A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:

Options

A. 1200K
B. 750 K
C. 600 K
D. 800 K

Explanation

To solve the problem, we will use the efficiency formula of a Carnot engine, which is given by:

\[ \eta = 1 - \frac{T_2}{T_1} \]

where:
- \(\eta\) is the efficiency of the engine,
- \(T_1\) is the temperature of the heat source,
- \(T_2\) is the temperature of the heat sink.

### Step 1: Calculate the sink temperature \(T_2\) for the initial engine

Given:
- Efficiency \(\eta = 40\% = 0.4\)
- Heat source temperature \(T_1 = 500 \, K\)

Using the efficiency formula:

\[ 0.4 = 1 - \frac{T_2}{500} \]

Rearranging the equation to find \(T_2\):

\[ \frac{T_2}{500} = 1 - 0.4 \]
\[ \frac{T_2}{500} = 0.6 \]
\[ T_2 = 0.6 \times 500 = 300 \, K \]

### Step 2: Calculate the intake temperature \(T_1'\) for the new engine

Now we want to find the new intake temperature \(T_1'\) for the desired efficiency of \(60\% = 0.6\), while keeping the same sink temperature \(T_2 = 300 \, K\).

Using the efficiency formula again:

\[ 0.6 = 1 - \frac{300}{T_1'} \]

Rearranging the equation to find \(T_1'\):

\[ \frac{300}{T_1'} = 1 - 0.6 \]
\[ \frac{300}{T_1'} = 0.4 \]
\[ T_1' = \frac{300}{0.4} = 750 \, K \]

### Final Answer

The intake temperature for the engine with \(60\%\) efficiency, while maintaining the same exhaust temperature, must be:

\[ \boxed{750 \, K} \]