Calculating Temperature Change in Materials with Specific Heat and Conductivity

In summary, the specific heat of a material is the amount of energy in joules required to raise the temperature of 1 kg of material by one degree Celsius. The density of a material is the mass in kilograms per cubic meter. If a current density exists inside a material for a time, the rise in temperature given the formula (σ is conductivity of the material) is: Delta T=(J2*Δt)/(s*σ*ρ).
  • #1
brad sue
281
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
  • #2
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:
  • #3
I tried but I cannot make it right. I cannot the time and temperature in the expression.
how can I introduce it?
 
Last edited:
  • #4
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.
 
  • #5
Ok I got it now
Thank you!
 
  • #6
you're very welcome.
 

Related to Calculating Temperature Change in Materials with Specific Heat and Conductivity

What is specific heat?

Specific heat is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius.

What is the difference between specific heat and temperature?

Specific heat is a physical property of a substance, while temperature is a measure of the average kinetic energy of particles in a substance. Specific heat is a measure of how much energy is needed to change the temperature of a substance, while temperature is a measure of the intensity of heat.

How is specific heat measured?

Specific heat is typically measured using a calorimeter, which is a device that measures the amount of heat absorbed or released during a chemical or physical process.

Why is specific heat important?

Specific heat is important because it helps us understand how heat energy is transferred and how substances respond to changes in temperature. It also plays a crucial role in many industrial processes, such as heating and cooling systems.

What factors affect specific heat?

The main factors that affect specific heat are the type of substance, its mass, and its temperature. Different substances have different specific heats, and the specific heat of a substance may also change depending on its phase (solid, liquid, or gas).

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
893
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
905
  • Introductory Physics Homework Help
Replies
3
Views
884
Back
Top