What mass of steam at 100¡ãC must be mixed with 388 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 45.0¡ãC? The specific heat of water is 4186 J/kg¡¤K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.

Answer 1

.388 kg ice
melt it and warm it to 45 C

.388 [ 333,000 + 45(4186) ]

= 202,292 Joules into what was ice

= m [2256,000 + 55(4186) ]

= 2,486,230 m

m = 202,292/2,486,230

= .0814 kg
= 81.4 grams