19 g of water at 30^{circ} C and 5 g of ice | vedclass

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Question

(C) Step $1$: Calculate heat required to bring $5 \ g$ of ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T = 5 \

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$5$

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(C) Step $1$: Calculate heat required to bring $5 \ g$ of ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T = 5 \ g 	imes 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1} 	imes 20^{\circ} C = 50 \ cal$.Step $2$: Calculate heat required to melt $5 \ g$ of ice at $0^{\circ} C$: $Q_2 = m_{ice} \cdot L_f = 5 \ g 	imes 80 \ cal \ g^{-1} = 400 \ cal$.Total heat required to convert ice to water at $0^{\circ} C$ is $Q_{total} = 50 + 400 = 450 \ cal$.Step $3$: Calculate heat released by $19 \ g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{released} = m_{water} \cdot c_{water} \cdot \Delta T = 19 \ g 	imes 1 \ cal \ g^{-1} (^{\circ} C)^{-1} 	imes 30^{\circ} C = 570 \ cal$.Since $Q_{released} > Q_{total}$, 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.$19 	imes 1 	imes (30 - T_f) = 450 + 5 	imes 1 	imes (T_f - 0)$.$570 - 19 T_f = 450 + 5 T_f$.$120 = 24 T_f$.$T_f = 5^{\circ} C$.

$19 \ g$ of water at $30^{\circ} C$ and $5 \ g$ of ice at $-20^{\circ} C$ are mixed together in a calorimeter. What is the final temperature of the mixture (in $^{\circ} C$)? Given specific heat of ice $= 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$.

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