Calculating Heat Transfer in a Two-Reservoir System: A Simple Problem

AI Thread Summary
The discussion revolves around calculating heat transfer between a hot reservoir at 653 K and a cold reservoir at 399 K, using a brass rod as a conductor. The calculated thermal resistance is 0.091743 K/W, leading to a thermal current of 2768.6 W. Over ten minutes, this results in an energy transfer of approximately 1,661,162.16 J. Participants suggest that the issue may stem from the homework software's requirements for significant figures, as the precision of the inputs does not warrant such a detailed output. There is also a suggestion to verify the units used in the final answer, as the software may expect kJ or MJ instead of J.
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Homework Statement


A hot reservoir with a temperature of 653 K is 0.7 m away from a cold reservoir with a temperature of 399 K. The two reservoirs are insulated from each other except for a rod of brass (k = 109 W/m-K) that has a cross-sectional area of 0.07 m2. The entire system is allowed to reach a steady-state condition.
1)How much energy is transferred between the hot reservoir and the cold reservoir by heat in ten minutes?

Homework Equations


thermal current=change in temp*thermal resistance, I=ΔT/R
R=|Δx|/kA

The Attempt at a Solution


R=.7/(109*.07)=.091743 K/W
I=(653-399)/.091743=2768.6 W
for 10 minutes: 2768.6 J/S * 60 S/min * 10 min=1,661,162.16 J
Sorry for the simple question, but the homework software says that answer is incorrect and I can't really see what could be wrong with such a straightforward problem, I imagine I'm missing something stupid but can't find it.
 
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Try to write the result in normal form, with three significant digits.
 
Hello and welcome to PF!

I think your work is correct. I get the same answer.
 
To expand on what ehild said: You are using way too many significant digits. Based on the precision of the inpot, there is no way that you can obtain an answer with that kind of precision. This means that most of your significant digits are junk and should not appear in the final answer. Keeping the digits throughout computation is a good thing to avoid rounding errors propagating, but in the final answer you simply do not have precision. Based on the input, giving any more than two significant digits would be overstating the precision.
 
Thanks for the replies, but it's smartphysics/flipitphysics. It doesn't mark things off for overdoing significant figures, it usually requires more than would be in line with the problem's parameters. I did try the answer to 2 and 3 figures just to be sure, but it seems like the problem is bugged.
 
Did you give the unit as J? What about using kJ or MJ?
 
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