A carnot engine absorbs `1000J` of heat energy from a reservoir at `127^(@)C` and rejecs `600J` of heat energy during each cycle. Calculate (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle.

Options

A. `70% and -10^@C`
B. `50% and -20^@C`
C. `40% and -33^@C`
D. `20% and -43^@C`

Answer

To solve the problem step by step, we will calculate the efficiency of the Carnot engine, the temperature of the sink, and the amount of useful work done per cycle.

### Step 1: Calculate the Efficiency of the Engine

The efficiency (\( \eta \)) of a Carnot engine is given by the formula:

\[ \eta = 1 - \frac{Q_2}{Q_1} \]

Where:
- \( Q_1 \) is the heat absorbed from the hot reservoir (1000 J)
- \( Q_2 \) is the heat rejected to the cold reservoir (600 J)

Substituting the values:

\[ \eta = 1 - \frac{600}{1000} \]

\[ \eta = 1 - 0.6 = 0.4 \]

To express this as a percentage, we multiply by 100:

\[ \eta = 0.4 \times 100 = 40\% \]

### Step 2: Calculate the Temperature of the Sink

To find the temperature of the sink (\( T_2 \)), we can use the relationship:

\[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \]

Where:
- \( T_1 \) is the temperature of the hot reservoir in Kelvin. First, we convert \( 127^\circ C \) to Kelvin:

\[ T_1 = 127 + 273 = 400 \, K \]

Now, substituting the values into the equation:

\[ \frac{600}{1000} = \frac{T_2}{400} \]

This simplifies to:

\[ 0.6 = \frac{T_2}{400} \]

Now, solving for \( T_2 \):

\[ T_2 = 400 \times 0.6 = 240 \, K \]

To convert \( T_2 \) back to Celsius:

\[ T_2 = 240 - 273 = -33^\circ C \]

### Step 3: Calculate the Amount of Useful Work Done per Cycle

The useful work done (\( W \)) by the Carnot engine is given by:

\[ W = Q_1 - Q_2 \]

Substituting the values:

\[ W = 1000 - 600 = 400 \, J \]

### Summary of Results

1. Efficiency of the engine: \( 40\% \)
2. Temperature of the sink: \( -33^\circ C \)
3. Amount of useful work done per cycle: \( 400 \, J \)