1,kg of ice at -10,^{circ}C is mixed with 4.4 | vedclass

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(D) Step $1$: Calculate heat required for ice to reach $0^{\circ}C$ and melt.Heat to raise ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m_{ice} c_{

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(D) Step $1$: Calculate heat required for ice to reach $0^{\circ}C$ and melt.Heat to raise ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m_{ice} c_{ice} \Delta T = 1 	imes 2100 	imes 10 = 21000\,J$.Heat to melt $1\,kg$ of ice at $0^{\circ}C$: $Q_2 = m_{ice} L_f = 1 	imes 336000 = 336000\,J$.Total heat required by ice to become water at $0^{\circ}C$: $Q_{total, ice} = 21000 + 336000 = 357000\,J$.Step $2$: Calculate heat released by water to reach $0^{\circ}C$.$Q_{released} = m_{water} c_{water} \Delta T = 4.4 	imes 4200 	imes 30 = 554400\,J$.Step $3$: Compare heat and find final temperature.Since $Q_{released} > Q_{total, ice}$,the ice melts completely and the mixture reaches a final temperature $T > 0^{\circ}C$.Remaining heat available to raise the temperature of the total mass ($1\,kg + 4.4\,kg = 5.4\,kg$ of water): $Q_{rem} = 554400 - 357000 = 197400\,J$.Using $Q = mc\Delta T$: $197400 = 5.4 	imes 4200 	imes (T - 0)$.$T = \frac{197400}{22680} \approx 8.7^{\circ}C$.

$1\,kg$ of ice at $-10\,^{\circ}C$ is mixed with $4.4\,kg$ of water at $30\,^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$ (specific heat of ice is $2100\,J/kg\cdot K$,latent heat of fusion of ice is $336000\,J/kg$,and specific heat of water is $4200\,J/kg\cdot K$).

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