Question

The efficiency of a heat engine is '$\eta$' and the coefficient of performance of a refrigerator is '$\beta$'. Then

• A. $\eta = \frac{1}{\beta}$
• B. $\eta = \frac{1}{\beta + 1}$
• C. $\eta\beta = \frac{1}{2}$
• D. $\eta = \frac{1}{\beta - 1}$

Hint:

Physics Tip : This relationship holds because both devices are limited by the same temperature differences of the reservoirs they operate between.

Answer 1

The Correct Option is B

Concept: Physics (Thermodynamics) - Heat Engines and Refrigerators.

Step 1: State the formula for Coefficient of Performance ($\beta$). For a refrigerator: $$\beta = \frac{T_{2}}{T_{1} - T_{2}}$$ where $T_1$ is the temperature of the hot reservoir and $T_2$ is the temperature of the cold reservoir.

Step 2: State the formula for Efficiency ($\eta$). For a heat engine: $$\eta = 1 - \frac{T_{2}}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$$

Step 3: Establish the relationship. The efficiency can be rewritten as: $$\eta = \frac{1}{\frac{T_{1}}{T_{1} - T_{2}}} = \frac{1}{\frac{(T_{1} - T_{2}) + T_{2}}{T_{1} - T_{2}}} = \frac{1}{1 + \frac{T_{2}}{T_{1} - T_{2}}}$$ Substituting $\beta$ into the equation: $$\eta = \frac{1}{1 + \beta}$$ $$ \therefore \text{The correct relationship is } \eta = \frac{1}{\beta + 1}. \text{} $$