Question

What mass of steam at 100°C must be added to 1.05 kg of ice at 0°C to yield liquid water at 18°C?

Answer 1

What are your 3 heat values? Heat of vaporization? Heat of fusion? Specific Heat?

Answer 2

All it gives me is the question... I believe it would be Heat of Vaporization

Answer 3

WATER
heat of fusion 79.7 cal /gm
vaporization 539 cal/gm
1 cal = 4.19 J
specific heat 1 cal/gm-oC

steam and ice have about half the specific heat of water, but won't need that here

Will do this in gm rather than kg. m is unknown mass of steam condensed

Heat loss by steam/water = m [(539 cal/gm) + (1 cal/gm-oC)(100-18) = m [621 cal/gm]

Heat gained by ice/water = (1050g)[(79.7 cal/gm) + (1 cal/gm)(18-0)] = 102585 cal

m [621 cal/gm] = 102585 cal

m = 165.2 g of steam needed condense and cool.

That's the method. Should check my arithmetic.