Heat of raise temperature of ice from −20∘C to 0∘C. Q1=m⋅Cice ⋅∆T =0.008×2100×20=336J Heat to melt ice at 0∘C to water Q2=m×Lf=0.008×336000=2,688J Heat to raise temperature of water from 0∘C to 100∘C Q3=m×Cwater ×∆T =0.008×4200×100=3360J Heat to convert water at 100∘C to steam Q4=M⋅Lv=0.008×2.268×106 =18144J Total heat required QTotal =Q1+Q2+Q3+Q4 =336+2688+3360+18144 =24,528J( or 24.5kJ) 1 calorie =4.18J Total heat in joules Q=
