Relationship between material resistance and temperature

[rede96]

For the material used in a heater plate to generate heat for example, for a given amount of power (say 4.5 kw) does the higher resistance of the material mean it will produce a higher temperature? Or is it the other way around?

 

[BvU]

Not completely clear to me what you mean. For a given voltage, the power dissipated in a heating wire is $P = I \times V = V^2/R$, so a higher resistance leads to a lower power.

 

[rede96]

Not completely clear to me what you mean. For a given voltage, the power dissipated in a heating wire is $P = I \times V = V^2/R$, so a higher resistance leads to a lower power.

Thanks for the reply. Basically I wanted to know for a given resistance in a material if I pass a current through it will materials of a higher resistance get hotter than those with lower resistance.

Taking $P = I \times V$ and Ohm's law $V = IR$ and eliminating the $V$ then $I^2R$ is the way the power is distributed. So heat (power) is proportional to resistance, i.e. the more resistance the more heat? But I haven't done any of this for such a long time I'm just not sure if that's right.

 

[CWatters]

You can write...

P=I^2R which suggests P is proportional to R
or
P=V^2/R which suggests P is proportional to 1/R

The apparent contradiction occurs because the variables are not independent.

The answer to your question is... It depends what you keep constant (I or V) when you change R. You cannot keep both constant.

 

[CWatters]

In most cases the voltage is constant (for example the mains voltage is fixed at 110V or 220V). In that case power is proportional to 1/R.

It a few cases the current is constant (for example some types of battery charger). In that case power is proportional to R.

 

[rede96]

In most cases the voltage is constant (for example the mains voltage is fixed at 110V or 220V). In that case power is proportional to 1/R.

It a few cases the current is constant (for example some types of battery charger). In that case power is proportional to R.

I might be getting mixed up with heat and power. Basically what I was interested in knowing, assuming a constant voltage, is if I have two heater plates one with a higher rated resistance than the other. Which would heat up more? I thought it’d be the one with the higher resistance?

 

[BvU]

See post #2. A lower resistance means a higher current. With the same voltage, the product I x V is then higher.

Power I x V is heat dissipated per unit time

 

[rede96]

See post #2. A lower resistance means a higher current. With the same voltage, the product I x V is then higher.

Power I x V is heat dissipated per unit time

Right, got it thanks. So for two heater plates given the same voltage the one with the lower resistance will heat to a higher temperature in the same time frame.

 

[morrobay]

Not completely clear to me what you mean. For a given voltage, the power dissipated in a heating wire is $P = I \times V = V^2/R$, so a higher resistance leads to a lower power.

Then with the conservation of energy principle this would be a case of electrical energy transferred to heat energy. The conduction electrons collide with the ionic lattice (resistance) this increases amplitude of thermal lattice vibrations corresponding to temperature increase.

 

[CWatters]

Right, got it thanks. So for two heater plates given the same voltage the one with the lower resistance will heat to a higher temperature in the same time frame.

That's correct (if we assume everything else is the same, such as the thermal properties of the heat plates).