Question

If the temperature of the source in a Carnot engine is increased while the sink temperature remains constant, the efficiency will:

• A. Increase
• B. Decrease
• C. Remain the same
• D. Become zero

Hint:

Important Points about Carnot Engine: Efficiency depends only on reservoir temperatures. Higher source temperature $\Rightarrow$ higher efficiency. Lower sink temperature $\Rightarrow$ higher efficiency.

Answer 1

The Correct Option is A

Concept: A Carnot engine is an ideal heat engine that operates between two heat reservoirs: a hot reservoir (source) at temperature $T_H$ and a cold reservoir (sink) at temperature $T_C$. The efficiency of a Carnot engine depends only on the temperatures of these reservoirs. The efficiency of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_C}{T_H} \] where $T_H$ = temperature of the hot reservoir (source)
$T_C$ = temperature of the cold reservoir (sink) Temperatures must be expressed in Kelvin scale.

Step 1: Observe the efficiency formula. \[ \eta = 1 - \frac{T_C}{T_H} \] If the sink temperature $T_C$ remains constant and the source temperature $T_H$ increases, the fraction $\frac{T_C}{T_H}$ becomes smaller.

Step 2: Effect on efficiency. As $\frac{T_C}{T_H}$ decreases, the value of $1 - \frac{T_C}{T_H}$ increases. Therefore, the efficiency $\eta$ increases.

Step 3: Physical interpretation. A higher source temperature means the engine receives heat at a higher thermal energy level, allowing more of that energy to be converted into useful work before being rejected to the sink. Thus, increasing the temperature of the source while keeping the sink temperature constant increases the efficiency of the Carnot engine.