Question

A Carnot engine, whose source is at $400K$, takes $200cal$ of heat and reflects $150cal$ to the sink. What is the temperature of the sink?

• A. $300K$
• B. $400K$
• C. $800K$
• D. none of these

Answer 1

The correct option is A $300K$

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}=\frac{Q_1-Q_2}{Q_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. $Q_1$ is the heat taken up from the source and $Q_2$ is the heat rejected to the sink.

Thus we get $\eta =1-\frac{T_2}{400}=\frac{200-150}{200}=\frac{1}{4}$

Thus we calculate $T_2$ as $300K$.