Calculating Temperature Change in Materials with Specific Heat and Conductivity

Click For Summary

Homework Help Overview

The discussion revolves around calculating the temperature change in materials based on specific heat, conductivity, and current density. The original poster seeks guidance on how to approach the problem involving the formula for temperature rise due to Joule heating.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between current density, Joule heating, and the specific heat of materials. Questions arise regarding the introduction of time and temperature into the expression for temperature change.

Discussion Status

Some participants provide insights into the relationship between current, resistance, and heat production, suggesting a method to calculate total heat energy. There is an exploration of the assumptions involved in the problem, particularly regarding the heating of a material without heat loss.

Contextual Notes

The original poster expresses uncertainty about how to begin the problem, while others clarify concepts related to Joule heating and the definitions of current density and conductivity. The discussion includes specific values and examples to illustrate the concepts being discussed.

brad sue
Messages
270
Reaction score
0
Hi ,please I need something to begin this problem:

The specific heat s of a material in [J/(kg deg C] is the amount of energy in joules required to raise the temperature of 1[kg] of material by one degree C. The density ρ of a material in [kg/m3] is the mass in [kg] per cubic meter. If a current density J exists inside a material for a time Δt, show that the rise in temperature Δ T in degree C given the formula (σ is conductivity of the material):

Δ T=(J2* Δ t)/(s* σ* ρ)

I don't know how to begin.
B
 
Physics news on Phys.org
Well I believe its pretty straight forward problem, assuming you know expression for Joule heating, normally expressed as I^2*R
where I=current. Now consider the second part, the mass that is being heated and the constant that relates mass and heat to increase 1 degree. Is this any help?

Incidentally as this may come up, the current density J is the same as I (total current) divided--or "normalized"--by the area thru which I flows thru. Conductivity (the reciprocal of resistance) is a normalized quantity. So the relation between J^2/conductivity is same as I^2*R
 
Last edited:
I tried but I cannot make it right. I cannot the time and temperature in the expression.
how can I introduce it?
 
Last edited:
NP. Let's just look at the case of a resistor and we can fix it later if you don't get the normalization in last part of post I mentioned.

I^2*R = rate of heat production (units of power). Multiplying by delta time gives total heat energy from Joule heating.

Lets just say it was 6A and 5 Ohms of resistance;
then heat production=180W Let's pick an arbitrary time of 10 seconds,
then heat=1800 Watt seconds.(Joules)

Let's say this was an aluminum block and assume no heat lost via transfer at surface so all energy goes into heating block.

Heat capacity=0.9J/C-g where K is degrees Kelvin and g is one gram (see here if confused by term:

http://www.iun.edu/~cpanhd/C101webnotes/matter-and-energy/specificheat.html


Lets say block is 50 grams, the total temperature rise delta T (big T)
Delta T=total heat energy/total heat capacity=1800/(50*0.9)

Or 4.5 degrees. In this problem your constants are provided in consistent units, so no worries there.
 
Ok I got it now
Thank you!
 
you're very welcome.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
1K