Question

1. what temperature is required to transfer waste heat to the environment for a heat engine to be 100 percent efficient? (1 point)

○ 0°c
○ 273 k
○ 212°f
○ 0 k

Explanation:

The efficiency of a heat engine is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_{c}}{T_{h}}$, where $T_{c}$ is the temperature of the cold reservoir (where waste heat is transferred) and $T_{h}$ is the temperature of the hot reservoir. For a heat engine to be 100% efficient ($\eta = 1$), we need $\frac{T_{c}}{T_{h}} = 0$. This occurs when $T_{c} = 0\ \text{K}$ (absolute zero), because at absolute zero, the entropy change associated with heat transfer to the cold reservoir would be zero, allowing all heat from the hot reservoir to be converted to work. $0^{\circ}\text{C}$ is $273\ \text{K}$, $212^{\circ}\text{F}$ is the boiling point of water (and is $373\ \text{K}$), and $273\ \text{K}$ is $0^{\circ}\text{C}$, none of which are absolute zero. So the temperature required for the waste heat reservoir (cold reservoir) to make the heat engine 100% efficient is $0\ \text{K}$.

Answer:

D. $0\ \text{K}$ (assuming the options are labeled A: $0^{\circ}\text{C}$, B: $273\ \text{K}$, C: $212^{\circ}\text{F}$, D: $0\ \text{K}$)