A $5 \ g$ piece of ice at $-20^oC$ is added into $50\ g$ of water at $20^oC$. Find the temperature of the mixture.($C_{ice}=0.5 \ cal/gm^oC$ and $L=80\ J/gm$)

Using the principle of heat exchange, the heat lost by the warm water equals the heat gained by the ice to warm up and melt. Setting 50 * 1 * (20 - T) equal to (5 * 0.5 * 20) + (5 * 80) yields 1000 - 50T = 500, meaning all the ice melts with excess heat available. Solving for the final liquid water temperature gives 50T = 500, so T = 10 degrees Celsius.

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