I Need Help With Chem Please!!!!!?

Question

How much heat (kJ) must be removed in order to change 3.70 moles of water at 62.0 oC to ice at -11.0 oC ?


The specific heat capacity of solid water is 2.03 J g-1 oC -1


Use a negative sign in your answer if heat is removed.

Answer 1

Well, there are three different steps to this.
First of all we need to go from 62oC to 0oC and see how much heat this will give off.
Secondly, during a phase change, you have to remove heat. During this process, the temperature will remain constant.
Thirdly, you have to make the ice go from 0oC to -12oC
Before we begin, it would be helpful to determine the mass of the water involved.
Mass H2O = 3.70 mol * 18.02 g/mol
= 66.67g
During it's liquid state, water has a specific heat capacity of 4.19 J/(goC). The change in temperature is (0.0 - 62.0 = -62 degrees).
To go from 62 to 0:
= 4.19J/(goC) * 66.67g * -62.0oC
= -17320J
= -17.3 kJ
Now for the phase change. The heat of fusion for ice is -6.02kJ/mol
= -6.02 kJ/mol * 3.70 mol
= -22.3kJ
Now from 0oC to -11.0oC Since we're cooling the water down, it is an exothermic process.
=2.03J /(goC) * - 11oC * 66.67g
=- 1489J
= - 1.5kJ
The change in enthalpy is the sum of these three
=-17.3 kJ+-22.3kJ+ - 1.5kJkJ
= - 41.1kJ
You can make sure my calculations are correct. I know that my procedure is. Good luck :)
-The Mad Scientist

Answer 2

1 mole of water =18g 3.7 moles =66.6g

first you must cold the water from 62 to zero

so the heat capacity of liquid water is 1cal/g

you must remove Q = -62*66.6 =-4129 cal
to freeze one gram of water you take 80 cal

so for66.6 g q =-80*66.6 =-5328 g

and to freeze to -11 you need if 1J =4.18cal

-66.6*11*2.03/4.18=- 355.7 =-9182 cal