Question

The efficiency of Car not engine is $50 \%$ and temperature of $\sin k$ is $500 \,K$. If the temperature of source is kept constant and its efficiency is to be raised to $60 \%$, then the required temperature of the sink will be:

• A. 600 K
• B. 400 K
• C. 500 K
• D. 100 K

Answer 1

Answer: (B)

Efficiency of the Carnot engine is given by
$\eta=1-\frac{T_{2}}{T_{1}} . . $(i)
where $T_{1}=$ temperature of source
$T_{2}=$ temperature of $\sin k$
Given, $\eta=50$
Substituting in relation (i), we have or
$0.5=1-\frac{500}{T_{1}} $ or $ \frac{500}{T_{1}}=0.5 $
$\therefore T_{1}=\frac{500}{0.5}=1000 \,K$
Now, the temperature of sink is changed to
$T_{2}$ and the efficiency becomes $60 \%$
i.e, $0.6$. Using relation (i), we get
$0.6=1-\frac{T_{2}}{1000} $
or $\frac{T_{2}}{1000}=1-0.6=0.4$
or $T_{2}=0.4 \times 100=400\, K$
Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which can practically not be obtained.