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 find the temperature of the source (T1) of a Carnot engine with a given efficiency and the temperature of the heat exhausted (T2), we can use the formula for the efficiency of a Carnot engine:

### Step-by-Step Solution:

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

2. **Convert Efficiency to Decimal**:
The efficiency is given as 20%. We convert this percentage into decimal form:
\[ \eta = \frac{20}{100} = 0.2 \]

3. **Substitute Known Values**:
We know \( T_2 = 240 \, K \) and we have the efficiency \( \eta = 0.2 \). Substitute these values into the efficiency formula:
\[ 0.2 = 1 - \frac{240}{T_1} \]

4. **Rearrange the Equation**:
Rearranging the equation to isolate \( T_1 \):
\[ \frac{240}{T_1} = 1 - 0.2 \]
\[ \frac{240}{T_1} = 0.8 \]

5. **Cross Multiply**:
Cross-multiplying gives:
\[ 240 = 0.8 \cdot T_1 \]

6. **Solve for \( T_1 \)**:
Now, divide both sides by 0.8 to find \( T_1 \):
\[ T_1 = \frac{240}{0.8} \]
\[ T_1 = 300 \, K \]

### Final Answer:
The temperature of the source (T1) is **300 K**.