How much heat (kJ) must be removed in order to change 3.11 moles of water at 41.0 oC to ice at -24.0

Question

How much heat (kJ) must be removed in order to change 3.11 moles of water at 41.0 oC to ice at -24.0 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.

7386.56 or - 7386.56 is wrong. I also got the same answer but it's wrong...

Answer 1

There are three heat calculations you have to do here:
1) the heat that must be removed to cool the liquid water to 0 deg C
2) the latent heat of fusion to turn the water to ice at 0 deg C
3) the heat that must be removed to cool the 0 deg C ice to -24 deg C

Add up the three heats and there is your answer.

1. The specific heat of water is 4.186 J/g C, so
h1 = (3.11 mol * 18 g/mol) * 41 deg C * 4.186 J/g C = 9607 J

2. The latent heat of fusion of water is 335.5 J/g, so
h2 = 335.5 J/g * (3.11*18) = 18781 J

3. The specific heat of ice is 2.03 J/g C, so
h3 = (3.11 mol * 18 g/mol) * 24 deg C * 2.03 J/g C = 2727 J


The total heat to be removed is then 31115 J = 31.115 kJ. More than half of the heat that needs to be removed is just used to accomplish the phase change at constant temperature!

Answer 2

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Answer 3

Hey the steps that "jstro" gave you are indeed correct, but what he forgot is that you need to divide it by 1000. So when you get that answer in J just divide it by 1000 and you should be good. The answer is going to be a negative since it is losing energy.

Answer 4

change in temp=41+24
=65
3.11 moles =3.11*molecular mass of water
=3.11*18 gms
=55.98 gms
heat removed for 1degree loss of temp for 1 gm water is
=2.03
hope u gt it
then the heat required=2.03*65*55.98=7386.56 j0ules