Solution
To melt the ice, we supply heat $Q$. By the First Law of Thermodynamics, the heat supplied equals the change in internal energy minus the work done by gravity (change in potential energy).
$$ Q = \Delta U_{\text{phase}} + \Delta PE_{\text{grav}} $$
Where $\Delta U_{\text{phase}} = mL$ (Latent heat of fusion), which is identical for both vessels since masses are equal.
Change in Gravitational Potential Energy ($\Delta PE$):
When ice melts into water, the density increases ($\rho_w > \rho_i$), so the volume decreases. Consequently, the center of mass (COM) of the system lowers. A lowering of COM means a decrease in potential energy ($\Delta PE < 0$). This release of gravitational potential energy contributes to the melting process.
For a cylindrical vessel of base area $S$:
- Height of ice column: $h_i = \frac{m}{\rho_i S}$ $\rightarrow$ COM height: $\frac{h_i}{2} = \frac{m}{2\rho_i S}$
- Height of water column: $h_w = \frac{m}{\rho_w S}$ $\rightarrow$ COM height: $\frac{h_w}{2} = \frac{m}{2\rho_w S}$
Change in Potential Energy:
$$ \Delta PE = mg(h_{\text{final}} – h_{\text{initial}}) = mg \left( \frac{m}{2\rho_w S} – \frac{m}{2\rho_i S} \right) $$
$$ \Delta PE = \frac{m^2 g}{2S} \left( \frac{1}{\rho_w} – \frac{1}{\rho_i} \right) $$
Note: Since $\rho_w > \rho_i$, the term $(\frac{1}{\rho_w} – \frac{1}{\rho_i})$ is negative, confirming potential energy decreases.
Heat Supplied ($Q$):
$$ Q = mL + \Delta PE = mL – \frac{m^2 g}{2S} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) $$
The term $\frac{m^2 g}{2S} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right)$ represents the gravitational energy assisting the melting. The narrower vessel (smaller $S$) has a larger height change, meaning more gravitational energy is released. Therefore, the narrower vessel requires less external heat.
The wider vessel (Area $2A$) requires a greater amount of heat because its center of mass lowers by a smaller distance.
Difference in Heat Supplied:
Let $S_1 = A$ (Narrow) and $S_2 = 2A$ (Wide).
$$ \Delta Q = Q_{2A} – Q_{A} $$
$$ \Delta Q = \left[ mL – \frac{m^2 g}{2(2A)} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) \right] – \left[ mL – \frac{m^2 g}{2A} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) \right] $$
$$ \Delta Q = \frac{m^2 g}{2A} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) – \frac{m^2 g}{4A} \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) $$
$$ \Delta Q = \frac{m^2 g}{2A} \left(1 – \frac{1}{2}\right) \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) $$