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raquelx226 · Nov 17, 2010, 05:14 PM

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?

Unknown008 · Nov 18, 2010, 06:39 AM

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.