Calculating the heat in the circuit when adding resistances

[satycorn]

Hi everyone! I've got to problems I need your help with, I would really appreciate it if you could help me.

1. Homework Statement

A conductor made out of copper (Cu) has a resistance of 4Ω in 20° celsius. When current flows in the conductor, it's temperature rises to 220°C. In parallel with the resistance, we connect a Rx resistance. How should the value of the Rx resistance be, compared to the original resistance, so that a greater heat can be released in the Rx? For Copper: α=0.0039, ρo=1.7x10^-8

Homework Equations

Q=I²Rt; Q=(U²/R)t

The Attempt at a Solution

So since the resistances are connected in parallel, I tried using the first one but I don't have the time, do I have to build the solution in form of a ratio?? And also, why do we need the ρo of Copper if the R=Ro(1+αΔt) formula is used, in my opinion? Can anyone please help me with this?

Thank you :)!

 

[haruspex]

You don't need the time - just treat it as a rate (Q/t). Quite possibly you don't need ρ_o, but if so it will just cancel later. Please post your attempt.

 

[satycorn]

Ughh, thanks for your reply, although I think I'm stuck again...

I separated R from the Q=I²Rt, which gives R=Q/(I²xt).
I treat the two resistances as a rate (Rx/R), which is Rx=Q1/(I²xt) and R=Q/(I²xt) but here's where I'm stuck, the problem doesn't ask for a specific heat, it just says: "So that the heat that would be released in Rx can be greater"... What do I do here? Do I just continue solving the rate, going to another rate of R=R_o(1+αΔt)? Wouldn't that give the same in both parts of the refraction??

 

[haruspex]

The answer will be in the form of an inequality. Rather than using <, >, and running the risk of getting them reversed somewhere, probably simplest to calculate the resistance that would make them equal, then figure out how to write the inequality.

 

[satycorn]

The answer will be in the form of an inequality. Rather than using <, >, and running the risk of getting them reversed somewhere, probably simplest to calculate the resistance that would make them equal, then figure out how to write the inequality.

Should I calculate the resistance from R=R_o(1+αΔt) or R=Q/(I²xt)?

Either way the symbols just cancel themselves and I am stuck with unknown ones again...
Q/(I²xt)=Q'/(I²xt). The current and time are constant, right? So they cancel themselves and I end up with Q=Q'... Same thing goes for the other formula...

So if I do it in form of an equation from the formula of heat: Q=I²xRxt, it will be:
Q'>Q -> I²xR'xt>I²xRxt, the I and t cancel themselves, so R'>R, I'm left with that, is that really it, to get a higher heat I need a higher resistance, so Rx has to be higher? It just sounds too easy :S..

 

[haruspex]

Should I calculate the resistance from R=R_o(1+αΔt) or R=Q/(I²xt)?

You don't know the current or the power, so R=R_o(1+αΔt) is the only option (for the hot copper resistor).

 

[satycorn]

Okay, I understand. Is this okay?

Ro'(1+αΔt)>Ro(1+αΔt)
Ro'x1,78>4x1,78
Ro'>4

I don't know how accurate I am, but the problem is asking me for the Rx resistance, which I haven't found, right?...