8.39

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joycelee1A
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Joined: Fri Sep 29, 2017 7:05 am

8.39

Postby joycelee1A » Fri Jan 19, 2018 4:05 pm

I didn't fully understand 8.39. The question asks for the amount of heat needed to convert 80.0g of ice at 0 °C into liquid water at 20°C. Would there be a calculation for phase change we need to take into consideration? I wasn't sure how to go about it.

Jenny Cheng 2K
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Joined: Fri Sep 29, 2017 7:05 am
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Re: 8.39

Postby Jenny Cheng 2K » Fri Jan 19, 2018 4:45 pm

For this problem, you need to consider the energy needed for the phase change (melting/fusion) as well as the energy needed for the temperature change. Looking at a heating curve for H2O, solid H2O will melt at 0.0°C. The sum of the energy needed for the phase change and the energy needed for the temperature change will provide the amount of heat needed:
(80 g)(1 mol H2O/18.02 g)(6.01 kJ/1 mol H2O) + (4.184 J/(g*°C))(80.0 g)(+20.0°C)
The portion of this calculation before the addition sign is the energy needed for the phase change (convert grams of water to moles of water, then multiply by the standard enthalpy of fusion for water). The second portion is the energy needed for the temperature change (use the specific heat capacity of liquid water in q=mCsΔT).

Luke Bricca 1H
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.39

Postby Luke Bricca 1H » Sat Jan 20, 2018 9:19 am

In addition, if the problem instead wanted the final phase to be steam, another phase change would be required, adding a third step to your calculations. For any phase change problem, the work can be chunked into sections: energy required to raise temp to 0 deg C, energy for fusion phase change, energy required to raise temp to 100 deg C, energy for vaporization phase change, energy required to raise temp to whatever specified steam temp. Of course, each problem may only use a few of those segments, but that's the general method.


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