Question

A Carnot engine absorbs $2000J$ of heat energy from a reservoir at $127°C$ and rejects $800J$ of heat energy during each cycle. The efficiency of engine and temperature of sink will be respectively

Answer 1

Step 1: Given

Heat absorbed, $Q_1=2000J$

Heat rejected, $Q_2=800J$

The temperature of the reservoir, $T_1=127°C=400K$

Let the temperature of the sink be $T_2$.

Step 2: Formula used

The efficiency of a Carnot engine is given by, $\eta =\frac{Q_1-Q_2}{Q_1}$ or $\eta =1-\frac{T_2}{T_1}$

Step 3: Determine the efficiency of the engine

The efficiency of a Carnot engine is given by,

$$\begin{gathered} \eta =\frac{Q_1-Q_2}{Q_1} \\ =\frac{2000-800}{2000} \\ =0.6 \\ =60\% \end{gathered}$$

Step 4: Determine the temperature of the sink

Let the temperature of the sink be $T_2$

The efficiency of the Carnot engine can also be written as

$$\begin{gathered} \eta =1-\frac{T_2}{T_1} \\ 0.6=1-\frac{T_2}{400} \\ {T}_2=(1-0.6)\times 400 \\ {T}_2=160K \end{gathered}$$

Therefore, the efficiency of the Carnot engine is $60\%$ and the temperature of the sink is $160K$.