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. efficiency of Carnot engine cannot be made larger than 50%
B. 1200 K
C. 750 K
D. 600 K

Answer

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

\[ \eta = 1 - \frac{T_L}{T_H} \]

where:
- \(\eta\) is the efficiency of the engine,
- \(T_H\) is the temperature of the hot reservoir (source),
- \(T_L\) is the temperature of the cold reservoir (sink).

### Step 1: Calculate the cold reservoir temperature \(T_L\) for the first engine

Given:
- Efficiency \(\eta = 40\% = 0.4\)
- Hot reservoir temperature \(T_H = 500 \, K\)

Using the efficiency formula:

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

Rearranging the equation to find \(T_L\):

\[ \frac{T_L}{500} = 1 - 0.4 = 0.6 \]

\[ T_L = 500 \times 0.6 = 300 \, K \]

### Step 2: Calculate the new hot reservoir temperature \(T_H'\) for the desired efficiency

Now, we want to find the new hot reservoir temperature \(T_H'\) for an efficiency of \(\eta' = 60\% = 0.6\).

Using the same efficiency formula:

\[ 0.6 = 1 - \frac{T_L}{T_H'} \]

We already found \(T_L = 300 \, K\), so substituting that in:

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

Rearranging to find \(T_H'\):

\[ \frac{300}{T_H'} = 1 - 0.6 = 0.4 \]

\[ T_H' = \frac{300}{0.4} = 750 \, K \]

### Final Answer

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

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