The efficiency of an ideal heat engine (Carnot heat engine) working between the freezing point and boiling poi

Question

The efficiency of an ideal heat engine (Carnot heat engine) working between the freezing point and boiling point of water is:

Accepted Answer

The efficiency ($\eta$) of a Carnot engine is given by the formula:
$$ \eta = 1 - \frac{T_2}{T_1} $$
where $T_1$ is the temperature of the source (hot reservoir) and $T_2$ is the temperature of the sink (cold reservoir) in Kelvin.

1.  **Identify Temperatures:**
    *   Freezing point of water ($T_2$) = $0^{\circ}\text{C} = 273.15 \text{ K} \approx 273 \text{ K}$.
    *   Boiling point of water ($T_1$) = $100^{\circ}\text{C} = 373.15 \text{ K} \approx 373 \text{ K}$.

2.  **Calculate Efficiency:**
    $$ \eta = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373} $$
    $$ \eta \approx 0.268 $$

3.  **Convert to Percentage:**
    Percentage Efficiency = $0.268 \times 100\% = 26.8\%$.

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