Help with an enthalpy lab?

Question

We did a lab about the enthalpy of the fusion of ice in chem 2 class where you add ice to water and watch the temperature change. data:
cup before 4.7 g, cup 1/3 full of water 180.5 g
the temperature started at 18 C and the final temp was 2 C

we need to calculate the heat lost, heat gained, heat of fusion, and the change in Hfus in kj/mol

would someone please help me figure out how to determine this assignment? thank you so much

please help soon! chemistry confuses me in the classroom

Answer 1

I don't see the mass of the ice, without which you cannot hope to complete your calculations.

mass of water = 180.5 - 4.7 = 175.8
Without knowing the cup material and thickness we can't calculate the heat gained/lost to and through the cup, so let's assume it is negligible.
The heat capacity of water is 4.184 J/g°C
The heat lost by the water is
(175.8 g)(4.184 J/g°C)(18 - 2)°C ≈ 11,768.7552 J or 11.7687552 kJ (1)
(Don't round until your calculations are complete, if possible.)
The heat gained by the ice must equal the heat lost by the water if there is no heat transfer in or out of the system, so
heat gained ≈ 11.7687552 kJ
This heat has to be distributed between the total heat of fusion and the heat required to raise that mass of water 2°C.
Let m = the mass of the ice in grams
mHf + (4.184 J/g°C)(2°C)m = 11.7687552 kJ
Hf = (11.7687552 kJ)/m - (0.008368 kJ/g)
(Note the conversion from J to kJ.)
To convert this to the desired units of kJ/mol you need to multiply by 18 g/mol.


If the mass of the ice is included in the 180.5 g
you will need to subtract it from the 175.8 g I used for the mass of the water and redo the calculation in equation (1).