm{~J}\)

**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 Q_{1}– Q_{2} = W, so Q_{2} = Q_{1}– 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 T_{1} = 273 + t_{1}, T_{2} = 273 + t_{2}, 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.**

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

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

During the adiabatic expansion,

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

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

**difference between heat engine heat pump and refrigerator**

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

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

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**

- \(\frac{5}{8}\)
- \(\frac{8}{5}\)
- \(\frac{5}{3}\)
- \(\frac{3}{5}\)

**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 Q_{1} = 800J,

Q_{2} = 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**

- 100 J
- 667 J
- 375 J
- 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,

Q_{2} = \(\frac{3}{4}\)

Q_{1} = \(\frac{3}{4}\) 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**

- 99°C, 37°C
- 124°C, 62°C
- 37°C, 99°C
- 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 T_{2} is reduced by 62°C, the efficiency becomes

⇒ \(\eta^{\prime}=2 \eta\)

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

**difference between heat engine heat pump and refrigerator**

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

⇒ \(\frac{T_2-62}{T_1}=\frac{2}{3}\)

Dividing (2) by (1),

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

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

=> 5T_{2} – 310 = 4T_{2}

=> T_{2} = 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 T _{1} (= 600 K) and rejects it to a reservoir at temperature T_{2}. Engine B receives heat rejected by Engine A and in turn, rejects heat to a heat reservoir at T_{3} (= 400 K). If the work outputs by the two engines are equal, the temperature T_{2} is equal to**

- 600 K
- 500 K
- 400 K
- 300 K

**Answer:** 2. 500 K

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

**For A:** Q_{1} = Q_{2} + W_{1},

So, W_{1} = Q_{1}– Q_{2}.

**For B:** Q_{2} = Q_{3} + W_{2},

So W_{2} = Q_{2}-Q_{3}.

Given that work, the output is the same for both A and B, hence W_{1} = W_{2}

Q_{1} – Q_{2} = Q_{2}-Q_{3}

=> Q_{1} + Q_{3} = 2Q_{2}

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

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

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

⇒ T_{2} = 500 K.