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The temperature of reservoir of Carnot's engine operating with an efficiency of 70% is 1000K. The temperature of its sink is -

300 K

400 K

500 K

700 K

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}}$ ⇒ $\frac{70}{100} = 1 - \frac{T_{2}}{1000}$

⇒ T₂ = 300 K

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In a Carnot engine, when T₂ = 0°C and T₁ = 200°C, its efficiency is η₁ and when T₁ = 0°C and T₂ = –200°C, its efficiency is $η_{2}$ , then what is η₁ /η₂ ?

0.577

0.733

0.638

Can not be calculated

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Explanation

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

$\Rightarrow η_{1} = \frac{(473 - 273)}{473} = \frac{200}{473}$ and $η_{2} = \frac{273 - 73}{273} = \frac{200}{273}$

So required ratio $\frac{η_{1}}{η_{2}} = \frac{273}{473} = 0 .577$

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The efficiency of Carnot's engine operating between reservoirs, maintained at temperatures 27°C and –123°C, is

50%

24%

0.75%

0.4%

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{(273 - 123)}{(273 + 27)} = 1 - \frac{150}{300} = \frac{1}{2} = 50 %$

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A Carnot engine operates between 227°C and 27°C . Efficiency of the engine will be -

$\frac{1}{3}$

$\frac{2}{5}$

$\frac{3}{4}$

$\frac{3}{5}$

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{500} = \frac{2}{5}$

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Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of outlet ?

200 K

400 K

600 K

800 K

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}}$ ⇒ $\frac{1}{2} = 1 - \frac{500}{T_{1}}$ ⇒ $\frac{500}{T_{1}} = \frac{1}{2}$ …..(i)

$\frac{60}{100} = 1 - \frac{T_{2}'}{T_{1}}$ ⇒ $\frac{T_{2}'}{T_{1}} = \frac{2}{5}$ …..(ii)

Dividing equation (i) by (ii), $\frac{500}{T_{2}'} = \frac{5}{4}$ ⇒ $T_{2} = 400 K$

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An ideal heat engine working between temperature T₁ and T₂ has an efficiency η, the new efficiency if both the source and sink temperature are doubled, will be

$\frac{η}{2}$

2η

3η

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Explanation

In first case $η_{1} = \frac{T_{1} - T_{2}}{T_{1}}$ = $η$

In second case $η_{2} = \frac{2 T_{1} - 2 T_{2}}{2 T_{1}}$ $= \frac{T_{1} - T_{2}}{T_{1}} = η$

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An ideal refrigerator has a freezer at a temperature of –13°C . The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be

325°C

325K

39°C

320°C

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Explanation

Coefficient of performance

$K = \frac{T_{2}}{T_{1} - T_{2}}$ ⇒ $5 = \frac{(273 - 13)}{T_{1} - (273 - 13)} = \frac{260}{T_{1} - 260}$

⇒ $5 T_{1} - 1300 = 260$ ⇒ $5 T_{1} = 1560$

⇒ $T_{1} = 312 K \to 39 ° C$

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An engine is supposed to operate between two reservoirs at temperature 727°C and 227°C. The maximum possible efficiency of such an engine is -

1/2

1/4

3/4

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Explanation

$η = \frac{T_{1} - T_{2}}{T_{1}} = \frac{(273 + 727) - (273 + 227)}{273 + 727}$

$= \frac{1000 - 500}{1000} = \frac{1}{2}$

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An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 10⁴ cal of heat at higher temperature. Amount of heat converted to work is -

2.4 × 10⁴ cal

6 × 10⁴ cal

1.2 × 10⁴ cal

4.8 × 10⁴ cal

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Explanation

$η = \frac{T_{1} - T_{2}}{T_{1}} = \frac{W}{Q}$ ⇒ $W = \frac{Q (T_{1} - T_{2})}{T_{1}}$

$= \frac{6 \times 10^{4} [(227 + 273) - (273 + 127)]}{(227 + 273)}$

$= \frac{6 \times 10^{4} \times 100}{500} = 1 .2 \times 10^{4} c a l$

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Which of the following processes is reversible ?

Transfer of heat by radiation

Electrical heating of a nichrome wire

Transfer of heat by conduction

Isothermal compression

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Explanation

Slow isothermal expansion or compression of an ideal gas is reversible process, while the other given process are irreversible in nature.

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