Question

A Carnot engine operating between temperatures \(600\,K\) and \(300\,K\), absorbs \(800\,J\) of heat from the source. The mechanical work done per cycle is

• A. \(400\,J\)
• B. \(650\,J\)
• C. \(750\,J\)
• D. \(600\,J\)

Hint:

For a Carnot engine, efficiency depends only on the temperatures of the hot and cold reservoirs: \[ \eta = 1-\frac{T_C}{T_H} \] Then use \(W=\eta Q_H\) to find the work done.

Answer 1

The Correct Option is A

Step 1: Recall the Carnot efficiency formula.
A Carnot engine is the most efficient possible heat engine. Its efficiency depends only on the temperatures of the hot and cold reservoirs:
\[ \eta = 1 - \frac{T_C}{T_H} \]

Step 2: Substitute the given temperatures.
$T_H = 600\,\text{K}$ (hot source) and $T_C = 300\,\text{K}$ (cold sink):
\[ \eta = 1 - \frac{300}{600} = 1 - \frac{1}{2} = \frac{1}{2} \]

Step 3: Find the heat rejected to the cold reservoir.
The heat rejected is proportional to temperatures:
\[ Q_C = Q_H \times \frac{T_C}{T_H} = 800 \times \frac{300}{600} = 800 \times \frac{1}{2} = 400\,\text{J} \]

Step 4: Apply the first law of thermodynamics to find work.
Energy conservation for the engine: all the heat absorbed is either converted to work or rejected:
\[ W = Q_H - Q_C = 800 - 400 = 400\,\text{J} \]

Step 5: Verify using the efficiency formula.
\[ \eta = \frac{W}{Q_H} = \frac{400}{800} = 0.5 \checkmark \]

Step 6: State the final answer.
\[ \boxed{400\,\text{J}} \]