A Carnot engine operates between two reservoirs at temperatures $600\text{ K}$ and $300\text{ K}$. If the engine performs $1000\text{ J}$ of work in each cycle, how much heat is rejected to the cold reservoir?

Solution

1. Identify the given temperatures: $T_H=600\text{ K}$ and $T_L=300\text{ K}$, and the work done $W=1000\text{ J}$.
2. Calculate the Carnot efficiency: $$\begin{array}{rl}\eta & =1-\frac{T_L}{T_H}\\ & =1-\frac{300}{600}\\ & =0.5\end{array}$$.
3. Use the efficiency formula $\eta =\frac{W}{Q_H}$ to find the heat absorbed: $$\begin{array}{rl}{Q}_H& =\frac{W}{\eta }\\ & =\frac{1000}{0.5}\\ & =2000\text{ J}\end{array}$$.
4. Apply the first law of thermodynamics $Q_L=Q_H-W$ to find the heat rejected: $$\begin{array}{rl}{Q}_L& =2000\text{ J}-1000\text{ J}\\ & =1000\text{ J}\end{array}$$.
5. Hence the answer is (B).