A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be:

A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. We need to find the efficiency of the engine and the temperature of the sink.

First, recall that the efficiency (η) of any heat engine is given by the ratio of the work done to the heat absorbed. The work done by the engine is the difference between the heat absorbed and the heat rejected. So:

Heat absorbed, $$Q_1 = 1000 \text{J}$$

Heat rejected, $$Q_2 = 600 \text{J}$$

Work done, $$W = Q_1 - Q_2 = 1000 - 600 = 400 \text{J}$$

Efficiency, $$\eta = \frac{W}{Q_1} = \frac{400}{1000}$$

Simplifying the fraction: $$\frac{400}{1000} = 0.4$$

To express as a percentage: $$0.4 \times 100 = 40\%$$

So, the efficiency is 40%.

For a Carnot engine, efficiency is also related to the absolute temperatures of the source and sink. The formula is:

$$\eta = 1 - \frac{T_2}{T_1}$$

where $$T_1$$ is the absolute temperature of the source (in Kelvin) and $$T_2$$ is the absolute temperature of the sink (in Kelvin).

The source temperature is given as 127°C. Convert this to Kelvin by adding 273:

$$T_1 = 127 + 273 = 400 \text{K}$$

We know $$\eta = 40\% = 0.4$$ (in decimal form). Substitute into the formula:

$$0.4 = 1 - \frac{T_2}{400}$$

Solve for $$T_2$$:

Rearrange: $$\frac{T_2}{400} = 1 - 0.4$$

$$\frac{T_2}{400} = 0.6$$

Multiply both sides by 400: $$T_2 = 0.6 \times 400$$

$$T_2 = 240 \text{K}$$

Convert $$T_2$$ from Kelvin to Celsius by subtracting 273:

$$T_2 (\text{in °C}) = 240 - 273 = -33°C$$

Therefore, the efficiency is 40% and the temperature of the sink is -33°C.

Comparing with the options:

A. 20% and -43°C

B. 40% and -33°C

C. 50% and -20°C

D. 70% and -10°C

Option B matches our calculated values.

Hence, the correct answer is Option B.