10 kg of ice at $-10^{\circ} \mathrm{C}$ is added to 100 kg of water to lower its temperature from 2

Question

10 kg of ice at $-10^{\circ} \mathrm{C}$ is added to 100 kg of water to lower its temperature from 25 ${ }^{\circ} \mathrm{C}$. Consider no heat exchange to surroundings. The decrement to the temperature of water is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
(specific heat of ice $=2100 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, specific heat of water $=4200 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, latent heat of fusion of ice $=3.36 	imes 10^5 \mathrm{~J} / \mathrm{Kg}$ )

Professor PC Thomas Classes · Answer

Let the final equilibrium temperature be $T^\circ$C. Heat lost by the water is $Q_L = 100 	imes 4200 	imes (25 - T)$. Heat gained by the ice to reach $0^\circ$C, melt, and warm to $T^\circ$C is $Q_G = 10 	imes 2100 	imes 10 + 10 	imes 3.36 	imes 10^5 + 10 	imes 4200 	imes (T - 0)$. Equating $Q_L = Q_G$, we have $420000(25 - T) = 210000 + 3360000 + 42000T$. Dividing by 42000 simplifies the equation to $10(25 - T) = 5 + 80 + T \implies 250 - 10T = 85 + T \implies 11T = 165 \implies T = 15^\circ$C. The decrement in the water's temperature is $25 - 15 = 10^\circ$C.