There are five heat transfers involved in this problem.
heat to warm ice + heat to melt ice + heat to warm water + heat to boil water + heat to warm steam
#q = q_1 + q_2 + q_3 +q_4 + q_5#
#q = mC_1ΔT_1 + mΔ_text(fus)H + mC_3ΔT_3 + mΔ_text(vap)H + mC_5ΔT_5#
We can calculate each of these separately and then add them up to get the total.
#q_1 = mC_1ΔT_1 = 50 color(red)(cancel(color(black)("g"))) × "2.09 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(0 -(-35))" color(red)(cancel(color(black)("°C"))) = "3658 J"#
#Δ_text(fus)H = "333.55 J/g"#
#q_2 = 50 color(red)(cancel(color(black)("g"))) × "333.55 J"/(1 color(red)(cancel(color(black)("g")))) = "16 678 J"#
#q_3 = mC_3ΔT_3 = 50 color(red)(cancel(color(black)("g"))) × "4.18 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(100 -0)" color(red)(cancel(color(black)("°C"))) = "20 900 J"#
#Δ_text(vap)H = "2260 J/g"#
#q_4 = 50 color(red)(cancel(color(black)("g"))) × "2260 J"/(1 color(red)(cancel(color(black)("g")))) = "113 000 J"#
#q_5 = mC_5ΔT_5 = 50 color(red)(cancel(color(black)("g"))) × "2.00 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(128 - 100)" color(red)(cancel(color(black)("°C"))) = "2800 J"#
#q = "(3658 + 16 678 + 20 900 + 113 000 + 2800) J" = "156 000 J" = "156 kJ"#
