A Carnot engine absorbs `6xx10^(5)cal`. At `227^(@)C`. Heat rejected to the sink at `127^(@)C` is

Options

A. `15xx10^(8)J`
B. `15xx10^(4)J`
C. `5xx10^(5)J`
D. `2xx10^(4)J`

Answer

To solve the problem, we need to find the heat rejected to the sink (Q2) by a Carnot engine that absorbs a certain amount of heat (Q1) at a given temperature (T1) and rejects heat at another temperature (T2).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Heat absorbed (Q1) = \(6 \times 10^5 \, \text{cal}\)
- Temperature of the heat source (T1) = \(227^\circ C\)
- Temperature of the heat sink (T2) = \(127^\circ C\)

2. **Convert Temperatures to Kelvin:**
- \(T1 = 227 + 273 = 500 \, K\)
- \(T2 = 127 + 273 = 400 \, K\)

3. **Use the Carnot Efficiency Formula:**
- The relationship between the heat absorbed and the heat rejected in a Carnot engine is given by:
\[ \frac{Q2}{Q1} = \frac{T2}{T1} \]
- Rearranging gives:
\[ Q2 = Q1 \times \frac{T2}{T1} \]

4. **Substitute the Values:**
- Substitute \(Q1\), \(T2\), and \(T1\) into the equation:
\[ Q2 = (6 \times 10^5) \times \frac{400}{500} \]

5. **Calculate Q2:**
- Simplify the fraction:
\[ \frac{400}{500} = 0.8 \]
- Now calculate:
\[ Q2 = (6 \times 10^5) \times 0.8 = 4.8 \times 10^5 \, \text{cal} \]

6. **Conclusion:**
- The heat rejected to the sink (Q2) is \(4.8 \times 10^5 \, \text{cal}\).