A 50 ,g ice cube at -10^{circ} C is added to | vedclass

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(B) Step $1$: Calculate heat required to bring ice from $-10^{\circ} C$ to $0^{\circ} C$ (ice): $Q_1 = m_i c_i \Delta T = 50 	imes 0.5 	imes 10 = 25

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$7$

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(B) Step $1$: Calculate heat required to bring ice from $-10^{\circ} C$ to $0^{\circ} C$ (ice): $Q_1 = m_i c_i \Delta T = 50 	imes 0.5 	imes 10 = 250 \,cal$.Step $2$: Calculate heat required to melt $50 \,g$ of ice at $0^{\circ} C$: $Q_2 = m_i L_f = 50 	imes 80 = 4000 \,cal$.Total heat required by ice to become water at $0^{\circ} C$ is $Q_{total} = 250 + 4000 = 4250 \,cal$.Step $3$: Calculate heat released by $200 \,g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{release} = m_w c_w \Delta T = 200 	imes 1 	imes 30 = 6000 \,cal$.Since $Q_{release} > Q_{total}$,the ice melts completely and the final temperature $T_f$ will be above $0^{\circ} C$.Step $4$: Apply the principle of calorimetry: Heat lost by water = Heat gained by ice.$200 	imes 1 	imes (30 - T_f) = 4250 + 50 	imes 1 	imes (T_f - 0)$.$6000 - 200 T_f = 4250 + 50 T_f$.$1750 = 250 T_f$.$T_f = 7^{\circ} C$.

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