Question

A Carnot engine operates between a heat source at $T_1 = 600\text{ K}$ and a sink at $T_2 = 300\text{ K}$. If the engine absorbs $1000\text{ J}$ of heat from the source per cycle, the work done per cycle is:

• A. $500\text{ J}$
• B. $1000\text{ J}$
• C. $250\text{ J}$
• D. $750\text{ J}$

Hint:

The temperature of the sink is exactly half of the source temperature, giving an efficiency of exactly $50\%$. Thus, half of the absorbed heat ($500\text{ J}$) is converted into useful work.

Answer 1

The Correct Option is A

Step 1: Understand a Carnot engine.
A Carnot engine takes heat from a hot source, does some useful work, and dumps the rest into a cold sink. Its efficiency tells what fraction of the heat turns into work.

Step 2: Write the efficiency formula.
For a Carnot engine, \[ \eta = 1 - \frac{T_2}{T_1} \] where $T_1$ is the source temperature and $T_2$ is the sink temperature, both in kelvin.

Step 3: Put in the numbers.
Here $T_1 = 600$ K and $T_2 = 300$ K. \[ \eta = 1 - \frac{300}{600} = 1 - \frac{1}{2} = \frac{1}{2} \] So the engine is fifty percent efficient.

Step 4: Link efficiency to work.
Efficiency is also the ratio of work done to heat absorbed. \[ \eta = \frac{W}{Q_1} \]

Step 5: Solve for the work.
The engine absorbs $Q_1 = 1000$ J. \[ W = \eta\, Q_1 = \frac{1}{2}\times 1000 = 500 \text{ J} \]

Step 6: State the answer.
Half of the absorbed heat becomes work. \[ \boxed{W = 500 \text{ J}} \]