When $1 ext{ kg}$ of ice at $0^{\circ} ext{C}$ melts into water at $0^{\circ} ext{C}$, the resulting chang

Question

When $1 	ext{ kg}$ of ice at $0^{\circ}	ext{C}$ melts into water at $0^{\circ}	ext{C}$, the resulting change in its entropy, taking the latent heat of ice to be $80 	ext{ cal/g}$, is:

Accepted Answer

The change in entropy ($\Delta S$) during a phase transition at a constant temperature is given by the formula:
$$\Delta S = \frac{q_{rev}}{T}$$
where $q_{rev}$ is the heat absorbed reversibly and $T$ is the absolute temperature .

1.  **Calculate Heat Absorbed ($q$):**
    The heat required to melt mass $m$ is given by $q = m 	imes L_f$, where $L_f$ is the latent heat of fusion.
    Given: $m = 1 	ext{ kg} = 1000 	ext{ g}$ and $L_f = 80 	ext{ cal/g}$.
    $$q = 1000 	ext{ g} 	imes 80 	ext{ cal/g} = 80,000 	ext{ cal}$$

2.  **Convert Temperature to Kelvin:**
    $$T = 0^{\circ}	ext{C} = 273 	ext{ K}$$

3.  **Calculate Entropy Change ($\Delta S$):**
    $$\Delta S = \frac{80,000 	ext{ cal}}{273 	ext{ K}} \approx 293 	ext{ cal/K}$$

Step-by-Step Solution

Practice All PHYSICS Questions