Problem 8.19

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Tyler_Ash_3C
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Joined: Wed Sep 21, 2016 2:57 pm
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Problem 8.19

Postby Tyler_Ash_3C » Mon Jan 16, 2017 4:13 pm

On problem 8.19 of the homework exercises, there are two terms for the total heat supplied, one to eat the copper kettle and one to heat the water. In both cases the temperature change used is 100-22 degrees Celsius. For these types of problems, is the temperature change of the container always the same as the temperature change for the substance inside?

vsyacoubian2A
Posts: 63
Joined: Fri Jul 22, 2016 3:00 am

Re: Problem 8.19

Postby vsyacoubian2A » Wed Jan 18, 2017 6:53 am

Yes, for these types of problems, you have to assume that the temperature change is the same for both.

David Sung 2H
Posts: 15
Joined: Wed Sep 21, 2016 2:56 pm

Re: Problem 8.19

Postby David Sung 2H » Wed Jan 18, 2017 8:47 pm

When I convert the copper into moles and use the molar specific heat for calculations, I end up with a different value for the heat than when I directly use the specific heat.

Using molar specific heat, I get (500.0g Cu) * (1 mol/63.55g Cu) * (33J/K*mol) * (78K) = 20251 J for the copper.
But using specific heat (grams), I get 14800 J.

Can anyone help me understand why there is this difference?

THANKS


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