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9677 lucky #13

Posted: Fri Nov 04, 2005 5:27 pm
by opticschmoptics
Ok. I have a little trouble with thermo normally, but this one seems to be missing a key bit of information, doesn't it?

I tried to reverse engineer it by saying the water cooled by blackbody radiation. Assuming I have Stefan's constant memorized, and that I can do the quartic of 373 K in my head, we get the correct answer if we assume that the liter of water has a an exposed surface of of 0.1m^2--which the word "pan" doesn't seem to explicitly endorse.

There's very little out there on the actual time-dependence of thermodynamics. Most of the stuff I read talks about equilibria which is frustrating (sounds more like statics to me). I never had a stat-mech class, maybe someone who has can help?

Posted: Fri Nov 04, 2005 11:49 pm
by yosofun
hi, the complete GR9677 solutions will be up sometimes this weekend on my website ... until then, please bear with the LaTeX syntax in my solution below:

Given $P=100$ W and $V=1L=1m^3=1kg$ for water, one can chunk out the specific heat equation for heat, $Q=mc\Delta T=Pt\Rightarrow 2200(1^\circ) = 100t\Rightarrow t \approx 40 s$, as in choice (B).

Posted: Sat Nov 05, 2005 7:17 am
by opticschmoptics
Thanks yosofun, I appreciate all the solutions in LaTeX. But here's what I don't get...we heated the water at 100 W, but how does that mean it will cool at 100 W?

With the heating source removed, it seems like the water could only cool by blackbody radiation, convection with the air above it, and conduction through the pan, none of which we know enough about to evaluate...

But, in the spirit of physics I'll take a guess: "Although the heating element is on for a long time, the water ... does not boil." So we have equilibrium at 373 K, right? Which means $P_{heat} = P_{cool}$. Tell me if I'm wrong, but I think that's the one logical piece of the puzzle I need to confirm.

Thanks again!