1, kg of ice at -10,^{circ}C is mixed with 4. | vedclass

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(D) Heat required to raise $1\, kg$ of ice from $-10\,^{\circ}C$ to $0\,^{\circ}C$: $Q_1 = m_i c_i \Delta T = 1 	imes 2100 	imes 10 = 21,000\, J$.He

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(D) Heat required to raise $1\, kg$ of ice from $-10\,^{\circ}C$ to $0\,^{\circ}C$: $Q_1 = m_i c_i \Delta T = 1 	imes 2100 	imes 10 = 21,000\, J$.Heat required to melt $1\, kg$ of ice at $0\,^{\circ}C$: $Q_2 = m_i L_f = 1 	imes 3.36 	imes 10^5 = 336,000\, J$.Total heat required to convert ice to water at $0\,^{\circ}C$: $Q_{total} = 21,000 + 336,000 = 357,000\, J$.Heat available from $4.4\, kg$ of water cooling from $30\,^{\circ}C$ to $0\,^{\circ}C$: $Q_{avail} = m_w c_w \Delta T = 4.4 	imes 4200 	imes 30 = 554,400\, J$.Since $Q_{avail} > Q_{total}$,the ice melts completely and the final temperature $	heta$ is above $0\,^{\circ}C$.Heat balance equation: $Q_{avail} - Q_{total} = (m_i + m_w) c_w (	heta - 0)$.$554,400 - 357,000 = (1 + 4.4) 	imes 4200 	imes 	heta$.$197,400 = 5.4 	imes 4200 	imes 	heta$.$197,400 = 22,680 	imes 	heta$.$	heta = \frac{197,400}{22,680} \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 = $2100\, J/kg\cdot K$,specific heat of water = $4200\, J/kg\cdot K$,latent heat of fusion of ice = $3.36 \times 10^5\, J/kg$)

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