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  • A Carnot’s engine whose sink is at a temperature of 300 K has an efficiency of 40%. By how much should the temperature of the source be increased so as to increase the efficiency to 60%?<

A Carnot’s engine whose sink is at a temperature of 300 K has an efficiency of 40%. By how much should the temperature of the source be increased so as to increase the efficiency to 60%?

Option: 1

250 K


Option: 2

275 K


Option: 3

300 K


Option: 4

325 K


Answers (1)

best_answer

\mathrm{T_2=300 \mathrm{~K} \text {. Now } \eta=1-T_2 / T_1 \text {. When } \eta=40 \%=0.4 \text {, the value of } T_1 \text { is given by }}

\mathrm{\frac{T_2}{T_1}=1-0.4=0.6}

\mathrm{\text { Or } T_1=\frac{T_2}{0.6}=\frac{300}{0.6} 500 \mathrm{~K} \text {. When } \eta=60 \%,=0.6 \text {, the value of } T_2 \text { should be }}

\mathrm{\begin{aligned} & T_2^{\prime}=\frac{300}{0.4}=750 \mathrm{~K} \\ & \therefore T_2^{\prime}-T_2=750-500=250 \mathrm{~K} \end{aligned}}

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Rakesh

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