Question

What is the heat required in Joules to raise the temperature of 100.0g ice at
-20 degrees C to steam at 120 degrees C?

Answer 1

to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed

now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from ice->water->steam
thus we need another data called it Latent Heat of Fusion and Latent Heat of Vaporization to calculate for the heat required to change its phase:
H1 = m(Lf)
H2 = m(Lv)

where
m = mass
Lf = Latent Heat of Fusion (fusion means melting)
Lv = Latent Heat of Vaporization (vaporization means from liquid, it becomes vapor)

thus for the problem,
H1 = 100*Lf
H2 = 100*Lv

also, the heat required to raise the ICE's temp (from -20 C to 0 C) is
Q1 = mc(0 -(-20)) = 100*20*c = 100*20*c

the heat required to raise the LIQUID's temp (from 0 C to 100 C) is
Q2 = mc(100 - 0) = 100*75*c = 100*100*c

the heat required to raise the STEAM's temp (from 100 to 135) is
Q3 = mc(120 - 100) = 100*20*c

thus the total heat needed is:
Q,total = Q1 + H1 + Q2 + H2 + Q3

note that can find c, Lf and Lv of water in books or you can just google them. remember and be careful of the units.

hope this helps~ :)