Efficiency of a Carnot engine is `50%` when temperature of outlet is `500 K`. In order to increase efficiency up to `60%` keeping temperature of intake the same what is temperature of outlet?

Options

A. `200 K`
B. `400 K`
C. `600 K`
D. `800 K`

Explanation

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

\[ \text{Efficiency} = 1 - \frac{T_2}{T_1} \]

where:
- \( T_1 \) is the temperature of the hot reservoir (intake temperature),
- \( T_2 \) is the temperature of the cold reservoir (outlet temperature).

### Step 1: Identify the given values
From the problem, we know:
- Initial efficiency \( = 50\% = \frac{1}{2} \)
- Initial outlet temperature \( T_2 = 500 \, K \)

### Step 2: Calculate \( T_1 \) using the initial efficiency
Using the efficiency formula:
\[ \frac{1}{2} = 1 - \frac{T_2}{T_1} \]

Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{2} = \frac{1}{2} \]

Substituting \( T_2 = 500 \, K \):
\[ \frac{500}{T_1} = \frac{1}{2} \]

Cross-multiplying:
\[ 500 = \frac{1}{2} T_1 \implies T_1 = 1000 \, K \]

### Step 3: Set up the equation for the new efficiency
Now, we want to find the new outlet temperature \( T_2' \) when the efficiency is increased to \( 60\% = 0.6 \).

Using the efficiency formula again:
\[ 0.6 = 1 - \frac{T_2'}{T_1} \]

Rearranging gives:
\[ \frac{T_2'}{T_1} = 1 - 0.6 = 0.4 \]

### Step 4: Substitute \( T_1 \) to find \( T_2' \)
Substituting \( T_1 = 1000 \, K \):
\[ \frac{T_2'}{1000} = 0.4 \]

Cross-multiplying gives:
\[ T_2' = 0.4 \times 1000 = 400 \, K \]

### Conclusion
The new outlet temperature \( T_2' \) required to achieve an efficiency of \( 60\% \) while keeping the intake temperature the same is:
\[ \boxed{400 \, K} \]