Question

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

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

Answer 1

The correct option is C 400 K

We know that,

%efficiency = (1-$\frac{T_2}{T_1})\times 100$ where T${}_2$ is temperature of sink and T${}_1$ is temperature of source.

From given data, we can calculate T${}_1$ = 1000K

And now using $T_1$ we can find T${}_2$=400K corresponding to 60% efficiency,