calculate the total amount of head absorbed (in kj) when 2.00 mol of ice at -30.0C is converted to steam at 140.0 C

Answer 1

I assume you want heat absorbed and not head absorbed.
You want to go through the following.
q1 = heat to move T of ice from -30 to zero C (note all solid phase)
q2 = heat to melt ice at zero to liquid at zero C(note phase change)
q3 = heat to move T of liquid water to boiling at 100 C.(note all liquid phase)
q4 = heat to convert liquid water at 100 C to steam at 100 C(phase change)
q5 = heat to move T of steam from 100 C to 140 C. (note all vapor phase).
You see here that you have two phase changes and three where there is no phase change. There is one formula to use for the phase changes and another one where the phase remains the same throughout the temperature change.
phase change formula is q = mass x heat fusion (at the melting point) or heat vaporization (at the boiling point).
Here are the first two steps for the above.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). mass ice is 2 mol (36 grams). You will need to look up the specific heat of ice but it is about 2 J/c). Tinitial is -30 C and Tfinal is 0 C.
q2 = mass ice x heat fusion
q2 = 2 mol ice x heat fusion. You will need to look up heat fusion if it isn't in the problem.
Now go through with q3, q4, q5 then add q1 + q2 + q3 + q4 + q5 to total q.
Post your work if you get stuck.