How many grams of ice could be melted by the energy obtained as 18.0 grams of steam is condensed at 100.0 degrees celcius and cooled to 0.0 degrees celcius?

Answer 1

The amount of heat liberated will be
Q = M*(Hv + 100 Cp)
where M = Mass (18.0g),
Hv = the heat of vaporization per gm (540 Cal/g), and
Cp is the specific heat of liquid water (1.00 Cal/g deg.C)

Q = 18*(540 + 100) = 11,520 Calories
Divide this by the heat of fusion of ice, 80 Cal/g, to get the amount of ice that will melt. This assumes that the ice started out at C.

I get 144 g. Check my numbers and method.