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

Click For Summary

Homework Help Overview

The problem involves calculating heat transfer between a hot reservoir at 653 K and a cold reservoir at 399 K, separated by a brass rod. The original poster seeks to determine the energy transferred in ten minutes, given the thermal properties of the system.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate thermal resistance and energy transfer but questions the correctness of their answer based on feedback from homework software. Some participants confirm the calculations while others suggest considering significant figures and unit presentation.

Discussion Status

The discussion is ongoing, with some participants validating the original poster's calculations and others providing insights on significant figures and potential issues with the homework software. There is no explicit consensus on the correctness of the answer, but productive suggestions have been made regarding presentation and precision.

Contextual Notes

Participants note that the homework software may have specific requirements for significant figures and units, which could affect the evaluation of the answer.

limper
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K