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