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A carnot engine whose heat sinks at \mathrm{27^{\circ}C}, has an efficiency of \mathrm{25%}. By now many degress should be the temperature of the source be changed to increase the efficiency by \mathrm{100%} of the original efficiencey ?

Option: 1

Increases by \mathrm{18^{\circ}C}


Option: 2

 Increases by \mathrm{200^{\circ}C}


Option: 3

 Increases by \mathrm{120^{\circ}C}


Option: 4

Increases by \mathrm{73^{\circ}C}


Answers (1)

best_answer

\mathrm{T _{sink}= 27^{\circ}C= 300 k}
\mathrm{\text{Initial efficiency}= \eta = 25\, %= \frac{1}{4}= 1-\frac{T_{sink}}{T_{source}}}
\mathrm{\frac{T_{sink}}{T_{source}}= \frac{3}{4}}
\mathrm{T_{source}= 400k= T}

\mathrm{\text{Final efficiency}= 2\eta = \frac{1}{2}= 1-\frac{300}{\left ( T+\Delta T \right )}}
 \mathrm{\frac{1}{2}= \frac{300}{\left ( T+\Delta T \right )}}

\mathrm{T+\Delta T= 600}
           \mathrm{\Delta T= 200^{\circ}C= 200k}

The correct option is (2)

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