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. `1200K`
C. `750K`
D. `600K`

Explanation

To solve the problem, we need to find the new intake temperature \( T_1' \) for a Carnot engine with an efficiency of 60%, given that the exhaust temperature \( T_2 \) remains the same at 300 K.

### Step-by-Step Solution:

1. **Understand the Efficiency Formula**:
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the heat source and \( T_2 \) is the temperature of the heat sink.

2. **Calculate \( T_2 \) from the First Engine**:
We know from the problem that the efficiency of the first engine is \( \eta_1 = 40\% = 0.4 \) and \( T_1 = 500 \, K \).
Using the efficiency formula:
\[ 0.4 = 1 - \frac{T_2}{500} \]
Rearranging gives:
\[ \frac{T_2}{500} = 1 - 0.4 = 0.6 \]
Therefore, we can find \( T_2 \):
\[ T_2 = 0.6 \times 500 = 300 \, K \]

3. **Set Up the Equation for the New Engine**:
Now, we want to find \( T_1' \) for the second engine with an efficiency \( \eta_2 = 60\% = 0.6 \) and the same \( T_2 = 300 \, K \).
Using the efficiency formula again:
\[ 0.6 = 1 - \frac{300}{T_1'} \]
Rearranging gives:
\[ \frac{300}{T_1'} = 1 - 0.6 = 0.4 \]

4. **Solve for \( T_1' \)**:
From the equation above, we can solve for \( T_1' \):
\[ T_1' = \frac{300}{0.4} \]
Calculating this gives:
\[ T_1' = 750 \, K \]

### Final Answer:
The intake temperature for the engine with 60% efficiency, while keeping the exhaust temperature the same, must be \( T_1' = 750 \, K \).

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