A Carnot engine takes `3xx10^6` cal of heat from a reservoir at `627^@C` and gives it to a sink at `27^@C`. The work done by the engine is:

Options

A. `4.2 xx 10^(6)J`
B. `8.4 xx 10^(6)J`
C. `16.8 xx 10^(6)J`
D. zero

Explanation

To solve the problem of finding the work done by a Carnot engine that takes \(3 \times 10^6\) calories of heat from a reservoir at \(627^\circ C\) and gives it to a sink at \(27^\circ C\), we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures need to be converted to Kelvin because the Carnot efficiency formula uses absolute temperatures.

- For the source temperature \(T_1\):
\[ T_1 = 627 + 273 = 900 \, K \]

- For the sink temperature \(T_2\):
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency \(\eta\) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substituting the values we found:
\[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Calculate the work done by the engine
The work done \(W\) by the engine can be calculated using the formula:
\[ W = \eta \times Q_1 \]
Where \(Q_1\) is the heat absorbed from the hot reservoir. Given \(Q_1 = 3 \times 10^6\) calories, we first convert this to joules:
\[ Q_1 = 3 \times 10^6 \, \text{cal} \times 4.2 \, \text{J/cal} = 12.6 \times 10^6 \, \text{J} \]

Now substituting the values into the work done formula:
\[ W = \frac{2}{3} \times 12.6 \times 10^6 \, \text{J} = 8.4 \times 10^6 \, \text{J} \]

### Final Answer
The work done by the engine is:
\[ W = 8.4 \times 10^6 \, \text{J} \]

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