Why does the temperature of a current carrying wire increase?

Quick question: Why does the temperature of a current carrying wire increase?

I am asking this because I am doing an experiment where I increase the temperature of a copper wire by increasing the voltage I pass through it. Except I don't fully understand the reason why the copper wires temperature increases.

I first thought it was due to the resistance of the wire, but shouldn't the resistance of the wire remain constant throughout the experiment? Which would mean the temperature would also remain constant, which is not the case.

Any help would be much appreciated!

Thanks

x)

Thank you!

So does the second explanation comply with V=IR? Voltage will increase and resistance will increase. Why exactly is heat given out due to the current and the resistance of the wire? more electron colliding with the atoms in the wire causing more vibrations and more heat?

From what I know you're right, heat is dissipated due to collisions with atoms (or some elements of atoms - I don't know exactly) .

Two things can happen to dissipated heat - either it will be taken by the surroundings or it can cause the wire's temperature $t_w$ to increase. If the difference between the $t_w$ and surrounding temperature $t_s$ is not high enough, some part of the heat will be given out to the surroundings, and some will contribute to increase in wire's temperature. When the difference in temperature $\Delta t=t_w-t_s$ has reached value high enough, thermodynamical equilibrium is reached, so that the temperature of the wire remains constant from this point on.

If you now increase voltage, more heat is dissipated. This "extra heat" cannot be taken by the surroundings because the difference in temperature is too small, so it "stays within the wire" and causes $t_w$ to increase. When $t_w$ has increased by an appropriate value, new thermodynamical equilibrium is reached, and a new, higher temperature $t_w$ is settled.

This explanation isn't influenced by the fact of increase in resistance qualitatively, but only quantitatively. However, quantitative treatment of the problem would be a bit complicated.

I'm not sure if I've managed to make it clearer, but I hope so As to whether it complies to $V=IR$ - yes, it does, but as I said mathematical treatment would be complicated. If you can specify your doubts about its complying with $V=IR$ more precisely, maybe then I could be of more help.

Well throughout my experiment the copper did not reach a constant temperature, it kept rising until I chose to end it. Therefore I think the first part of your reply doesn't exactly apply here?

It didn't stop rising for given voltage? This probably means that you hadn't waited long enough




All right, I now see what you mean

When you increase voltage, both current and resistance increase (although the increase of resistance is not a direct consequence of increasing voltage - it's because of the rise in temperature). So when you increase voltage, more power is dissipated because of the increase of the current but not because of the increase of resistance - if resistance didn't increase, even more power would be dissipated! It's not so easily seen from the formula V=IR, because I is dependent on R. I suggest that you look at P=V^2/R. The increase in dissipated power is mostly due to the increase in current - electrons have greater velocities, so when they collide, more energy is dissipated.

Ah, wow okay, it took a while but I've got it now. Thanks a lot for your help!!!