Question

A heat engine is supplied with 250 kJ/s of heat at a constant temperature of 227°C; the heat is rejected at 27°C. If the cycle is reversible, then the amount of heat rejected is:

• A. 50 kJ/s
• B. 150 kJ/s
• C. 200 kJ/s
• D. 250 kJ/s

Hint:

For reversible heat engines (Carnot cycles), the efficiency depends only on the temperatures of the hot and cold reservoirs. Also, the ratio of heat exchanged with the reservoirs is equal to the ratio of their absolute temperatures.

Answer 1

The Correct Option is B

Step 1: Identify the given parameters.
Heat supplied to the engine, \( Q_H = 250 \, \text{kJ/s} \).
Temperature at which heat is supplied, \( T_H = 227^\circ\text{C} = 227 + 273 = 500 \, \text{K} \).
Temperature at which heat is rejected, \( T_C = 27^\circ\text{C} = 27 + 273 = 300 \, \text{K} \).
The cycle is reversible, which implies it is a Carnot cycle operating between these two temperatures.
Step 2: Recall the efficiency of a reversible (Carnot) heat engine.
The efficiency \( \eta \) of a Carnot heat engine is given by: \[ \eta = 1 - \frac{T_C}{T_H} = \frac{W_{net}}{Q_H}, \] where \( W_{net} \) is the net work done by the engine. Step 3: Calculate the efficiency of the Carnot cycle.
Using the given temperatures: \[ \eta = 1 - \frac{300 \, \text{K}}{500 \, \text{K}} = 1 - 0.6 = 0.4. \] So, the efficiency of the reversible heat engine is 40%. Step 4: Relate the heat rejected \( Q_C \) to the heat supplied \( Q_H \) and the net work done \( W_{net} \).
From the first law of thermodynamics for a cycle: \[ Q_H - Q_C = W_{net}. \] Also, we know that \( \eta = \frac{W_{net}}{Q_H} \), so \( W_{net} = \eta Q_H \). Step 5: Calculate the net work done by the engine.
\[ W_{net} = \eta Q_H = 0.4 \times 250 \, \text{kJ/s} = 100 \, \text{kJ/s}. \] Step 6: Calculate the amount of heat rejected \( Q_C \).
Using the first law of thermodynamics: \[ Q_C = Q_H - W_{net} = 250 \, \text{kJ/s} - 100 \, \text{kJ/s} = 150 \, \text{kJ/s}. \] Alternatively, for a reversible heat engine, the ratio of heat transferred is proportional to the ratio of absolute temperatures: \[ \frac{Q_C}{Q_H} = \frac{T_C}{T_H}. \] \[ Q_C = Q_H \times \frac{T_C}{T_H} = 250 \, \text{kJ/s} \times \frac{300 \, \text{K}}{500 \, \text{K}} = 250 \times 0.6 = 150 \, \text{kJ/s}. \] Step 7: Select the correct answer.
The amount of heat rejected is 150 kJ/s, which corresponds to option 2.