A system comprising of 2 kg of fluid undergoes a reversible process at 500 K during which it receives 1 MJ of heat. The change in specific entropy of system for this case and for the other case where heat supply had been doubled for the same initial and final states are

Concept:

To determine the change in specific entropy of a system undergoing a reversible process, we use the relationship between heat transfer and temperature.

Calculation:

Given:

• Mass of fluid, \( m = 2 \, \text{kg} \)
• Temperature, \( T = 500 \, \text{K} \)
• Heat received in the first case, \( Q_1 = 1 \, \text{MJ} = 1000 \, \text{kJ} \)
• Heat received in the second case, \( Q_2 = 2 \times Q_1 = 2000 \, \text{kJ} \)

The formula for the change in entropy is:

\( \Delta S = \frac{Q_{\text{rev}}}{m \cdot T} \)

Calculation:

For the first case, where the system receives 1000 kJ of heat:

\( \Delta S_1 = \frac{1000 \, \text{kJ}}{2 \, \text{kg} \times 500 \, \text{K}} = \frac{1000}{1000} = 1 \, \text{kJ/kg·K} \)

For the second case, where the heat supply is doubled to 2000 kJ:

\( \Delta S_2 = \frac{2000 \, \text{kJ}}{2 \, \text{kg} \times 500 \, \text{K}} = \frac{2000}{1000} = 2 \, \text{kJ/kg·K} \)

Therefore, the change in specific entropy for the two cases is:

\( 1 \, \text{kJ/kg·K} \text{ and } 2 \, \text{kJ/kg·K} \text{ respectively.} \)