An ideal engine whose low-temperature reservoir is at `27^@C` has an efficiency of 40%. By how much should the temperature of the high-temperature reservoir be increased so as to increase the efficiency to 50% ?

Options

A. `100^@C`
B. `600^@C`
C. `375K`
D. `100K`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the efficiency formula
The efficiency (η) of an ideal engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_2 \) is the temperature of the low-temperature reservoir and \( T_1 \) is the temperature of the high-temperature reservoir.

### Step 2: Convert the low-temperature reservoir temperature to Kelvin
The low-temperature reservoir is given as \( 27^\circ C \). To convert this to Kelvin:
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 3: Calculate the initial high-temperature reservoir temperature (T1)
Given that the initial efficiency is \( 40\% \) (or \( 0.4 \)):
\[ 0.4 = 1 - \frac{300}{T_1} \]
Rearranging the equation gives:
\[ \frac{300}{T_1} = 1 - 0.4 = 0.6 \]
Now, solving for \( T_1 \):
\[ T_1 = \frac{300}{0.6} = 500 \, K \]

### Step 4: Set up the equation for the new efficiency
We want to increase the efficiency to \( 50\% \) (or \( 0.5 \)). Let the new high-temperature reservoir temperature be \( T_1 + \Delta T \):
\[ 0.5 = 1 - \frac{300}{T_1 + \Delta T} \]
Rearranging gives:
\[ \frac{300}{T_1 + \Delta T} = 0.5 \]
Now, solving for \( T_1 + \Delta T \):
\[ T_1 + \Delta T = \frac{300}{0.5} = 600 \, K \]

### Step 5: Calculate the change in temperature (ΔT)
Now, substituting the value of \( T_1 \):
\[ 500 + \Delta T = 600 \]
Solving for \( \Delta T \):
\[ \Delta T = 600 - 500 = 100 \, K \]

### Conclusion
The temperature of the high-temperature reservoir should be increased by \( 100 \, K \) to achieve an efficiency of \( 50\% \).