Carnot engine is an ideal heat engine, which converts heat energy into mechanical energy. Efficiency of Carnot engine is given by `eta =1 - (T_2 // T_1)`, where `T_1` is temperature of source and `T_1`, is temperature of sink. If `Q_1` is the amount of heat absorbed/cycle from the source, `Q_2` is the amount of heat rejected/cycle to the sink and W is the amount of useful work done/cycle, then
` W= Q_1 -Q_2 and (Q_2)/(Q_1) = (T_2)/(T_1)`
Work done/cycle by the engine in the above question is. if source and sink temperature respectively 227 and 127 . heat absorbed by engine is given 6 .

Options

A. `5.04 xx 10^5 ` J
B. `5.04 xx 10^5 ` cal
C. `4.8 xx 10 ^5 ` cal
D. `4.8 xx 10^5 J`

Answer

To solve the problem regarding the work done per cycle by a Carnot engine, we will follow these steps:

### Step 1: Identify the given values
- Temperature of the source \( T_1 = 227^\circ C \)
- Temperature of the sink \( T_2 = 127^\circ C \)
- Heat absorbed from the source \( Q_1 = 6 \times 10^5 \) calories

### Step 2: Convert temperatures to Kelvin
To use the Carnot efficiency formula, we need to convert the temperatures from Celsius to Kelvin:
\[ T_1 = 227 + 273 = 500 \text{ K} \]
\[ T_2 = 127 + 273 = 400 \text{ K} \]

### Step 3: Calculate the efficiency of the Carnot engine
The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substituting the values:
\[ \eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2 \]

### Step 4: Calculate the work done per cycle
The work done \( W \) by the engine can be calculated using the formula:
\[ W = Q_1 \cdot \eta \]
Substituting the values:
\[ W = 6 \times 10^5 \text{ calories} \cdot 0.2 = 1.2 \times 10^5 \text{ calories} \]

### Step 5: Convert work done from calories to Joules
Since the problem states that the answer should be in Joules, we convert calories to Joules using the conversion factor \( 1 \text{ calorie} = 4.2 \text{ Joules} \):
\[ W = 1.2 \times 10^5 \text{ calories} \times 4.2 \text{ Joules/calorie} = 5.04 \times 10^5 \text{ Joules} \]

### Final Answer
The work done per cycle by the Carnot engine is:
\[ W = 5.04 \times 10^5 \text{ Joules} \]

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