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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Mass of ice: $m_i = 1.5\,\text{kg}$
• Initial temperature of ice: $T_{i,\text{ice}} = 0^\circ\text{C}$
• Mass of water: $m_w = 2\,\text{kg}$
• Initial temperature of water: $T_{i,\text{water}} = 70^\circ\text{C}$
• Final temperature of mixture: $T_f = 5^\circ\text{C}$
• Specific heat of water: $s_w = 4186\,\text{J\,kg}^{-1}\text{K}^{-1}$
• Latent heat of fusion of ice: $L_f$ (unknown)
Step 2: Heat Lost by the Hot Water
The water cools from $70^\circ\text{C}$ to $5^\circ\text{C}.$ The heat lost by the water,
$Q_{\text{lost by water}},$ is:
$$
Q_{\text{lost by water}}
= m_w \times s_w \times \bigl(T_{i,\text{water}} - T_f\bigr).
$$
Substituting values:
$$
Q_{\text{lost by water}}
= 2\,\text{kg} \times 4186\,\text{J\,kg}^{-1}\text{K}^{-1} \times (70 - 5)\,\text{K}.
$$
$$
= 2 \times 4186 \times 65\,\text{J}
= 544180\,\text{J}.
$$
Step 3: Heat Gained by the Ice
Initially, the ice at $0^\circ\text{C}$ must fully melt, then the resulting water warms up from $0^\circ\text{C}$ to $5^\circ\text{C}.$
1) Heat required to melt the ice:
$$
Q_{\text{melt}} = m_i \times L_f.
$$
2) Heat required to warm the melted water from $0^\circ\text{C}$ to $5^\circ\text{C}$:
$$
Q_{\text{heat up}}
= m_i \times s_w \times (T_f - 0^\circ\text{C}).
$$
Substituting:
$$
Q_{\text{heat up}}
= 1.5\,\text{kg} \times 4186\,\text{J\,kg}^{-1}\text{K}^{-1} \times 5\,\text{K}
= 31395\,\text{J}.
$$
Therefore, total heat gained by ice:
$$
Q_{\text{gained by ice}}
= m_i \times L_f + 31395\,\text{J}.
$$
Step 4: Apply Conservation of Energy (Calorimetry)
Assuming no heat is lost to the environment, the heat lost by the hot water equals the heat gained by the ice:
$$
Q_{\text{lost by water}}
= Q_{\text{gained by ice}}.
$$
Hence:
$$
544180\,\text{J}
= (1.5\,\text{kg} \times L_f) + 31395\,\text{J}.
$$
Rearranging to solve for $L_f$:
$$
544180\,\text{J} - 31395\,\text{J}
= 1.5\,\text{kg} \times L_f,
$$
$$
512785\,\text{J}
= 1.5\,\text{kg} \times L_f,
$$
$$
L_f
= \frac{512785\,\text{J}}{1.5\,\text{kg}}
= 341856.67\,\text{J\,kg}^{-1}.
$$
Step 5: Final Answer
In scientific notation,
$$
L_f \approx 3.42 \times 10^5\,\text{J\,kg}^{-1}.
$$
