Heating Curve

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

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makenzie2K
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Heating Curve

In the heating curve diagram, in general is the amount of heat released during exothermic phase changes (melting and vaporization) EXACTLY equal to the heat used to break the bonds? I just noticed that the lines in the heating curve depicted in phase changes had no slope which indicates that the positive heat and negative heat cancel each other out.

Shubham Rai 2C
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

Re: Heating Curve

Melting and vaporizing do have to do with breaking small intermolecular bonds, however, I think that the majority of the heat is transferred to the molecules themselves and is translated into kinetic energy. I infer this as the molecules in the liquid state move around at greater speeds than those at solid. Hopefully, this is helpful even though it might not answer your question completely.

Kevin ODonnell 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

Re: Heating Curve

I would assume so yes just based off the assumption that you can't create heat without any work being done. So any heat released from the breaking of molecules would be from the exact energy stored in those bonds.

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