Answer:
7479 cal.
31262.2 joules
Explanation:
This is a calorimetry problem where water in its three states changes from ice to vapor.
We must use, the calorimetry formula and the formula for latent heat.
Q = m . C . ΔT
Q = Clat . m
First of all, let's determine the heat for ice, before it melts.
10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal
Now, the ice has melted.
Q = Clat heat of fusion . 10 g
Q = 79.7 cal/g . 10 g → 797 cal
We have water at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.
Q = m . C . ΔT
Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal
Water is ready now, to become vapor so let's determine the heat.
Q = Clat heat of vaporization . m
Q = 539.4 cal/g . 10 g → 5394 cal
Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°
Q = m . C . ΔT
Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal
Now, we have to sum all the heat that was added in all the process.
100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.
We can convert this unit to joules, which is more acceptable for energy terms.
1 cal is 4.18 Joules.
Then, 7479 cal are (7479 . 4.18) = 31262.2 joules
(where Q is heat, m is mass, c is specific heat and 
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature