How many joules in total would be required to boil 11.8 g of water if i begin heating it at a temperature of 65.5 c ? ( two steps required)

In this solution we will need two steps, 1) calculate the heat required to get water to the temp of 100 C and 2) find the latent heat of vapourization required. 

First, change the 11.8g to kg: = 0.0118kg
Equations: Q = mcT, Q=mL

c for water = 4186 J/(kg C)
L for water = 22.6 x 10^5 J/kg

1) Q = mcT
Q = 0.0118 kg x 4186 J/(kg C) x (100 C - 65.5 C)
Q = 1704.12 J

2) Q = mL
Q = 0.0118 kg x 22.6 x 10^5 
Q = 26668 J

Qf = 1704.12J + 26668J
Qf = 28372.12J

With sig figs: 2.84 x 10^4 J