Physics: heat and gases

Let T be the final temperature of the mixture.
Heat released by water
= 0.2 x 4200 x (25-T)
Heat absorbed by ice cubes
= (50/1000) x [2050 x 15 + 334000 + 4200T]
where 334000 J/kg is the latent heat of fusion of water

Assume no heat loss to the surroundings,
0.2 x 4200 x (25-T) = (50/1000) x [2050 x 15 + 334000 + 4200T]
solve for T gives T = 2.63'C






2011-11-06 11:05:18 補充：
O...sorry that I have overlooked there are 2 cubes. The correct solution is as follws:
Heat given out by water when cooled to 0'C = 0.2 x 4200 x 25 J = 21,000 J

2011-11-06 11:05:30 補充：
Heat absorbed by the 2 ice cubes when they turns into water at 0'C
= (2x0.05) x [2050 x 15 + 334000] J = 36,475 J
In the above, the first term account for the temperture rise from -15'C to0'C and the second term accounts for the change of state from ice to water.

2011-11-06 11:07:36 補充：
The heat supllied by the 0.2kg water is therefore not sufficient to melt the 2 ice cubes. Therefore, the final mixture contains ice and water in equilibrium. The temperature of which is apparently at 0'C.

2011-11-06 11:14:59 補充：
In fact, you can calculate the amount of ice melted.
Let m be the mass of melted melted,
21000 = m.[2050 x 15 + 334000]
this gives m = 0.0576 kg
Ice remained = (2x0.5 - 0.0576) kg = 0.0424 kg
Hence, the final mixture contains (0.2+0.0576) kg = 0.2576 kg of water and 0.0424 kg of ice, both at 0'C.