Question

A carnot engine, whose source is at 400 K, takes 100 cal of heat and reflects 75 cal to the sink. What is temperature of the sink?

• A. 300 K
• B. 400 K
• C. 200 K
• D. 100 K

Answer 1

Answer: (A)

300 K

Answer 2

The efficiency of a Carnot engine is given by:

$$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}$$

Given:

• $T_{\text{source}} = 400\,K$
• $Q_{\text{in}} = 100\,cal$
• $Q_{\text{out}} = 75\,cal$

Calculate efficiency using heat quantities:

$$\eta = 1 - \frac{75}{100} = 1 - 0.75 = 0.25$$

Now,

$$1 - \frac{T_{\text{sink}}}{400} = 0.25 \quad \Longrightarrow \quad \frac{T_{\text{sink}}}{400} = 0.75$$ $$T_{\text{sink}} = 400 \times 0.75 = 300\,K$$