Question

A carnot engine, whose efficiency is $40\%$ takes in heat from a source maintained at a temperature of $500$ K, It is desired to have an engine of efficiency $60\%$ . Then, the intake temperature for the same exhaust (sink) temperature must be

• A. Efficiency of Carnot engine cannot be made larger than $50\%$
• B. $1200$ K
• C. $750$ K
• D. $600$ K

Answer 1

The correct option is C $750$ K

Efficiency , $\eta =1-\frac{T_{\text{sink}}}{T_{\text{source}}}$
A Carnot engine whose efficiency is $40\%$.
Use the given values in $\eta$
Now, $0.4=1-\frac{T_{\text{sink}}}{500K}$
when efficiency $60\%$
$\Rightarrow T_{\text{sink}}=0.6\times 500K=300K$
Thus, $0.6=1-\frac{300K}{T_{\text{source}}^{\prime }}$
$T_{source}^{\prime }=\frac{300K}{0.4}=750K$