What mass of ice at 0 oC must you add to 0.5 kg of water at 22 oC to bring the final temperature of the water to 5 oC? (cice = cvapor = 0.5 cliq , cliq = 1 cal/(g oC), hfusion = 80 cal/g, hvaporiztion = 540 cal/g)
A. 5.9
B.71
C 100
D106
E1.7
The 0.5 kg of water must lose 17*500 = 8500 calories. The mass M (grams) of ice must gain that amount of heat while melting and heating from 0 to 5 C.
M*(80 + 5) = 8500
M = 100 grams
You don't need the heat of vaporization nor the specific heat of vapor and ice.