A carnot engine operates with source at `127^(@)C` and sink at `27^(@)C`. If the source supplies `40kJ` of heat energy. The work done by the engine is

Explanation

To solve the problem regarding the Carnot engine, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures given are:
- Source temperature (T1) = 127°C
- Sink temperature (T2) = 27°C

To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating:
- \( T1 = 127 + 273 = 400 \, K \)
- \( T2 = 27 + 273 = 300 \, K \)

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T2}{T1} \]

Substituting the values we calculated:
\[ \eta = 1 - \frac{300}{400} \]
\[ \eta = 1 - 0.75 = 0.25 \]

### Step 3: Use the efficiency to find the work done
The efficiency can also be expressed in terms of work done (W) and heat supplied (Q):
\[ \eta = \frac{W}{Q} \]
Where:
- \( Q = 40 \, kJ \) (heat supplied)

Rearranging the formula to find work done:
\[ W = \eta \times Q \]
Substituting the values:
\[ W = 0.25 \times 40 \, kJ \]
\[ W = 10 \, kJ \]

### Conclusion
The work done by the engine is \( 10 \, kJ \).

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