what amount of energy in calories would it take to bring 20 grams of solid ice at negative twelve degrees celsius to liquid water at 100 degrees celsius?

Answer 1

Q = ΔU1 + ΔU2 + ΔU3 =
=c1•m •ΔT1 +λ•m + c2•m •ΔT2 =
=20•10^-3•(2.060•10^3• 12 + 335•10^3 + 4.183•10^3•100) =
=15560 J
(c1, c2 –specific heats of ice and water, λ –heat of fusion of ice)