A heat engine has an efficiency `eta`.Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine

Options

A. Increases
B. Decreases
C. Remains constant
D. Becomes 1

Explanation

To solve the problem, we need to analyze how the efficiency of a heat engine changes when the temperatures of the source and sink are both decreased by 100 K.

### Step-by-Step Solution:

1. **Understanding Efficiency of a Heat Engine**:
The efficiency (η) of a heat engine is given by the formula:
\[ \eta = \frac{T_1 - T_2}{T_1} \]
where \( T_1 \) is the temperature of the source and \( T_2 \) is the temperature of the sink.

2. **Initial Conditions**:
Let’s denote the initial temperatures of the source and sink as \( T_1 \) and \( T_2 \) respectively. The initial efficiency can be expressed as:
\[ \eta = \frac{T_1 - T_2}{T_1} \]

3. **Change in Temperatures**:
According to the problem, both temperatures are decreased by 100 K. Therefore, the new temperatures will be:
\[ T_1' = T_1 - 100 \]
\[ T_2' = T_2 - 100 \]

4. **New Efficiency Calculation**:
The new efficiency \( \eta' \) can be calculated using the new temperatures:
\[ \eta' = \frac{T_1' - T_2'}{T_1'} \]
Substituting the new temperatures:
\[ \eta' = \frac{(T_1 - 100) - (T_2 - 100)}{T_1 - 100} \]
Simplifying this gives:
\[ \eta' = \frac{T_1 - T_2}{T_1 - 100} \]

5. **Comparison of Efficiencies**:
Now, we need to compare the original efficiency \( \eta \) and the new efficiency \( \eta' \). We have:
\[ \eta = \frac{T_1 - T_2}{T_1} \]
\[ \eta' = \frac{T_1 - T_2}{T_1 - 100} \]
Since \( T_1 - 100 < T_1 \), the denominator of \( \eta' \) is smaller than that of \( \eta \), which means \( \eta' \) will be greater than \( \eta \).

6. **Conclusion**:
Therefore, the efficiency of the heat engine increases when both the source and sink temperatures are decreased by 100 K.

### Final Answer:
The efficiency of the heat engine increases.