Question

The efficiency of a heat engine that works between the temperatures where Celsius-Fahrenheit scales coincide and Kelvin-Fahrenheit scales coincide is (approximately):

• A. \( 45% \)
• B. \( 35% \)
• C. \( 60% \)
• D. \( 50% \)

Hint:

The efficiency of a heat engine operating between two temperatures is given by \( \eta = 1 - \frac{T_C}{T_H} \). Ensure temperature values are converted to Kelvin before substitution.

Answer 1

The Correct Option is C

Step 1: Understanding the Given Temperatures - The Celsius-Fahrenheit coincidence occurs at: \[ T_1 = -40^\circ C = 233 \text{ K}. \] - The Kelvin-Fahrenheit coincidence occurs at: \[ T_2 = 574.25 \text{ K}. \] 
Step 2: Applying Carnot’s Efficiency Formula The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_C}{T_H}. \] Substituting values: \[ \eta = 1 - \frac{233}{574.25}. \] \[ \eta = 1 - 0.4059. \] \[ \eta = 0.594 \approx 60%. \] Thus, the correct answer is:  60%. 

Answer 2

Step 1: Identifying the Special Temperatures
We are given two specific temperature points that are notable due to their unique relationships across different temperature scales:

- The Celsius-Fahrenheit coincidence occurs at:
    \( T_1 = -40^\circ C \)
To convert this to the Kelvin scale:
    \( T(K) = T(°C) + 273.15 \)
    \( T_1 = -40 + 273.15 = 233.15 \text{ K} \approx 233 \text{ K} \)

- The Kelvin-Fahrenheit coincidence point is given directly as:
    \( T_2 = 574.25 \text{ K} \)

These values are used as the temperatures for the cold reservoir (\( T_C = 233 \text{ K} \)) and the hot reservoir (\( T_H = 574.25 \text{ K} \)) of a Carnot heat engine.

Step 2: Understanding the Carnot Engine Efficiency
The efficiency of a Carnot engine operating between two thermal reservoirs is given by the formula:
    \( \eta = 1 - \frac{T_C}{T_H} \)
where:
- \( \eta \) is the efficiency (expressed as a decimal)
- \( T_C \) is the cold reservoir temperature in Kelvin
- \( T_H \) is the hot reservoir temperature in Kelvin

Step 3: Substituting the Values
Plugging in the known values:
    \( \eta = 1 - \frac{233}{574.25} \)
    \( \eta = 1 - 0.4059 \)
    \( \eta = 0.5941 \)

Converting to a percentage:
    \( \eta \times 100 = 59.41\% \)
Rounding to the nearest whole number:
    \( \eta \approx 60\% \)

Final Answer:
The ideal efficiency of the Carnot engine under these conditions is approximately 60%.