How much energy is needed to heat 3 kg of water through the following changes?
1. heat 3 kg ice from -10 c to 0 c
2.melt 3 kg ice at 0 c
3. heat 3 kg water from 0 c to 100 c
4. vaporize 3 kg water at 100 c
5. heat 3 kg steam from 100 c to 110

1. Heat 3 kg ice from -10 C to 0 C
is sensible heating so
Q = m cp (T2 - T1)
Q = 3 kg (2.09 kJ/kg C) (0 - (-10)) C
Q = 62.7 kJ

2.Melt 3 kg ice at 0 C
requires latent heat so,
Q = m Hfus
Q = 3 kg (333.55 kJ/kg)
Q = 1000.65 kJ

3. Heat 3 kg water from 0 C to 100 C
requires
Q = 3 kg (4.18 kJ/kg C) (100 - 0) C
Q = 1254 kJ

4. Vaporize 3 kg water at 100 C
requires latent heat, so
Q = m Hvap
Q = 3 kg (2258.33 kJ/kg)
Q = 6775 kJ

5. Heat 3 kg steam from 100 C to 110
requires 
Q = 3 kg (2.02 kJ/kg C) (110 - 100) C
Q = 60.6 kJ

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way