Calculate temperature of a conductor

[Themadflower]

Hello !
1. Homework Statement
i study for my exams. here is my current problem:
a non-specific conductor is connected to an ideal battery (surrounding temperature: 20°C) and reaches a temperature of 24 °C . after cut in half it is again connected to the battery. what temperature does it reach now?

Homework Equations

that is all.

The Attempt at a Solution

i looked for equations to deal with the demand.. i used an equation from stackexchange:

P=I2∗R(T)
E(t)=∫Pdt
T=T0+dT
dT=E(t)m∗C
m=V∗density
V=l∗A
R(T)=l/A∗r(T)

The above can be condensed into a linear approximation:
R(T) =l/A∗(r+T∗α)−>R(dT) =l/A∗(r0+dT∗α)

combining all this: dT =∫I2∗l/A∗(r0+dT∗α)dt/(l∗A∗density∗C)=I2/(A2∗density∗C)∗∫r0+dT∗αdt

if dT∗α<<r0

then dT =I2∗r0∗dt/(A2∗density∗C)

. i know the difference in temperatur is 4 K. so i transposed to get dT for which i put in the 4K then i figured out that the current should be the same so i took the same equation again without exchanging dt with 4 and took l/2 instead of l and well, than equated it. now i have an equation with lots of d´s (diameter) and l´s. so maybe that does not work.

i consider another way to deal with the temperature coefficient. i used that the resistance of the conductor changes with the temperature and is half less than before if we cut the conductor in half.. i used:

and just claimed that Rk (the R at the beginning) is 100 Ohm and dR is 50 Ohm ( because it is 1/2 Rk) and dT = 4K . then i have alpha.. i do not know how to continue, or if it is even right what i did.. could you help?

kind regards!

 

[Chestermiller]

Suppose that the resistance did not depend on temperature. How would you solve the problem?

 

[Themadflower]

well, maybe i would it now is half as long as before so the temperatur rise will also be half dT1? that makes 2 K?
and a total temperatur of 26?

another idea would be to say, it is first: 20 + 1/5 *20 = 24
and now 24 + 1/5 * 1/2 *24 = 26,4

 

[NascentOxygen]

maybe i would it now is half as long as before so

If its original resistance was R, then once its length is halved what will its new resistance be?

How will its electrical watts compare with before?

 

[Themadflower]

R/2? could it be?

 

[Themadflower]

than m2 = m1 /2 .
if i take Q=c*m*dT = I^2 *R
-> dT = (I^2 *R)/(c*m) and dT2= (I^2 *R/2)/(c*m/2) which is the same as dT ...

 

[NascentOxygen]

If its original resistance was R, then once its length is halved what will its new resistance be?

How will its electrical watts compare with before?

Also, how will the current change?

 

[Themadflower]

it will double

 

[Chestermiller]

than m2 = m1 /2 .
if i take Q=c*m*dT = I^2 *R
-> dT = (I^2 *R)/(c*m) and dT2= (I^2 *R/2)/(c*m/2) which is the same as dT ...

You should not be using the term involving the heat capacity since, when the resistor reaches a steady state temperature, it will not be storing any more thermal energy. All the generated energy will be transferring to the surroundings. Are you familiar with heat transfer mechanism of Newton's cooling?

 

[Themadflower]

yes, does that mean it's not getting hotter at all? or can i simply use dT = I^2 * R and than see how the parameters are affected by cutting the conducter in half? than dt2 would be 8K

 

[Chestermiller]

yes, does that mean it's not getting hotter at all? or can i simply use dT = I^2 * R and than see how the parameters are affected by cutting the conducter in half? than dt2 would be 8K

Correct. $\Delta T=k(I^2R)$, where k is a constant.

 

[Themadflower]

than thank you for your help! now i understood !

 

[CWatters]

I agree the power doubles but what effect does halving the length have on the thermal conductivity?

 

[Themadflower]

So dt is twice as high. Here dt2 is 8k

 

[Chestermiller]

I agree the power doubles but what effect does halving the length have on the thermal conductivity?

The thermal conductivity is, of course, an intensive property of the material, so, when the amount of material changes, the thermal conductivity of the doesn't. In this problem, they expect the person to assume that the conductive heat transfer is very high, such that the dominant resistance to heat transfer is the convective heat transfer outside the resistor.

You make an important point. Since the heat transfer area is halved, the constant k doubles, and this brings about an additional factor of two increase in the temperature. I forgot to take this into account when I said that 4C was correct. The actual temperature rise should be 8 C.

 

[CWatters]

That's what I was trying to hint at but I make the answer 36C. The temperature delta was 4C, doubling the power makes it 8C and doubling the thermal resistance makes it 16C. Then 20+16=36C.

 

[Chestermiller]

That's what I was trying to hint at but I make the answer 36C. The temperature delta was 4C, doubling the power makes it 8C and doubling the thermal resistance makes it 16C. Then 20+16=36C.

Yes. Thanks for the correction.

Chet