The temperature of source of a Carnot engine of efficiency 20% when the heat exhausted is at 240 K is

Options

A. 1200 K
B. 600 K
C. 540 K
D. 300 K

Answer

To solve the problem of finding the temperature of the source of a Carnot engine with an efficiency of 20% and a heat exhaust temperature of 240 K, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Given Values**:
- Efficiency (η) = 20% = 0.20
- Temperature of the cold reservoir (T_C) = 240 K

3. **Set Up the Equation**:
Plug the known values into the efficiency formula:
\[ 0.20 = 1 - \frac{240}{T_H} \]

4. **Rearrange the Equation**:
To isolate \(T_H\), rearrange the equation:
\[ \frac{240}{T_H} = 1 - 0.20 \]
\[ \frac{240}{T_H} = 0.80 \]

5. **Solve for \(T_H\)**:
Cross-multiply to solve for \(T_H\):
\[ 240 = 0.80 \cdot T_H \]
\[ T_H = \frac{240}{0.80} \]

6. **Calculate \(T_H\)**:
\[ T_H = \frac{240}{0.80} = 300 \text{ K} \]

### Final Answer:
The temperature of the source (T_H) is **300 K**.

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