# Specific heat and temperature

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

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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

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I tried but I cannot make it right. I cannot the time and temperature in the expression.
how can I introduce it?

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NP. Lets 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 Lets pick an arbitrary time of 10 seconds,
then heat=1800 Watt seconds.(Joules)

Lets 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.