Question

The efficiency of a Carnot engine working between temperatures \(T_1\) and \(T_2\) (where \(T_1 > T_2\)) is given by:

• A. \( \eta = 1 - \frac{T_1}{T_2} \)
• B. \( \eta = 1 - \frac{T_2}{T_1} \)
• C. \( \eta = \frac{T_1}{T_2} - 1 \)
• D. \( \eta = \frac{T_2}{T_1} \)

Hint:

Always remember: efficiency depends only on temperature ratio for Carnot engines, not on working substance.

Answer 1

The Correct Option is B

Step 1: Understand a Carnot engine.
A Carnot engine is an ideal heat engine. It takes heat from a hot source at temperature $T_1$ and dumps some heat into a cold sink at temperature $T_2$, with $T_1 > T_2$. It gives the highest possible efficiency.

Step 2: Define efficiency.
Efficiency is the useful work divided by the heat taken in: \[ \eta = \frac{W}{Q_1} \]

Step 3: Use the first law.
The work done equals the heat taken in minus the heat given out: \[ W = Q_1 - Q_2 \] So the efficiency becomes: \[ \eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} \]

Step 4: Apply the Carnot condition.
For a reversible Carnot engine, the heat ratio equals the temperature ratio: \[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \]

Step 5: Put this into the efficiency.
Replace the heat ratio with the temperature ratio: \[ \eta = 1 - \frac{T_2}{T_1} \]

Step 6: Check the form.
Note the cold temperature is on top and the hot one on the bottom, so the efficiency stays between $0$ and $1$. This matches the second option. \[ \boxed{\eta = 1 - \dfrac{T_2}{T_1}} \]