An engine has an efficiency of 0.25 when temperature of sink is reduced by `58^(@)C` , If its efficiency is doubled, then the temperature of the source is

Options

A. `150 ^(@)C`
B. `222 ^(@)C`
C. `242 ^(@)C`
D. `232 ^(@)C`

Explanation

To solve the problem step by step, we will follow the principles of thermodynamics related to the efficiency of a heat engine.

### Step 1: Understand the efficiency formula
The efficiency (η) of a heat engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where:
- \(T_1\) is the temperature of the source (hot reservoir),
- \(T_2\) is the temperature of the sink (cold reservoir).

### Step 2: Set up the initial condition
We are given that the initial efficiency is \(0.25\):
\[ \eta = 0.25 = 1 - \frac{T_2}{T_1} \]
This can be rewritten as:
\[ \frac{T_2}{T_1} = 1 - 0.25 = 0.75 \]
Thus, we can express this as:
\[ T_2 = 0.75 T_1 \quad \text{(Equation 1)} \]

### Step 3: Analyze the change in efficiency
The problem states that the efficiency is doubled when the temperature of the sink is reduced by \(58^\circ C\). Therefore, the new efficiency (\(\eta'\)) is:
\[ \eta' = 2 \times 0.25 = 0.5 \]
Using the efficiency formula again for the new conditions:
\[ \eta' = 1 - \frac{T_2 - 58}{T_1} \]
Substituting \(\eta' = 0.5\):
\[ 0.5 = 1 - \frac{T_2 - 58}{T_1} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \frac{T_2 - 58}{T_1} = 0.5 \]
This can be rewritten as:
\[ T_2 - 58 = 0.5 T_1 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, we know \(T_2 = 0.75 T_1\). Substituting this into Equation 2:
\[ 0.75 T_1 - 58 = 0.5 T_1 \]

### Step 6: Solve for \(T_1\)
Now, simplify the equation:
\[ 0.75 T_1 - 0.5 T_1 = 58 \]
\[ 0.25 T_1 = 58 \]
\[ T_1 = \frac{58}{0.25} = 232 \text{ K} \]

### Conclusion
The temperature of the source \(T_1\) is \(232 \text{ K}\).