A reversible engine has an efficiency of `1/4` If the temperature of the sink is reduced by `58^(@)` C, its efficiency becomes double. Calculate the temperature of the sink :

Options

A. `280^(@)` C
B. `382^(@)`C
C. `174^(@)` C
D. `180.4^(@)`C

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the efficiency formulas
The efficiency \( \eta \) of a reversible engine is given by:
\[ \eta = 1 - \frac{T_s}{T_h} \]
where \( T_s \) is the temperature of the sink and \( T_h \) is the temperature of the hot reservoir.

### Step 2: Set up the initial condition
Given that the initial efficiency \( \eta_1 = \frac{1}{4} \), we can substitute this into the efficiency formula:
\[ \frac{1}{4} = 1 - \frac{T_s}{T_h} \]
Rearranging gives:
\[ \frac{T_s}{T_h} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, we have:
\[ T_s = \frac{3}{4} T_h \quad \text{(Equation 1)} \]

### Step 3: Set up the second condition
When the temperature of the sink is reduced by \( 58^\circ C \), the new efficiency becomes:
\[ \eta_2 = 2 \times \frac{1}{4} = \frac{1}{2} \]
Using the efficiency formula again:
\[ \frac{1}{2} = 1 - \frac{T_s - 58}{T_h} \]
Rearranging gives:
\[ \frac{T_s - 58}{T_h} = 1 - \frac{1}{2} = \frac{1}{2} \]
Thus, we have:
\[ T_s - 58 = \frac{1}{2} T_h \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we know \( T_s = \frac{3}{4} T_h \). Substituting this into Equation 2:
\[ \frac{3}{4} T_h - 58 = \frac{1}{2} T_h \]

### Step 5: Solve for \( T_h \)
Rearranging gives:
\[ \frac{3}{4} T_h - \frac{1}{2} T_h = 58 \]
Finding a common denominator (which is 4):
\[ \frac{3}{4} T_h - \frac{2}{4} T_h = 58 \]
This simplifies to:
\[ \frac{1}{4} T_h = 58 \]
Multiplying both sides by 4:
\[ T_h = 232 \, \text{°C} \]

### Step 6: Calculate \( T_s \)
Now substituting \( T_h \) back into Equation 1 to find \( T_s \):
\[ T_s = \frac{3}{4} \times 232 = 174 \, \text{°C} \]

### Step 7: Final answer
Thus, the temperature of the sink \( T_s \) is:
\[ \boxed{174 \, \text{°C}} \]

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