Question

The efficiency of a Carnot heat engine operating between two reservoirs at temperatures TH (Hot source) and TC (Cold sink) is given by:

• A. \(T_C / T_H\)
• B. \(T_H / T_C\)
• C. \(1 - T_C / T_H\)
• D. \(1 - T_H / T_C\)

Hint:

Efficiency is always a positive number less than 1. Since \(T_C \lt T_H\), the term \(T_C / T_H\) is less than 1, making \(1 - T_C/T_H\) a valid positive efficiency.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Carnot engine is a theoretical thermodynamic cycle that operates on reversible processes, establishing the upper limit of efficiency for any heat engine operating between two temperatures.

Step 2: Key Formula or Approach:
The thermal efficiency \(\eta\) of any heat engine is:
\[ \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = 1 - \frac{Q_C}{Q_H} \] For a reversible (Carnot) engine, the ratio of heat transfer is equal to the ratio of absolute temperatures:
\[ \frac{Q_C}{Q_H} = \frac{T_C}{T_H} \]

Step 3: Detailed Explanation:
Substituting the temperature relationship into the general efficiency equation:
\[ \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \] where \(T_C\) is the temperature of the cold sink (in Kelvin) and \(T_H\) is the temperature of the hot source (in Kelvin).

Step 4: Final Answer:
The correct option is 3, which corresponds to \(1 - \frac{T_C}{T_H}\).