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

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(A) The heat required to melt $540\, g$ of ice at $0\,^{\circ}C$ is given by $Q_{melt} = m \cdot L_f$,where $L_f = 80\, cal/g$.$Q_{melt} = 540\, g 	i

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$0\,^{\circ}C$

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(A) The heat required to melt $540\, g$ of ice at $0\,^{\circ}C$ is given by $Q_{melt} = m \cdot L_f$,where $L_f = 80\, cal/g$.$Q_{melt} = 540\, g 	imes 80\, cal/g = 43200\, cal$.Now,calculate the heat lost by $540\, g$ of water when it cools from $80\,^{\circ}C$ to $0\,^{\circ}C$:$Q_{lost} = m \cdot c \cdot \Delta T = 540\, g 	imes 1\, cal/g^{\circ}C 	imes (80\,^{\circ}C - 0\,^{\circ}C) = 43200\, cal$.Since the heat lost by the water is exactly equal to the heat required to melt all the ice,the final state of the mixture is water at $0\,^{\circ}C$.

$540\, g$ 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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