Thermal dependancy of conductance

AI Thread Summary
The discussion centers on the relationship between temperature and conductivity, particularly for metals like 304 and 316 steel. It is established that for normal metals, resistance increases with temperature, leading to a decrease in conductivity. The formula presented indicates that conductivity should increase with temperature if interpreted correctly, but confusion arises regarding the terms used in the equation. Clarification is provided that for metals, conductivity typically decreases as temperature rises, while semiconductors behave differently. Ultimately, the consensus is that for the specified steels, conductivity will decrease with increasing temperature.
NEWO
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I am struggling to work out out whether the conductance will increase or decrease with an increase in temperature. This I know sounds so basic yet i can't grasp something. I know that resistance increases with temperature so I would assume that conductivity will decrease. However a formula I have doesn't show this trend. below is the equation in question;

\sigma ^' = \frac {\sigma} {1+ \alpha \delta T}

{my latex command wouldn't work so deleted the tex command to show the equation}

where

\delta{T}=T-T^{'}

\sigma^{'}= conductivity at common temperature = 293K

\sigma= the conductivity at the measured temperature
T^{'}= the common temperature
T=Measured Temperature


which will then mean that;

\sigma= \sigma^{'}\beta

where
\beta =1+\alpha \delta{T}

which says that the conductivity will increase with temperature, from what I understand this doesn't make sense to me!

Please Help

Thanks

n
 
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Do you know what type of material this is for?

Note that for normal metals, it would be true that one would expect the conductivity to drop with increasing temperature. But for a semiconductor, this is not the case. Since increasing the temperature will promote more charge carriers into the conduction band, you will in fact increase its conductivity as the temperature increases.

Zz.
 
ZapperZ said:
Do you know what type of material this is for?

Note that for normal metals, it would be true that one would expect the conductivity to drop with increasing temperature. But for a semiconductor, this is not the case. Since increasing the temperature will promote more charge carriers into the conduction band, you will in fact increase its conductivity as the temperature increases.

Zz.

its for the following material

304, 316, 533, and 508 steel!

thanks for your relply,
 
NEWO said:
its for the following material

304, 316, 533, and 508 steel!

thanks for your relply,
I don't know exactly what kind of steels are labeled 304,316,533,508 ,but if Fe is highly dominant element in the alloy I would always expect that resistance increases as the temperature increases in the vast range of temperatures where techical applications of steels are present.
 
As ZapperZ wrote, in common metals, conductivity decreases with temperature. The problem is in the interpretation of the terms of your formula. \sigma is for T, \sigma' is for T', and \delta T is (T'– T). And all is OK if \alpha >0.

PS: Conductivity of carbon filaments (as in Edison electrical bulbs) increase with temperature.
 
So therefore am I right in saying that,

\sigma=\frac{ \sigma'}{ (1+\alpha\lbrackT-T'))

Which implies that conductivity will decrease with temperature as long as alpha is less than 0

by the way for 304 steel alpha is 0.00172K^{-1}

Thanks for you inputs!

Newo

p.s why won't my latex command work??
 
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(As tex do not works I broke the tags with an space)

When [t ex]\alpha>0[/t ex], conductivity decreases with temperature:

[t ex]\sigma'={\sigma\over 1+\alpha(T'-T)}[/t ex]
If [t ex]\alpha>0[/t ex], when [t ex]T'[/t ex] increases, [t ex]\sigma'[/t ex] decreases.
 
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