10, g of ice at -20^{circ}C is dropped into a | vedclass

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(C) Let the specific heat of ice be $s_{ice} = 0.5\, cal/g^{\circ}C$ and water be $s_{water} = 1.0\, cal/g^{\circ}C$.$1$. Heat released by $10\, g$ of

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$10\, g$ of ice and $10\, g$ of water

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(C) Let the specific heat of ice be $s_{ice} = 0.5\, cal/g^{\circ}C$ and water be $s_{water} = 1.0\, cal/g^{\circ}C$.$1$. Heat released by $10\, g$ of water cooling from $10^{\circ}C$ to $0^{\circ}C$:$Q_{release} = m \cdot s_{water} \cdot \Delta T = 10\, g 	imes 1.0\, cal/g^{\circ}C 	imes (10^{\circ}C - 0^{\circ}C) = 100\, cal$.$2$. Heat required by $10\, g$ of ice to warm from $-20^{\circ}C$ to $0^{\circ}C$:$Q_{warm} = m \cdot s_{ice} \cdot \Delta T = 10\, g 	imes 0.5\, cal/g^{\circ}C 	imes (0^{\circ}C - (-20^{\circ}C)) = 10\, g 	imes 0.5 	imes 20 = 100\, cal$.$3$. Since the heat released by water $(100\, cal)$ is exactly equal to the heat required to bring the ice to $0^{\circ}C$ $(100\, cal)$,no further heat is available to melt the ice.$4$. Therefore,the final state of the system is $10\, g$ of ice and $10\, g$ of water at $0^{\circ}C$.

$10\, g$ of ice at $-20^{\circ}C$ is dropped into a calorimeter containing $10\, g$ of water at $10^{\circ}C$. The specific heat of water is twice that of ice. When equilibrium is reached,the calorimeter will contain:

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