Resistance vs. Temperature, etc.

[kkurutz]

Hi, I have two homework questions that I'm stuck on. I've worked my way through both of them, but and coming up w/ the wrong answers. Here they are:

1) A copper wire has a resistance of 0.501 ohms at 20.0 degrees C, and an iron wire has a resistance of 0.486 ohms at the same temperature. At what temperature are their resistances equal?

First, I found the resistivities of each of the materials and filled in the following equation for each, then setting them eqaul: R = R(1 + alpha(T-T(ini.))

I then set the equality equal to zero: 0 = (R(copper) - R(iron)) + ([R(copper)*alpha(copper)] - [R(iron)*alpha(iron)]) * (T - T(ini.))

I then solved for T: T = T(ini.) + ( R(copper) - R(iron) ) \ ( [R(copper)*alpha(copper)] - [R(iron)*alpha(iron)])

After plugging in all the values, I'm coming up w/ -11.2 degree C though this is the wrong answer


2) In baking a cake, an electric oven uses an average of 19 A of electricity at 230 V for 45 minutes. A personal computer uses only 1.5 A at 115 V. With the same amount of electrical energy used in baking the cake, how long could you surf the internet on the computer?


Starting out, I found the power of the oven: P = IV

I then plugged to the power and other known values into the equation: P = (Q \ t)V to find the energy.

Then, I found the power of the computer: again, P = IV

Lastly, I plugged in the known values (P, Q, V) into: P = (Q \ t)V

I'm coming up w/ 570 minutes, which is the wrong answer.




If anyone can help me out and let me know what I'm doing wrong, I'd greatly appreciate it ... thanks in adavance.

-Keith

 

[andrevdh]

2) The amount of electrical energy used in both processes are the same which can be calculated from
$$I_1V_1\Delta t_1=I_2V_2\Delta t_2$$
the time shoud be in seconds for S.I. units, but factor 60 appears both sides, so you can drop it and work in minutes. I got 19 hours.

 

[kkurutz]

Thanks a lot for your help andrevhd. Anyone have any ideas for the first problem?

 

[andrevdh]

1)Try it this way with symbols first
$$R_{copper}=R_{iron}$$
so we are going to have all copper quantities on the lhs and iron on the rhs
$$R_c(1+{\alpha}_c\Delta T)=R_i(1+{\alpha}_i\Delta T)$$
next the ratio
$$a=\frac{R_c}{R_i}$$
changing the above to
$$a+a{\alpha}_c\Delta T=1+{\alpha}_i\Delta T$$
...

 

[kkurutz]

So am I solving for T in the following equation then: $$a+a{\alpha}_c\Delta T=1+{\alpha}_i\Delta T$$

I tried doing that, but apparently my algebra sucks and I didn't rearrange the equation correctly. Sometimes I think math is my biggest problem w/ this class.

 

[andrevdh]

$$\Delta T=T-T_{ini}$$
so you want
$$\Delta T$$
on the lhs of the equation, hopefully all the quantities on the rhs are known at that stage.