5 g of ice at -30^{circ} C and 20 g of wate | vedclass

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(D) Given: Mass of ice $m_{	ext{ice}} = 5 \ g$,initial temperature $T_{	ext{ice}} = -30^{\circ} C$. Mass of water $m_w = 20 \ g$,initial temperature

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$9$

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(D) Given: Mass of ice $m_{	ext{ice}} = 5 \ g$,initial temperature $T_{	ext{ice}} = -30^{\circ} C$. Mass of water $m_w = 20 \ g$,initial temperature $T_w = 35^{\circ} C$.Specific heat of ice $s_{	ext{ice}} = 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $s_w = 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion $L_f = 80 \ cal \ g^{-1}$.Heat required to raise ice from $-30^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{	ext{ice}} s_{	ext{ice}} \Delta T = 5 	imes 0.5 	imes 30 = 75 \ cal$.Heat required to melt ice at $0^{\circ} C$: $Q_2 = m_{	ext{ice}} L_f = 5 	imes 80 = 400 \ cal$.Total heat required to convert ice at $-30^{\circ} C$ to water at $0^{\circ} C$ is $Q_{	ext{total}} = 75 + 400 = 475 \ cal$.Heat available from water cooling from $35^{\circ} C$ to $0^{\circ} C$: $Q_{	ext{avail}} = m_w s_w \Delta T = 20 	imes 1 	imes 35 = 700 \ cal$.Since $Q_{	ext{avail}} > Q_{	ext{total}}$,the final temperature $T$ will be above $0^{\circ} C$.Heat lost by water = Heat gained by ice:$m_w s_w (35 - T) = Q_{	ext{total}} + m_{	ext{ice}} s_w (T - 0)$$20 	imes 1 	imes (35 - T) = 475 + 5 	imes 1 	imes T$$700 - 20T = 475 + 5T$$225 = 25T$$T = 9^{\circ} C$.

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