Heat Transfer
Posted by Nathaniel.
I’m competing for the geekiest post of the week prize, but this is something that I’ve been thinking about for a while.
When you take a cold can of soda out of the fridge and put it on a table on a warm humid day, it obviously heats up. Heat enters the can through conduction from the table, conduction to the surrounding air, and by giving absorbing energy from the latent heat of water condensing on the can surface.
The question is this, how much of the warming of the can is due to that latent heat? Is it significant or insignificant?
July 17th, 2007 at 1:42 pm Using
Ok, I just answered this one myself.
It takes about 36,000 J to heat a can of soda from fridge temperature (36 F) to “warm” (80 F).
Q=355 ml * 1 g/ml * 4.186 J/g/C * (80-36) F * 5/9 C/F = 36,325 J
The latent heat of vaporization for water is 540 C/g or 2260 J/g.
So, if the warming was solely due to the latent heat, you’d end up condensing 16 g of water onto the table. 16 g = 16 ml =16 cc = a cube of water almost exactly one inch on a side.
Since the puddle on the table is nothing close to that much water, the latent heat isn’t significant in the warming.
QED
July 18th, 2007 at 1:08 pm Using
this was not a competition. you win hands down.