- #1
Themadflower
- 14
- 0
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?
that is all.
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!
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!