In a Carnot engine , the temperature of the reservoir is `927^@C ` and that of sink is `127^@C` . If the work done by the engine when it transfers heat from the reservoir to sink is `12.6xx10^6 J` . the quantity of heat absorbed by the engine from the reservoir is

Options

A. `18.9xx10^6J`
B. `20.5xx10^6J`
C. `15.7xx10^6J`
D. `12.6xx10^6J`

Answer

To find the quantity of heat absorbed by the Carnot engine from the reservoir, we can use the relationship between the work done by the engine, the heat absorbed from the hot reservoir, and the heat rejected to the cold sink.

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- The temperature of the hot reservoir (T1) is given as \(927^\circ C\).
- The temperature of the cold sink (T2) is given as \(127^\circ C\).

Convert these temperatures to Kelvin:
\[ T_1 = 927 + 273 = 1200 \, K \]
\[ T_2 = 127 + 273 = 400 \, K \]

2. **Use the Efficiency Formula**:
The efficiency (\(\eta\)) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substitute the values of \(T_1\) and \(T_2\):
\[ \eta = 1 - \frac{400}{1200} = 1 - \frac{1}{3} = \frac{2}{3} \]

3. **Relate Efficiency to Work and Heat**:
The efficiency can also be expressed in terms of work done (W) and heat absorbed (Q1):
\[ \eta = \frac{W}{Q_1} \]
Rearranging this gives:
\[ Q_1 = \frac{W}{\eta} \]

4. **Substitute the Values**:
Given that the work done (W) is \(12.6 \times 10^6 \, J\):
\[ Q_1 = \frac{12.6 \times 10^6}{\frac{2}{3}} = 12.6 \times 10^6 \times \frac{3}{2} \]

5. **Calculate Q1**:
\[ Q_1 = 12.6 \times 10^6 \times 1.5 = 18.9 \times 10^6 \, J \]

Thus, the quantity of heat absorbed by the engine from the reservoir is \(18.9 \times 10^6 \, J\).