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 xx10^6 J`
B. `B.4 xx 10^6 J`
C. `16.B xx10^6 J`
D. zero

Explanation

To solve the problem, we need to find the work done by a Carnot engine that operates between two thermal reservoirs. The steps to find the work done are as follows:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures need to be in Kelvin for the Carnot efficiency formula.

- **T1 (hot reservoir)** = 627°C + 273 = 900 K
- **T2 (cold reservoir)** = 27°C + 273 = 300 K

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:

\[ \eta = 1 - \frac{T2}{T1} \]

Substituting the values we found:

\[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.67 \]

### Step 3: Relate efficiency to work done
The efficiency can also be expressed in terms of work done (W) and heat absorbed (Q1):

\[ \eta = \frac{W}{Q1} \]

Rearranging this gives:

\[ W = \eta \times Q1 \]

### Step 4: Substitute the values to find work done
We know that \( Q1 = 3 \times 10^6 \) cal. First, we need to convert calories to joules (1 cal = 4.184 J):

\[ Q1 = 3 \times 10^6 \text{ cal} \times 4.184 \text{ J/cal} = 12.552 \times 10^6 \text{ J} \]

Now substituting the values into the work done formula:

\[ W = \frac{2}{3} \times 12.552 \times 10^6 \text{ J} \approx 8.368 \times 10^6 \text{ J} \]

### Final Answer
The work done by the engine is approximately \( 8.368 \times 10^6 \) J.

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