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 of finding the work done by a Carnot engine operating between two temperatures, we can follow these steps:

### Step 1: Convert Temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin. The conversion formula is:
\[ T(K) = T(°C) + 273.15 \]

- For the source temperature \( T_1 = 127°C \):
\[ T_1 = 127 + 273.15 = 400.15 \, K \]

- For the sink temperature \( T_2 = 27°C \):
\[ T_2 = 27 + 273.15 = 300.15 \, K \]

### Step 2: Calculate the Efficiency of the Carnot Engine
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]

Substituting the values we found:
\[ \eta = 1 - \frac{300.15}{400.15} \]

Calculating the fraction:
\[ \frac{300.15}{400.15} \approx 0.750 \]

Thus,
\[ \eta = 1 - 0.750 = 0.250 \]

### Step 3: Calculate the Work Done by the Engine
The work done \( W \) by the engine can be calculated using the efficiency formula:
\[ \eta = \frac{W}{Q} \]
Where \( Q \) is the heat supplied by the source.

Rearranging gives:
\[ W = \eta \times Q \]

Given that \( Q = 40 \, kJ \):
\[ W = 0.250 \times 40 \]

Calculating the work done:
\[ W = 10 \, kJ \]

### Final Answer
The work done by the Carnot engine is **10 kJ**.

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