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(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.According to the principle of calorimetry,Heat los

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$50$

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(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.According to the principle of calorimetry,Heat lost by water = Heat gained by ice.Heat lost by $100 \,g$ of water cooling from $80^{\circ} C$ to $0^{\circ} C$ is $Q_{lost} = m_w \cdot s_w \cdot \Delta T = 100 	imes 1 	imes (80 - 0) = 8000 \,cal$.Heat required to melt $m$ grams of ice at $0^{\circ} C$ is $Q_{gained} = m \cdot L_f = m 	imes 80$.Equating the two: $80m = 8000$.Solving for $m$: $m = 100 \,g$.Since the total mass of ice is $150 \,g$,the amount of ice that does not melt is $150 \,g - 100 \,g = 50 \,g$.

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

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