In Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency 50% , what will be the temperature of hot reservoir?

Options

A. `(2T)/5`
B. `6T`
C. `(6T)/5`
D. `T/5`

Answer

To solve the problem regarding the Carnot engine's efficiency and the relationship between the temperatures of the hot and cold reservoirs, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Efficiency in a Carnot Engine**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]
where \( T_H \) is the temperature of the hot reservoir and \( T_L \) is the temperature of the cold reservoir.

2. **First Situation**:
In the first situation, we know that the efficiency is 40% (or 0.4) at the hot reservoir temperature \( T \).
\[ \eta_1 = 0.4 \]
Substituting this into the efficiency formula:
\[ 0.4 = 1 - \frac{T_L}{T} \]
Rearranging gives:
\[ \frac{T_L}{T} = 1 - 0.4 = 0.6 \]
Thus, we can express \( T_L \) in terms of \( T \):
\[ T_L = 0.6T \]

3. **Second Situation**:
In the second situation, we want to find the new temperature \( T' \) of the hot reservoir when the efficiency is 50% (or 0.5).
\[ \eta_2 = 0.5 \]
Using the efficiency formula again:
\[ 0.5 = 1 - \frac{T_L}{T'} \]
Rearranging gives:
\[ \frac{T_L}{T'} = 1 - 0.5 = 0.5 \]
Thus, we can express \( T' \) in terms of \( T_L \):
\[ T' = \frac{T_L}{0.5} = 2T_L \]

4. **Substituting \( T_L \)**:
Now substitute \( T_L = 0.6T \) into the equation for \( T' \):
\[ T' = 2(0.6T) = 1.2T \]

5. **Final Result**:
Therefore, the temperature of the hot reservoir for an efficiency of 50% is:
\[ T' = 1.2T \]