Hard physics q...can anyone solve?

Question

A toaster uses a Nichrome heating wire. When the toaster is turned on at 20°C, the initial current is 1.50 A. A few seconds later, the toaster warms up and the current has a value of 1.23 A. The average temperature coefficient of resistivity for Nichrome wire is 4.5 multiplied by 10-4 (C°)^-1. What is the temperature of the heating wire?

Answer 1

If I am reading this right, I think this is relatively simple.

@20°C, I = 1.5A

@?°C    I = 1.23A

If the temp coefficient of resistivity for Nichrome is 4.5x10^-4 A/°C  (read Amps per degree C), then subtract the ending current from the starting current.  This gives you 0.27 A.

Dividing this by the temp coefficient will give you the temperature increase based off of the current change.

0.27 / (4.5x10^-4) = 600.

Add this to your starting temperature of 20°C and you get a toaster temp of :

620°C

Answer 2

The equation containing the temperature coefficient is:-

    Rt  =  Ro * ( 1  +  a*t)

where Ro is the resistance at 0 deg C.

           Rt  is the resistance at  t deg C

             t is the temparature deg C.

            a is the temparature coefficient.

If the voltage across the wire was "V" and V = I*R so R = V / I

Then  the   resistance in both cases was   V/1.5  and V/1.23

R20  =  V/1.5   and Rt  =   V/1.23

We can apply the above equation , once for each

temperaure, and we get:-

     Rt  =  Ro*(1  +  a*t)

   V/1.5  =  Ro*(1  +  a * 20)

and  V/1.23  =  Ro*(1  +  a * t)

So  V  =  1.5 * Ro*(1 + a * 20)

  and  V  =  1.23 * Ro*(1 + a * t)

Therefore     1.5*Ro*(1 + 20*a)  =  1.23*Ro*(1 + t*a)

              1.5*(1 + 20*a)  =  1.23*(1 + t*a)

   1.5  + 30*a  =  1.23 + 1.23*t*a

Simplifying, and entering value of "a" gives:-

           t  =  512 deg C.