Calculating Heat Loss from an Insulated Cup

Click For Summary

Homework Help Overview

The discussion revolves around calculating heat loss from an insulated cup containing coffee, focusing on the principles of thermal conduction. The problem involves determining the rate of heat loss using the thermal conductivity of the cup material and the temperature difference between the coffee and the surroundings.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the formula for heat transfer by conduction, questioning the correct use of area and whether to consider heat loss from all sides of the cup. There is an exploration of unit conversion from calories to joules and the implications of using the outer dimensions of the cup.

Discussion Status

The conversation is ongoing, with participants providing guidance on the correct application of the heat transfer equation and clarifying the need to account for all sides of the insulated cup. There is a recognition of differing interpretations regarding the dimensions used in calculations.

Contextual Notes

Participants are navigating assumptions about the geometry of the cup and the thermal properties, including whether to consider heat loss from all six sides of the cube. The original poster references feedback from a teacher indicating an error in their calculations.

jdog6
Messages
17
Reaction score
0
Help! Heat Loss!

A cup of coffee (with a lid) is enclosed in an insulated cup 0.5cm thick in the shape of a cube 13.1cm on a side. The thermal conductivity of the cup is 0.0002 cal/s*cm*DegreeC. The temperature of the coffee is 87C and the temperature of the surroundings is 14C. Find the heat loss due to conduction. Answer in units of J/s.

I've tried P=kA(Th-Tc)/L ------> Energy transfer by conduction but somehow it's wrong?

I see no other way!

I know that:
k = 0.0002cal/s*cm*DegC which = 0.0008372J/s*cm*DegC
L=0.5cm
A=13.1cm*13.1cm (because cube)
A=171.61 cm^2
Th=87C
Tc=14C

ANY IDEAS?!
 
Last edited:
Physics news on Phys.org
You're actually using kA(Th - Tc)/L , right?

Are you leaking Heat thru all 6 sides?

Did you convert from cal to J?

Are you using 13.1 cm as inside measure, or outside?
How do you know it's wrong?
 
I know its wrong because my teacher said so :frown:

Yes I am using that equation.

I think I'm leaking on all sides

I did convert from cal to J.

And I'm using 13.1cm as an outside measure I believe.
 
Last edited:
Did you multiply the area of each side by 6 sides?
 
Why x6 we are looking for overvall heat loss from this thermos like apparatus.
 
you think Heat only leaks out the top?
How many sides does a cube have?
 
Ok i understand but how do i fit it into the equation? k(6A)(Th-Tc)/L ?
 
Last edited:
yea, or A = 6 s^2
 
Correct! Thank you very much!
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
49
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 47 ·
2
Replies
47
Views
6K
  • · Replies 1 ·
Replies
1
Views
5K