Steam of mass 11.0 g at 100°C is added to 126.7 g of ice at 0.00°C. Find the final temperature of the system.

Answer 1

Find these values in textbooks or google:
Hv = latent heat of vaporization
Hf = latent heat of fusion
c,water = specific heat of liquid water

Formula for energy: Q = m*c*(T2 - T1)
Summation of all energy is zero:
m,steam*Hv + m,steam(c,water)(T2-100) + m,ice(c,water)(T2-0) + m,ice*Hf = 0

m,steam = 11.0 g
m,ice = 126.7 g
T1 (ice) = 0 °C
T1 (steam) = 100 °C
Look first for the latent heats and specific heat values. Then everything is given so you can solve for T2. Units in °C.