4C.13 7th Ed.

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Summer de Vera 2C
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

4C.13 7th Ed.

Postby Summer de Vera 2C » Sat Feb 09, 2019 4:05 pm

Hi! I understand how to do the problem, however the small part that I'm confused about is which specific heat of water I should use to calculate q(ice cube). Shouldn't Csp=2.03 J/gC instead of 4.184 since it's changing temperature in the solid state still?

Hovik Mike Mkryan 2I
Posts: 95
Joined: Fri Sep 28, 2018 12:25 am

Re: 4C.13 7th Ed.

Postby Hovik Mike Mkryan 2I » Sat Feb 09, 2019 4:47 pm

Hello,
I believe that for this problem both the heat capacity for ice and liquid water is used because it is required to melt the ice first than raise the temperature from 0Celsius to the desired temperature. Hope this helped!

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: 4C.13 7th Ed.

Postby Matthew Choi 2H » Sat Feb 09, 2019 11:45 pm

If you look at josephyim1H's post in this same thread you will see an explanation for this problem. Basically, the gist of what I said was that the ice will first melt (use delta H) and then increase in temperature to the exact same temperature as the rest of the water that was already in the glass. That heat gained will equal the amount of heat lost by the water which is a simple temperature change problem (q=mCAT).


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