Question

A reversible engine is being operated with the temperature limits of 800 K and 300 K. If it takes heat 560 kJ, then the available energy and unavailable energy are

• A. \( 350 \text{ kJ and } 210 \text{ kJ} \)
• B. \( 300 \text{ kJ and } 260 \text{ kJ} \)
• C. \( 210 \text{ kJ and } 350 \text{ kJ} \)
• D. \( 260 \text{ kJ and } 300 \text{ kJ} \)

Hint:

For a reversible (Carnot) engine, the thermal efficiency is determined by the temperature limits \( \eta_{th} = 1 - T_L/T_H \). Available energy is the work output \( W = \eta_{th} Q_H \), and unavailable energy is the heat rejected \( Q_L = Q_H - W \) (or \( Q_L = Q_H (T_L/T_H) \)).

Answer 1

The Correct Option is A

Step 1: Identify the given parameters for the reversible engine (Carnot engine).
High temperature reservoir \( T_H = 800 \text{ K} \).
Low temperature reservoir \( T_L = 300 \text{ K} \).
Heat taken by the engine (heat supplied) \( Q_H = 560 \text{ kJ} \).
Step 2: Calculate the thermal efficiency of the reversible engine.
For a reversible engine (Carnot engine), the thermal efficiency \( \eta_{th} \) is given by: \[ \eta_{th} = 1 - \frac{T_L}{T_H} \] \[ \eta_{th} = 1 - \frac{300 \text{ K}}{800 \text{ K}} = 1 - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8} \] \[ \eta_{th} = 0.625 \]
Step 3: Calculate the available energy (work done) by the engine.
Available energy is the maximum possible work that can be obtained from the heat supplied, which for a reversible engine is the work output \( W \). \[ W = \eta_{th} \times Q_H \] \[ W = 0.625 \times 560 \text{ kJ} = \frac{5}{8} \times 560 \text{ kJ} \] \[ W = 5 \times 70 \text{ kJ} = 350 \text{ kJ} \] So, the available energy is 350 kJ.
Step 4: Calculate the unavailable energy (heat rejected) by the engine.
Unavailable energy is the portion of the heat supplied that cannot be converted into useful work, which is the heat rejected to the low-temperature reservoir \( Q_L \).
According to the first law of thermodynamics for a cyclic process: \( Q_H = W + Q_L \)
Therefore, \( Q_L = Q_H - W \). \[ Q_L = 560 \text{ kJ} - 350 \text{ kJ} = 210 \text{ kJ} \] Alternatively, for a reversible engine, the ratio of heat rejected to heat supplied is equal to the ratio of absolute temperatures: \[ \frac{Q_L}{Q_H} = \frac{T_L}{T_H} \] \[ Q_L = Q_H \times \frac{T_L}{T_H} = 560 \text{ kJ} \times \frac{300 \text{ K}}{800 \text{ K}} \] \[ Q_L = 560 \times \frac{3}{8} = 70 \times 3 = 210 \text{ kJ} \] So, the unavailable energy is 210 kJ. The available energy is 350 kJ and the unavailable energy is 210 kJ. The final answer is $\boxed{\text{1}}$.