0.1kg of ice initially at a temperature of -10 degrees Celsius is added to a cup with .5kg of water initially at 20 degrees Celsius. The water and ice are isolated thermally from their surroundings. The specific heat of ice is 2000 J/kgoC, the specific heat of water is 4186 J/kgoC and the latent heat of fusion of water is 33.5x104 J/kg

What is the final temperature of the water?

The heat energy gained by the ice is equal to the energy lost by the water, before the ice starts melting.

Hence, we have:

-Q_{gained} = Q_{lost}

-m_{ice}c_{ice}(0 - -10) = m_{water}c_{water}(T_1 - 20)

When the ice melts, another equation is set up.

-m_{ice}L_{fusion} = m_{water}c_{water}(T_2 - T_1)

Now that the ice is water, the cold water will cool down the warmer water.

-m_{cold\ water}c_{water}(T_3 - 0) = m_{warm\ water}c_{water}(T_3 - T_2)

Work out T3, the final temperature of the water.