Change in temp using specific heat capacity?

Question

If 100g of water at 20C absorbs 5 kJ of heat by what amount will the temp of the water increase?

Answer 1

specific heat 4.2 J/gC

4.2 J/gC x 100g x ΔT = 5000J

ΔT = 11.9 deg C

final temp = 20 + 11.9 deg C = 31.9 C

Answer 2

0.100kg x 4.184kJ/kg/°C x ΔT = 5kJ.

ΔT = 5kJ / 0.4184kJ/°C = 12°C temp. increase. (As requested).

(Check: 100g x 4.184J/g/°C x 12°C = 5,021J = 5kJ)