You need three steps to get liquid water at 0 °C to vapor at 167 °C:

1. Heat it from 0 °C to 100 °C

2. Turn the liquid water to vapor, all at 100 °C

3. Heat the vapor from 100 °C to 167 °C

I split the entire heating process up because there are different heat capacities involved.

1. The specific heat capacity of water* is around 4.2 J/(g K).

So energy to heat the specified amount of water from 0 °C to 100 °C amounts to:

Q₁ = 4.2 J/(g K) × 457 g × 100 K ≈ 192 kJ

2. The heat of evaporation of is** 2,257 J/g

Q₂ = 2,257 J/g × 457 g ≈ 1,031 kJ

3. The specific heat capacity of vapor* is around 2.1 J/(g K)

Q₃ = 2.1 J/(g K) × 457 g × 67 K ≈ 64 kJ

For the entire process we simply add Q₁, Q₂, and Q₃:

Q = 192 kJ + 1031 kJ + 64 kJ = 1,287 kJ

This figure is rounded, you can get better precision if you take more precise values for the specific heat capacities if necessary.

Hope that helped.

* I use the figure from this table: https://en.wikipedia.org/wiki/Heat_capacity#Table_...

** I use the figure from this paragraph: https://en.wikipedia.org/wiki/Water#Polarity_and_h...