Given that the specific heat capacities of ice

and steam are 2.06 J/g ·
◦C and 2.03 J/g·
◦C,
respectively, calculate the total quantity of
heat necessary to melt 36.1 g of ice at −50.0
◦C and heat it to steam at 200◦C. The molar
heats of fusion and vaporization for water are
6.02 kJ/mol and 40.6 kJ/mol, respectively.
Assume water freezes at 0◦C and boils at
100◦C.
Answer in units of J.

In steps.

q1 to change ice @ -50 C to ice @ 0 C.
q1 = mass ice x specific heat ice x Tfinal-Tinitial (I call Tf-Ti = dT which stands for delta T) where Tfinal is 0 and Tinitial is -50 so Tf - Ti = 0 -(-50) = 50 so you will have
q1 = 36.1 g x 2.06 J/g*C x 50 = ?

q2 = heat to melt ice @ 0 C to liquid @ 0 C.
q2 = mass ice x heat fusion = ?

q3 = heat to change T of liquid water @ 0 C to 100 C
q3 = mass H2O x specific heat H2O x delta T

q4 = heat to vaporize H2O @ 100 C to steam @ 100 C.
q4 = mass H2O x heat vaporization = ?

q5 = heat to change T of steam @ 100 C to steam @ 200 C
q5 = mass H2O(steam) x specific heat steam x delta T.

Then total Q = q1 + q2 + q3 + q4 + q5.
Note here that I have not been consistent with the units BECAUSE the problemis not consistent with units for specific heat. For example, q1 comes out in J because specific heat is in J/g*C/ Also true for q3 and q5. But the two places where there is a phase change (q2 and q4) the heat fusion and heat vaporization are given in kJ/mol so you must change that 36.1 g ice to mols (mols = 36.1/18.01 = ?) AND then the answer is in kJ and not J so you will need to correct that too.
Post your work if you get stuck.