Question

A new absolute temperature scale is proposed based on a Carnot engine operating between hot and cold reservoirs of temperatures \(T_L\) and \(T_H\) respectively. Let \(Q_L\) and \(Q_H\) be the respective heat transfers, with the relation given by \( \frac{T_L}{T_H} = \frac{Q_L}{Q_H} \). On the new scale, the difference between the steam and ice points of water is 500 units and the efficiency of the engine is 0.268. The steam point of water on this scale is .......... units (rounded off to the nearest integer).

Hint:

For problems involving Carnot engine efficiencies: 1. Use \( \eta = 1 - \frac{T_L}{T_H} \) to relate temperatures and efficiency.
2. Ensure temperature differences are consistent with the scale provided.
3. Verify all units and steps to avoid errors in conversions.

Answer 1

Step 1: Recall the formula for the efficiency of a Carnot engine.
The efficiency of a Carnot engine (\( \eta \)) is given by: \[ \eta = 1 - \frac{T_L}{T_H}, \] where: - \( T_L \) is the temperature of the cold reservoir, - \( T_H \) is the temperature of the hot reservoir. Step 2: Express \( T_L \) in terms of \( T_H \) and \( \eta \).
Rearranging the formula: \[ \frac{T_L}{T_H} = 1 - \eta. \] Substitute \( \eta = 0.268 \): \[ \frac{T_L}{T_H} = 1 - 0.268 = 0.732. \] Step 3: Determine the temperatures on the new scale.
The difference between the steam and ice points is 500 units: \[ T_H - T_L = 500. \] Let \( T_H = x \). Then \( T_L = 0.732 \cdot x \). Substituting this into the temperature difference: \[ x - 0.732 \cdot x = 500. \] Simplify: \[ 0.268 \cdot x = 500. \] Solve for \( x \): \[ x = \frac{500}{0.268} \approx 1864 \, \text{units}. \] Thus, \( T_H = 1864 \, \text{units}. \) Conclusion: The steam point of water on the new scale is \( 1864 \, \text{units}. \)