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.

Options

A. `6.25xx10^(5) J`
B. `3xx10^(5) J`
C. `2.19xx10^(5) J`
D. `4xx10^(4) J`

Explanation

To solve the problem step by step, we will find the efficiency of the heat engine, the work done per cycle, and the heat rejected to the sink per cycle.

### Step 1: Calculate the Efficiency of the Engine
The efficiency (η) of a heat engine can be calculated using the formula:

\[ η = 1 - \frac{T_2}{T_1} \]

Where:
- \(T_1\) is the temperature of the source (500 K)
- \(T_2\) is the temperature of the sink (375 K)

Substituting the values:

\[ η = 1 - \frac{375}{500} \]

Calculating the fraction:

\[ η = 1 - 0.75 = 0.25 \]

Thus, the efficiency of the engine is:

\[ η = 0.25 \text{ or } 25\% \]

### Step 2: Calculate the Work Done per Cycle
The work done (W) by the engine can be calculated using the efficiency and the heat absorbed from the source (Q1):

\[ W = η \times Q_1 \]

Given that \(Q_1 = 25 \times 10^5 \, J\):

Substituting the values:

\[ W = 0.25 \times (25 \times 10^5) \]

Calculating:

\[ W = 6.25 \times 10^5 \, J \]

Thus, the work done per cycle is:

\[ W = 6.25 \times 10^5 \, J \]

### Step 3: Calculate the Heat Rejected to the Sink per Cycle
The heat rejected to the sink (Q2) can be calculated using the relationship:

\[ Q_2 = Q_1 - W \]

Substituting the values:

\[ Q_2 = (25 \times 10^5) - (6.25 \times 10^5) \]

Calculating:

\[ Q_2 = 18.75 \times 10^5 \, J \]

Thus, the heat rejected to the sink per cycle is:

\[ Q_2 = 18.75 \times 10^5 \, J \]

### 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: \(18.75 \times 10^5 \, J\)

---