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  • 70. A carnot’s engine whose sink is at has an efficiency of 25%

70. A carnot’s engine whose sink is at 27^{\circ} \mathrm{C} has an efficiency of 25%

Option: 1

The temperature of the source is 400 K


Option: 2

b) To increase efficiency by 20%, the temperature of the source should be increased by 28.6^{\circ} \mathrm{C}


Option: 3

If the heat energy supplied to the engine is 800 J per cycle the work output per cycle is 200 J


Option: 4

All of the above


Answers (1)

best_answer

\mathrm{\text { (a) } T_2=27^{\circ} \mathrm{C}=300 \mathrm{~K} \text {, efficiency } \eta=0.25}

\mathrm{\eta=1-\frac{T_2}{T_1} \Rightarrow 0.25=1-\frac{300}{T_1} \text { which gives }}

\mathrm{T_1=400 \mathrm{~K} \text {. So choice (a) is correct }}
(b) Increase in effieciency = 20% of 0.25=0.05. So new efficiency is \mathrm{\eta^{\prime}=0.25+0.05=0.30} The new temperature of the source should be\mathrm{T_1^{\prime}} so that

\mathrm{0.30=1-\frac{T_2}{T_1^{\prime}}=1-\frac{300}{T_1^{\prime}}}

\mathrm{\text { Which gives } T_1^{\prime}=428.6 \mathrm{~K} \text {. So, increase in temperature of the source }=T_1^{\prime}-T_1=}

\mathrm{428.6-400=28.6 \mathrm{~K}^{\circ}{ }^{\circ} \mathrm{C} \text {. So choice (b) is also correct }}

\mathrm{\text { (c) Work output }=Q \eta=800 \times 0.25=200 \mathrm{~J}}

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Riya

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