540, g of ice at 0^circ C is mixed with 540, | vedclass

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(A) Heat required by ice to melt completely at $0^\circ C$ is given by $Q_1 = m_i L_f$,where $L_f = 80\, cal/g$.$Q_1 = 540 	imes 80 = 43200\, cal$.He

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$0^\circ C$

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(A) Heat required by ice to melt completely at $0^\circ C$ is given by $Q_1 = m_i L_f$,where $L_f = 80\, cal/g$.$Q_1 = 540 	imes 80 = 43200\, cal$.Heat released by water when cooling from $80^\circ C$ to $0^\circ C$ is given by $Q_2 = m_w c_w \Delta T$.$Q_2 = 540 	imes 1 	imes (80 - 0) = 43200\, cal$.Since the heat released by the water $(Q_2)$ is exactly equal to the heat required to melt the ice $(Q_1)$,all the ice melts and the final temperature of the mixture remains $0^\circ C$.

$540\, g$ of ice at $0^\circ C$ is mixed with $540\, g$ of water at $80^\circ C$. The final temperature of the mixture is

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