Q.A copper block of mass $2.5\ \text{kg}$ is heated in a furnace to a temperature of $500\ ^\circ\text{C}$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $= 0.39\ \text{J g}^{-1}\ \text{K}^{-1}$; heat of fusion of water $= 335\ \text{J g}^{-1}$).

As the copper cools from $500^\circ$C to $0^\circ$C (the large ice block holds it there), it releases $Q=m_{\text{Cu}}c_{\text{Cu}}\Delta T = 2500\,\text{g}\times0.39\times500 = 487{,}500$ J. …

When a hot object is placed on a large block of ice, energy flows from the hot object into the ice until the object reaches $0^\circ$C - the ice's melting point - because the ice, being large, can absorb heat and melt without its own temperature rising above $0^\circ$C.

Step 1 - Heat released by the copper as it cools

The copper cools from $500^\circ$C down to $0^\circ$C, a change of $\Delta T = 500$ K. Converting mass to grams to match the given specific heat:

Step 2 - Mass of ice this energy can melt

All of this energy goes into melting ice at $0^\circ$C (a phase change, no further temperature rise needed): …

Shortcut — think in terms of a 'melting ratio' per unit mass of copper. Instead of computing total heat released and then dividing by $L_f$, find directly what mass of ice each gram of copper can melt as it cools: $\frac{m_{\text{ice}}}{m_{\text{Cu}}} = \frac{c_{\text{Cu}}\,\Delta T}{L_f} = \frac{0.39\times500}{335} \approx 0.582.$ This says: every gram of copper cooling from $500\,^\circ\text{C}$ to $0\,^\circ\text{C}$ melts about $0.58\ \text{g}$ of ice. Scaling up: $m_{\text{ice}} = 0.582 \times 2.5\ \text{kg} \approx 1.455\ \text{kg}$ — same answer, without carrying large intermediate joule values. …