In a heat engine, the temperature of the source and sink are 500 K and 375 K. If the engine consumes `25xx10^5J` per cycle, find(a) the efficiency of the engine, (b) work done per cycle, and (c) heat rejected to the sink per cycle.

Explanation

To solve the problem step by step, we will use the principles of thermodynamics, particularly focusing on the efficiency of a heat engine and the relationships between heat absorbed, work done, and heat rejected.

### Step 1: Calculate the Efficiency of the Engine
The efficiency (\( \eta \)) of a heat engine operating between two temperatures can be calculated using the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where:
- \( T_1 \) is the temperature of the source (500 K)
- \( T_2 \) is the temperature of the sink (375 K)

Substituting the values:
\[ \eta = 1 - \frac{375}{500} = 1 - 0.75 = 0.25 \]
Thus, the efficiency of the engine is:
\[ \eta = 0.25 \text{ or } 25\% \]

### Step 2: Calculate the Work Done per Cycle
The work done (\( W \)) by the engine can be calculated using the efficiency formula:
\[ \eta = \frac{W}{Q_1} \]
where:
- \( Q_1 \) is the heat absorbed (given as \( 25 \times 10^5 \) J)

Rearranging the formula to find work done:
\[ W = \eta \times Q_1 \]
Substituting the values:
\[ W = 0.25 \times 25 \times 10^5 = 6.25 \times 10^5 \text{ J} \]

### Step 3: Calculate the Heat Rejected to the Sink per Cycle
The heat rejected to the sink (\( Q_2 \)) can be calculated using the relationship:
\[ Q_2 = Q_1 - W \]
Substituting the values:
\[ Q_2 = 25 \times 10^5 - 6.25 \times 10^5 = 18.75 \times 10^5 \text{ J} \]

### Summary of Results
(a) Efficiency of the engine: **25%**
(b) Work done per cycle: **\( 6.25 \times 10^5 \) J**
(c) Heat rejected to the sink per cycle: **\( 18.75 \times 10^5 \) J**