Document

1. ⇒ (NEET 2023 )

A Carnot engine has an efficiency of $50 %$ when its source is at a temperature $327^{\circ} C$ . The temperature of the sink is :-

(A) $15^{\circ} C$

(B) $100^{\circ} C$

(C) $200^{\circ} C$

(D) $27^{\circ} C$

Correct answer is (D)

Efficiency of carnot engine

$\begin{aligned} % η = (1 - \frac{T_{sink}}{T_{source}}) \times 100 \\ T_{source} = 327^{\circ} C = 600 K \\ 50 = (1 - \frac{T_{sink}}{600}) \times 100 \\ \frac{1}{2} = 1 - \frac{T_{sink}}{600} \\ T_{Sink} = 300 K \end{aligned}$

So temp. of sink is $^{\circ} C = 300 - 2763 = 27^{\circ} C$

2. ⇒ (NEET 2018 )

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

(A) 26.8%

(B) 20%

(C) 6.25%

(D) 12.5%

Correct answer is (A)

Efficiency of ideal heat engine,

$η = (1 - \frac{T_{2}}{T_{1}})$

Freezing point of water = 0^oC = 273 K

Boiling point of water = 100^oC = (100 + 273) K = 373 K

$∴$ Sink temperature, T₂ = 0^oC = 0 + 273 = 273 K

$∴$ Source temperature, T₁ = 100^oC = 100 + 273 = 373 K

Percentage efficiency, % $η = (1 - \frac{T_{2}}{T_{1}}) \times 100$

= $(1 - \frac{273}{373}) \times 100$

= $(\frac{100}{373}) \times 100$ = 26.8%

3. ⇒ (NEET 2017 )

A carnot engine having an efficiency of $\frac{1}{10}$ as 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 lower temperature is

(A) 90 J

(B) 99 J

(C) 100 J

(D) 1 J

Correct answer is (A)

Given, efficiency of engine, $η = \frac{1}{10}$

work done on system W = 10J

Coefficient of performance of refrigerator

$β = \frac{Q_{2}}{W} = \frac{1 - η}{η} = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}} = 9$

Energy absorbed from reservoir

$Q_{2} = β w$

$Q_{2} = 9 \times 10 = 90 J$

4. ⇒ (NEET 2016 Phase 2 )

The temperature inside a refrigerator is t₂ ^oC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

(A) $\frac{t_{1}}{t_{1} - t_{2}}$

(B) $\frac{t_{1} + 273}{t_{1} - t_{2}}$

(C) $\frac{t_{2} + 273}{t_{1} - t_{2}}$

(D) $\frac{t_{1} + t_{2}}{t_{1} + 273}$

Correct answer is (B)

Temperature inside refrigerator = t₂ °C

Room temperature = t₁ °C

For refrigerator,

$\frac{H e a t g i v e n t o h i g h t e m p e r a t u r e (Q_{1})}{H e a t t a k e n f r o m l o w e r t e m p e r a t u r e (Q_{2})} = \frac{T_{1}}{T_{2}}$

$\frac{Q_{1}}{Q_{2}} = \frac{t_{1} + 273}{t_{2} + 273}$

$\Rightarrow \frac{Q_{1}}{Q_{1} - W} = \frac{t_{1} + 273}{t_{2} + 273}$

$\Rightarrow 1 - \frac{W}{Q_{1}} = \frac{t_{2} + 273}{t_{1} + 273}$

$\Rightarrow \frac{W}{Q_{1}} = \frac{t_{1} - t_{2}}{t_{1} + 273}$

The amount of heat delivered to the room for each joule of electrical energy (W = 1 J)

$Q_{1} = \frac{t_{1} + 273}{t_{1} - t_{2}}$

5. ⇒ (NEET 2016 Phase 1 )

A refrigerator works between 4^oC and 30^oC. 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 (Take 1 cal = 4.2 Joules)

(A) 236.5 W

(B) 2365 W

(C) 2.365 W

(D) 23.65 W

Correct answer is (A)

Now $C O P = \frac{L_{Q}}{P o w e r} = \frac{T_{L}}{T_{H} - T_{L}}$

$\Rightarrow P o w e r = \frac{L_{Q} \times (T_{H} - T_{L})}{T_{L}}$

= (600 × 4.2) × 26/277 = 236.53 W

6. ⇒ (AIPMT 2015 )

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $-$ 20^oC, the temperature of the surroundings to which it rejects heat is

(A) 11^oC

(B) 21^oC

(C) 31^oC

(D) 41^oC

Correct answer is (C)

Coefficient of performance,

$C o p = \frac{T_{2}}{T_{1} - T_{2}}$

$5 = \frac{273 - 20}{T_{1} - (273 - 20)} = \frac{253}{T_{1} - 253}$

5T₁ – (5 × 253) = 253

5T₁ = 253 + (5 × 253) = 1518

$∴$ $T_{1} = \frac{1518}{5} = 303.6$

$\Rightarrow$ T₁ = 303.6 – 273 = 30.6 $≅$ 31°C

7. ⇒ (AIPMT 2015 Cancelled Paper )

A Carnot engine, having an efficiency of as 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 lower temperature is

(A) 90 J

(B) 1 J

(C) 100 J

(D) 99 J

Correct answer is (A)

Given, efficiency of engine, $η = \frac{1}{10}$

work done on system W = 10J

Coefficient of performance of refrigerator

$β = \frac{Q_{2}}{W} = \frac{1 - η}{η} = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}} = 9$

Energy absorbed from reservoir

$Q_{2} = β w$

$Q_{2} = 9 \times 10 = 90 J$

8. ⇒ (NEET 2013 (Karnataka) )

Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T₁ and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T₂. For what values of T the efficiencies of the two engines are equal

(A) $\frac{T_{1} - T_{2}}{2}$

(B) $T_{1} T_{2}$

(C) $\sqrt{T_{1} T_{2}}$

(D) $\frac{T_{1} + T_{2}}{2}$

Correct answer is (C)

Efficiency of engine A, $η_{1} = 1 - \frac{T}{T_{1}}$

Efficiency of engine B, $η_{2} = 1 - \frac{T_{2}}{T}$

Here, $η_{1} = η_{2}$

$∴ \frac{T}{T_{1}} = \frac{T_{2}}{T} \Rightarrow T = \sqrt{T_{1} T_{2}}$

9. ⇒ (AIPMT 2007 )

An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62^oC, its efficiency is doubled. Temperatures of the source is

(A) 37^oC

(B) 62^oC

(C) 99^oC

(D) 124^oC.

Correct answer is (C)

Since efficiency of engine is $η = 1 - \frac{T_{2}}{T_{1}}$

According to problem,

$\frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$ ...(i)

When the temperature of the sink is reduced by 62°C, its efficiency is doubled

$2 (\frac{1}{6}) = 1 - \frac{T_{2} - 62}{T_{1}}$ ...(ii)

Solving (i) and (ii)
T₂ = 372 K

T₁ = 99°C = Temperature of source.

10. ⇒ (AIPMT 2006 )

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

(A) 380 K

(B) 275 K

(C) 325 K

(D) 250 K

Correct answer is (D)

We know that efficiency of Carnot Engine

$= \frac{T_{1} - T_{2}}{T_{1}}$

where, T₁ is temp. of source & T₂ is temp. of sink

$∴ 0.40 = \frac{T_{1} - 300}{T_{1}} \Rightarrow T_{1} - 300 = 0.40 T_{1}$

$0.6 T_{1} = 300 \Rightarrow T_{1} = \frac{300}{0.6} = \frac{3000}{6} = 500 K$

Now efficiency to be increased by 50%

$∴$ $0.60 = \frac{T_{1} - 300}{T_{1}} \Rightarrow T_{1} - 300 = 0.6 T_{1}$

$0.4 T_{1} = 300 \Rightarrow T_{1} = \frac{300}{0.4} = \frac{300 \times 10}{4} = 750$

Increase in temp = 750 – 500 = 250 K

11. ⇒ (AIPMT 2005 )

An ideal gas heat engine operates in Carnot cycle between 227^oC and 127^oC. It absorbs 6 $\times$ 10⁴ cal of heat at higher temperature. Amount of heat converted to work is

(A) 4.8 $\times$ 10⁴ cal

(B) 6 $\times$ 10⁴ cal

(C) 2.4 $\times$ 10⁴ cal

(D) 1.2 $\times$ 10⁴ cal.

Correct answer is (D)

$1 - \frac{T_{2}}{T_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$

$\Rightarrow 1 - \frac{400}{500} = 1 - \frac{Q_{2}}{6 \times 10^{4}}$

$\Rightarrow \frac{4}{5} = \frac{Q_{2}}{6 \times 10^{4}} \Rightarrow Q_{2} = 4.8 \times 10^{4} c a l$

Net heat converted into work

$= 6.0 \times 10^{4} - 4.8 \times 10^{4} = 1.2 \times 10^{4} c a l$

12. ⇒ (AIPMT 2003 )

An ideal gas heat engine operates in a Carnot cycle between 227^oC and 127^oC. It absorbs 6 kcal at the height temperature. The amount of heat (in kcal) converted into work is equal to

(A) 4.8

(B) 3.5

(C) 1.6

(D) 1.2

Correct answer is (D)

Efficiency of Carnot engine

$= \frac{W}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$

$\frac{W}{6} = 1 - \frac{400}{500} = \frac{1}{5} \Rightarrow W = \frac{6}{5} = 1.2 k c a l$

13. ⇒ (AIPMT 2002 )

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

(A) 100 K

(B) 600 K

(C) 400 K

(D) 500 K

Correct answer is (C)

Efficiency ( $η$ ) of a carnot engine is given by $η = 1 - \frac{T_{2}}{T_{1}}$ , where T₁ is the temperature of the source and T₂ is the temperature of the sink.

Here, T₂ = 500 K

$∴$ $0.5 = 1 - \frac{500}{T_{1}} \Rightarrow T_{1} = 1000 K$

Now, $η = 0.6 = 1 - \frac{{T_{2}}^{'}}{1000}$ (T₂' is the new sink temperature)

$\Rightarrow {T_{2}}^{'} = 400 K$

14. ⇒ (AIPMT 2001 )

A scientist says that the efficiency of his heat engine which work at source temperature 127^oC and sink temperature 27^oC is 26%, then

(A) it is impossible

(B) it is possible but less probable

(C) it is quite probable

(D) data are incomplete.

Correct answer is (A)

Efficiency is maximum in Carnot engine which is an ideal engine.

$∵ η = \frac{400 - 300}{400} \times 100 % = 25 %$

$∴$ efficiency 26% is impossible for his heat engine.

15. ⇒ (AIPMT 2000 )

The (W/Q) of a Carnot engine is 1/6, now the temperature of sink is reduced by 62^oC, then this ratio becomes twice, therefore the initial temperature of the sink and source are respectively

(A) 33^oC, 67^oC

(B) 37^oC, 99^oC

(C) 67^oC, 33^oC

(D) 97 K, 37 K

Correct answer is (B)

Initially the efficiency of the engine was $\frac{1}{6}$ which increases to $\frac{1}{3}$ when the sink temperature reduces by 62º C.

$η = \frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$ when T₂ = sink temperature , T₁ = source temperature

$\Rightarrow T_{2} = \frac{5}{6} T_{1}$

Secondly,
$\frac{1}{3} = 1 - \frac{T_{2} - 62}{T_{1}} = 1 - \frac{T_{2}}{T_{1}} + \frac{62}{T_{1}}$

$= 1 - \frac{5}{6} + \frac{62}{T_{1}}$

$\Rightarrow T_{1} = 62 \times 6 = 372 K = 372 - 273 = 99^{o} C$

& $T_{2} = \frac{5}{6} \times 372 = 310 K = 310 - 273 = 37^{o} C$

⇒Thermodynamics

Topic 03: Second Law of Thermodynamic Heat Engine and Heat Refrigerator