If 180 grams of ice at 0.0 degrees C is added to 180 grams of water at 100 degrees C

what is the final temperature?
specific heat of ice is 2.03 j/gc
sp water = 4.18
sp steam = 1.99
delta hvap =40.79 kj/mol
delta hfus = 6.01 kj/mol

If you add ice to water, the temperature of the ice is increasing (gaining energy), while the temperature of the water is decreasing (losing energy).

qgain = -qlost

qgain consists of two parts: melting the ice and raising the temperature of the resulting liquid ice.

Melting the ice:

q = Hf * m = 6.01 kJ/mol * 180 g (1 mol / 18 g) = 60.1 kJ

Raising the temperature of ice from 0 C to final temperature of x:

q = c*m*dT = 4.184 * 180 * (x - 0) = 753.1x

So:

qgain = 60100 + 753.1x


On the other side, adding ice to the water lowers is temperature. Since it is not specified that the water is steam, there is no phase change from gas to liquid. The temperature is going from 100 C to final temperature x.

qlost = c*m*dT = 4.184 * 180 * (x - 100)


Put the two equations together in the formula and solve:

qgain = -qlost

60100 + 753.1x = - (753.1x - 75310)

x = 10.1 C