Q.When $0.15\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ is mixed with $0.30\ \text{kg}$ of water at $50\ ^\circ\text{C}$ in a container, the resulting temperature is $6.7\ ^\circ\text{C}$. Calculate the heat of fusion of ice. ($s_{\text{water}} = 4186\ \text{J kg}^{-1}\ \text{K}^{-1}$)

Concept: Specific Heat Capacity & Latent Heat — The heat lost by warm water equals the heat gained by ice (to melt) plus the heat gained by the melted water (to warm to final temperature).

Step 1: Heat lost by warm water

Step 2: Heat gained by ice

Ice first melts: $Q_{\text{melt}} = m_i L_f$

Then the resulting water warms: $Q_{\text{warm}} = m_i s_w (T_f - 0) = 0.15 \times 4186 \times 6.7$

$Q_{\text{warm}} = 0.15 \times 4186 \times 6.7 = 4206.93\ \text{J}$ …

Why this approach works

The core idea is calorimetry: in an isolated system, the total heat exchange is zero. Heat flows from the warmer water to the colder ice until thermal equilibrium is reached. The ice first absorbs heat to melt at $0^\circ\text{C}$ (this is the latent heat of fusion, $L_f$), and then the resulting water warms up to the final temperature. Meanwhile, the original warm water cools down. By equating the heat lost and gained, we can solve for the unknown $L_f$.

A common mistake is forgetting that the melted ice also needs to be warmed from $0^\circ\text{C}$ to the final temperature — that extra heat comes from the warm water too.

Step-by-step solution

1. Identify the heat exchanges

   The system has three thermal processes:

   • Ice melts at $0^\circ\text{C}$: heat absorbed $Q_1 = m_{\text{ice}} L_f$
   • The melted ice (now water at $0^\circ\text{C}$) warms to $T_f = 6.7^\circ\text{C}$: heat absorbed $Q_2 = m_{\text{ice}} s_{\text{water}} (T_f - 0)$
   • The original warm water cools from $50^\circ\text{C}$ to $T_f$: heat released $Q_3 = m_{\text{water}} s_{\text{water}} (50 - T_f)$

   No heat is lost to the container (assumed negligible or included in the given data).

2. Apply conservation of energy

   Heat gained by ice = Heat lost by warm water:

3. Substitute the known values $m_{\text{ice}} = 0.15\ \text{kg}$, $m_{\text{water}} = 0.30\ \text{kg}$, $s_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$, $T_f = 6.7^\circ\text{C}$.

4. Simplify numerically Compute the known terms:
   • $Q_2 = 0.15 \times 4186 \times 6.7 = 0.15 \times 28046.2 = 4206.93\ \text{J}$ …

Instead of naming $Q_{\text{lost}}$ as an intermediate quantity, view the ice's journey as a single energy budget: to go from solid at $0^\circ$C to liquid at $6.7^\circ$C it needs $m_i\big(L_f + s_w\times6.7\big)$. Setting this equal to the heat the warm water gives up, $m_w s_w(50-6.7)$, and solving directly for $L_f$ skips a step. As a sanity check, the recovered latent heat $L_f\approx3.34\times10^5\ \text{J/kg}$ matches the standard textbook value …