Question

A carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

• A. 200°C
• B. 27°C
• C. 15°C
• D. 100°C

Answer 1

Answer: (B)

27°C

Answer 2

The correct option is (B): 27°C

The efficiency of Carnot engine,

%$\eta$=(1-$\frac{T_{sink}}{T_{source}}\times100$)

$T_{source}=327^{\circ}C=600K$

$50=(1- \frac{T_{sink}}{600})\times100$

$\frac{1}{2}=1- \frac{T_{sink}}{600}$

$T_{sink}=300K$

So the temperature of the sink is= 327-300=27$^{\circ}$C