Reef Chemistry Question of the Day #147 Heat Dissipation

Randy Holmes-Farley · Nov 4, 2015

beaslbob said:
can we ignore the evaporative effect of condensate on the outside of the container?

Yes. Would there be any on a hot container?

beaslbob · Nov 4, 2015

Randy Holmes-Farley said:
Yes. Would there be any on a hot container?

just trying to cover all bases.

heatdissipation · Nov 4, 2015

Are you ever going to tell us your answer on this randy?

cee · Nov 4, 2015

Really a heat transfer problem. The higher the temperature difference, the quicker the cooling. However, the ice has less surface area than the surface of the glass and both cool primarily by convection so I think we would really need more information (q=hA(deltaT)). Given enough time, all options will reach the same equilibrium of 70 degrees but I don't know the volume of water, the diameter/height of the glass, the thickness of the glass, the dimensions of the ice cube, etc.

s2nhle · Nov 4, 2015

I would said d

Cory · Nov 4, 2015

Id say c because your allowing time to cool it for 30 minutes and the ice cube.

Randy Holmes-Farley · Nov 5, 2015

heatdissipation said:
Are you ever going to tell us your answer on this randy?

Wow, with a name like that you should have the answer already.

Randy Holmes-Farley · Nov 5, 2015

And the answer is...C. Wait 25 minutes, then add one 10 mL ice cube which melts within 5 minutes.

The rate of heat flow between any two objects is related to the difference in temperature between them. Think of heat like a pressure of temperature. The higher the temp difference, the higher the pressure pushing heat into the cooler object. The higher the difference, the faster the warmer object cools (all other things being the same).

So heat leaves the supplement (or coffee or whatever) fastest when it is hottest. For that reason, if you want it to loose heat fast, you want it to stay warmer for more of the time period. Consequently, adding the ice cube at the end will result in the coolest temperature.

Happy Reefing!