Question

At 1 atm, how much energy is required to heat 37.0 g of H2O(s) at –18.0 °C to H2O(g) at 125.0 °C?

Answer 1

for this you have to know enthalpy of fusion and vaporization for water, and heat capacities of water in solid, liquid and gaseous phase.
i found this values on internet:
enthalpy of fusion - Hf = 0,334 kJ/g = 334 J/g
enthalpy of vaporization - Hvap = 2,26 kJ/g = 2260 J/g
heat capacity of solid phase Cp,s = 2,11 J/ (g K)
heat capacity of liquid phase Cp,l = 4,18 J/(g K)
heat capacity of gaseous phase Cp,g = 2,08 J/(g K)

now you have few processes of heating ice to vapor:
1. heating ice to 0°C
delta T = 0°C - 18°C = 18K
Q1 = m * Cp,s * deltaT = 37g * 2,11 J/(g K) * 18K = 140500 J = 140.5 kJ
2. melting the ice
Q2= m * Hf = 37g * 334 J/g = 12360 J = 12.36 kJ
3. heating water to 100°C
delta T = 100°C - 0°C = 100K
Q3= m * Cp,l * deltaT = 37g * 4,18 J/(g K) * 100K = 1546600 J = 1546.6 kJ
4. vaporization of water
Q4= m * Hvap = 37g * 2260 J/g = 83620 J = 83.62 kJ
5. heating of vapor
delta T = 125°C - 100°C = 25 K
Q5= m * Cp,g * deltaT = 37g * 2,08 J/(g K) * 25K = 192400 J = 192.4 kJ

total heat is Q = Q1 + Q2 + Q3 + Q4 + Q5 = 140.5 kJ + 12.36 kJ + 1546.6 kJ + 83.62 kJ + 192.4 kJ = 1975.5 kJ = 1975500 J