At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at –12.0 °C to H2O(g) at 169.0 °C?

Answer 1

q1 = heat to raise T of solid to zero C.
q1 = mass ice x specific heat ice x delta T.

q2 = heat to melt ice at zero C to water at zero C.
q2 = mass ice x heat fusion

q3 = heat to raise T of liquid water from zero C to liquid water at 100 C.
q3 = mass water x specific heat H2O x delta T.

q4 = heat to boil water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization.

q5 = heat to raise T of steam from 100 C to 169 C.
q5 = mass steam x specific heat steam x delta T.

Total = q1 + q2 + q3 + q4 + q5.

Answer 2

To elaborate on DrBob's answer (nearly a decade later):
There are five things to calculate here:
Q1: the energy needed to raise the ice to 0°C
Q2: the energy needed to melt the ice
Q3: the energy needed to raise the water to 100°C
Q4: the energy needed to boil the water
Q5: the energy needed to raise the steam to 169°C
Whatever homework you're using should give you the heat transfer constants so I'll be using the following but adjust your calculations accordingly if your constants are different:
Enthalpy of fusion at 0°C: 333.6J/g
Enthalpy of vaporization of at 100°C: 2257 J/g
Specific heat of solid H2O (ice): 2.087 J/(g*°C)
Specific heat of liquid H2O (water): 4.184 J/(g*°C)
Specific heat of gas H2O (steam): 2.000 J/(g*°C)
Q1 [heating the ice]: (49g)(2.087 J/g*°C)(12°C) = 1227.156 J
Q2 [melting the ice]: (49g)(333.6 J/g) = 16346.4 J
Q3 [heating the water to 100°C]: (49g)(4.184 J/g*°C)(100°C) = 20501.6 J
Q4 [boiling the water]: (49g)(2257 J/g) = 110593 J
Q5 [heating the steam to 169°C]: (49g)(2 J/g*°C)(69) = 6762
Q1 + Q2 + Q3 + Q4 + Q5 = 155430.156 J or 155.4 kJ