Question

A Carnot engine operates between the temperatures 850 K and 300 K. The engine performs 1200 J of work each cycle, which takes 0.25 s. The efficiency of this engine is :

• A. $\simeq 50\%$
• B. $\simeq 65\%$
• C. $\simeq 85\%$
• D. $\simeq 90\%$

Hint:

Examiners frequently include "distractor" data in physics problems. If a question explicitly asks for the efficiency of a Carnot engine, completely ignore power, work, time, or heat flow variables if you are already given the two operating temperatures.

Answer 1

The Correct Option is B

Step 1: Concept:
The problem asks for the thermodynamic efficiency of an ideal Carnot heat engine operating between two given temperature reservoirs. Extra information (work done and cycle time) is provided as distractors.

Step 2: Key Formula or Approach:
The theoretical maximum efficiency ($\eta$) of any heat engine operating between a hot reservoir at absolute temperature $T_H$ and a cold reservoir at absolute temperature $T_C$ is given by Carnot's theorem:
\[ \eta = 1 - \frac{T_C}{T_H} \]
This value can be multiplied by 100 to express it as a percentage.

Step 3: Step-by-step Explanation:

• Identify the given variables:
Hot reservoir temperature, $T_H = 850 \text{ K}$
Cold reservoir temperature, $T_C = 300 \text{ K}$
(The work $W = 1200 \text{ J}$ and time $t = 0.25 \text{ s}$ are not needed to calculate Carnot efficiency, as Carnot efficiency depends entirely and exclusively on the reservoir temperatures).

• Substitute the temperature values into the efficiency formula:
\[ \eta = 1 - \frac{300}{850} \]

• Simplify the fraction:
\[ \frac{300}{850} = \frac{30}{85} = \frac{6}{17} \]

• Calculate the final decimal value:
\[ \eta = 1 - \frac{6}{17} = \frac{11}{17} \]
\[ \eta \approx 1 - 0.3529 \approx 0.6471 \]

• Convert to a percentage:
\[ \eta \approx 64.71\% \]

• Looking at the options, $64.71\%$ is closest to $\simeq 65\%$.

Step 4: Final Answer:
The efficiency is approximately $65\%$, which matches option (B).