## Heat Release from ice (0C) to water (20C)

Chem_Mod
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### Heat Release from ice (0C) to water (20C)

Hi Professor Lavelle.
I have a question, "How much heat is needed to convert 80.0 g of ice at 0.0 degrees Celsius into liquid water at 20.0 degrees Celsius?" Why do you use the specific heat capacity for liquid water instead of solid water in this problem?

Chem_Mod
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### Re: Heat Release from ice (0C) to water (20C)

The question requires two equations: $Q_{fusion} = mC_{fusion}$ to turn the ice into water. And Q = mCΔT for heating the water from $0^{o}C$ to $20^{o}C$. The specific heat for water is used because the temperature is not below $0^{o}C$. The specific hear for ice is only used till $0^{o}C$ then in order to convert to liquid the heat of fusion must be added.
$Q_{fusion} = (80.0 g)(334.0 \frac{J}{g}) = 26700 J$
$Q_{water} = (80.0 g)(4.1813\frac{J}{g^{o}C})(20.0^{o}C - 0.0^{o}C) = 6694 J$
$Q_{total} = 26700 J + 6694J = 3340 J$ (assuming 3 sig figs)

Edward Suarez 1I
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Joined: Fri Sep 28, 2018 12:27 am

### Re: Heat Release from ice (0C) to water (20C)

Chem_Mod wrote:The question requires two equations: $Q_{fusion} = mC_{fusion}$ to turn the ice into water. And Q = mCΔT for heating the water from $0^{o}C$ to $20^{o}C$. The specific heat for water is used because the temperature is not below $0^{o}C$. The specific hear for ice is only used till $0^{o}C$ then in order to convert to liquid the heat of fusion must be added.
$Q_{fusion} = (80.0 g)(334.0 \frac{J}{g}) = 26700 J$
$Q_{water} = (80.0 g)(4.1813\frac{J}{g^{o}C})(20.0^{o}C - 0.0^{o}C) = 6694 J$
$Q_{total} = 26700 J + 6694J = 3340 J$ (assuming 3 sig figs)

i understand errythang but how did u get 334.0 for the Cfusion? when i checked it on the table it was like 6. something lol

1K Kevin
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### Re: Heat Release from ice (0C) to water (20C)

The one that starts with 6 is 6.01 kJ*mol^-1. This is per mole and the one Lavelle used is per gram and isn't given to us on the constant sheet. The two options here are to remember and use that one for this problem or convert the grams of ice to moles and use the 6.01kJ/mol (If you convert to moles and follow through using 6.01kJ/mol you should get the right answer.

Edward Suarez 1I
Posts: 75
Joined: Fri Sep 28, 2018 12:27 am

### Re: Heat Release from ice (0C) to water (20C)

1K Kevin wrote:The one that starts with 6 is 6.01 kJ*mol^-1. This is per mole and the one Lavelle used is per gram and isn't given to us on the constant sheet. The two options here are to remember and use that one for this problem or convert the grams of ice to moles and use the 6.01kJ/mol (If you convert to moles and follow through using 6.01kJ/mol you should get the right answer.

and i did; thank you for the help!