your answer:Given that:10 gm of ice at 0°C is kept in calorimeter of water 10 gm.To find,The amount of HEAT that must be supplied.Solution:First we must calculate the energy needed for ice at 0°C to Convert into water at same temperature. So,  We know that,Q_{(ice - water)} = amount\ of\ ice \times L_{f}\ of \ waterHere , L v - Latent Heat of FUSION of water = 80 cal/gm ⇒  Q_{(ice - water)} = 10 g \times 80\ cal/gm = 800 calNow it has to convert from 0°C water to 100°C of water.Then,⇒ Q_{(0-100)} = [Amount\ of \ Water + Equivalent\ of\ water](S) (T_{f} - T_{i})Here , S- Specific Heat of water = (1 cal /gm k) , And CHANGE in temperature is (T f) - (T i)Then,⇒ Q_{(0-100)} = (10+10)(1)(100) = 20 \times 100 = 2000\ cal.Now, From 100°C water to Vapor at same temperature. Q_{(Water-Vapour)} = (Amount\ of\ water)\times L_{v}Here Lv = 540 cal/gm.⇒ Q_{(Water-Vapour)}= 10 \times 540 = 5400\ calThen,⇒ Total\ Heat\ needed = Sum\ of\ all\ energy\ used.⇒ 800+2000+5400 = 8200\ cal\bold{THEREFORE\ total\ heat\ required\ is\ 8200\ cal}