# Can anyone figure out this electricity problem?

#### ToTalk

1. The problem statement, all variables and given/known data -

Consider a copper wire 1 mm in diameter providing the power to run an appliance drawing 4.8 kW at 12 V. Assuming no heat radiated away from the wire while the currecnt flows, (a) what will the temperature of the wire be after the current has run for 1 second through the wire? (b) what will the physical condition of the wire be at that time?

2. Relevant equations -

I = P/V
R = V/I
R = pl/A?

I also know the Temperature coefficient of copper is .0068 and its resistivity is 1.68E-8

3. The attempt at a solution -

I solved I = P/V to get 4800 / 12 = 400 A in wire

I also solved R=V/I to get .03 ohms of resistance

At this point I do not know what to do. I am not sure how to turn this info into a heat increase in the wire??? - Thanks very much for any help.

Last edited:

#### berkeman

Mentor
The 400A looks correct. Don't go R=V/I like that -- that would be calculating the input impedance of the load, which is of no help. You want to use the current number and the wire characteristics to calculate how much power is being dissipated by a unit length of wire, and use that to estimate the temperature of the wire after 1s. Use the resistivity of the wire and the diameter to calculate the power. And then calculate what the delta-Temp is for that volume of copper after 1s at that power level. What is the melting point of copper?

#### ToTalk

Thank you for your help! :)

Ok, I understand what you are saying, but I am not sure how to get power from what I am given? Do I multiply the resistivity by the area

1.68E-8 x 7.85E-7 = 1.319E-14 ohms / meter?

1.319E-14 ohms / meter x 400 A = 5.278E-12?

#### berkeman

Mentor
You want the total resistance for a chunk of wire, like say for a 1cm long piece. You listed the relevant equation "R = pl/A", so that's what you will use to figure out the total resistance of that wire piece. What equation do you use to calculate power from current and resistance alone?

Once you have the power, then you need to calculate the delta-Temperature for a 1 second time period. Certainly the change in energy is the power multiplied by the time interval, but to be honest, I'd have to hunt for the right relation to use to figure out the delta-Temp. Probably has to do with the specific heat or something?

#### ToTalk

Thanks again for your help.

So if I use R = (1.68E-8) * (1m / 7.85E-7 m^2) = .021 ohms / meter

Then I take P = I^2R = 400^2 * .021 = 3360 Watts / Meter?

Energy = 3360 Joules / second / meter

Copper Specific Heat = 390 J / kg *C

Volume of 1 meter = 7.85E-7 * 1 = 7.85E-7 cubic meters

Density of copper = 8.9E3 kg/cubic meter

volume * density = .00699 kg

energy / specific heat = 8.62 C / kg

8.62 / .00699 = 1233.19 C ?

Does that seem right? Thank you very much for your help.

#### berkeman

Mentor
The units of resistance are Ohms, not Ohms/meter. So your first line should have units of Ohms, and you can parenthetically say (total R for a 1m piece of wire).

The rest looks reasonable, but you need to be careful about what temperature thing you are calculating. It's probably the rise in temperature above ambient for that energy input? So you might need to add in room temp if you are calculating the delta-temp. And what is the melting point of copper? Is this wire close?

#### ToTalk

Thanks for the corrections. In the question it says the wire starts at 20 C, I just forgot to include that on here. I find the melting point of copper to be 1084.62 C, so this wire would be mush!

Thanks so much for your help, I really learned a lot, and I really appreciate it.

#### berkeman

Mentor
You're welcome, glad to help. It makes things a lot easier when people use the Homework Posting Template and show their work, as you did.

BTW, to avoid small mistakes that can lead to larger ones, I like to carry units along in my calculations, and keep checking for unit consistency. Just as you multiply and divide quantities, do the same with their units, and keep the overall units next to the quantities as you go. I like to use square brackets to delineate my units in the equations. So for each term, you will have the units for that term sitting next to them. The units of each term to be added must be the same, and the units of the left hand side and the RHS must be the same. If you make an algebra mistake, the units will very often show it up right away. Handy trick that I learned a long time ago, and it's so valuable that I still use it every day in my work.

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