Question

How much energy (heat) is required to convert 52.0 g of ice at -10.0 C to steam at 100 C?

Specific heat of ice: 2.09 J/g * C DHfus = 6.02 kJ/mol

Specific heat of water: 4.18 J/g * C DHvap = 40.7 kJ/mol

Specific heat of steam: 1.84 J/g * C

Answer 1

You need to break it down. Energy when not in a phase change is Q=mc(deltaT) whereas energy at a phase change is Q=deltaH(m)

So from -10C to 0C, no phase change occurs and it is ice:
Q=52g*2.09 J/g C*(10C)= 1086.8J=1.087kJ

At 0C, melting occurs, so you use deltaH fusion:
Make sure your units work out properly:
Q=6.02kJ/mol(52g)(1mol/18g) = 17.39 kJ

From 0C to 100C, no phase change occurs when it's liquid:
Q=52g*4.18 J/g C*(100C)= 21736J=21.736 kJ

At 100C, phase change occurs as water evaporates:
Q=40.7kJ/mol*52g*(1mol/18g) = 117.6 kJ

Add these values of heat up: 1.087 + 17.39+21.736+117.6=157.8 kJ