Resistance vs. Temperature, etc.

AI Thread Summary
The discussion revolves around two homework problems related to resistance and electrical energy calculations. For the first problem, the user struggles to determine the temperature at which the resistances of copper and iron wires are equal, initially arriving at an incorrect temperature of -11.2 degrees Celsius. The second problem involves calculating how long a personal computer can run using the same energy consumed by an electric oven, with the user incorrectly concluding 570 minutes instead of the correct 19 hours. Participants suggest using symbolic representation and algebraic manipulation to solve the first problem, while emphasizing the importance of maintaining consistent units. The conversation highlights common challenges in applying mathematical concepts to physics problems.
kkurutz
Messages
5
Reaction score
0
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
 
Physics news on Phys.org
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.
 
Thanks a lot for your help andrevhd. Anyone have any ideas for the first problem?
 
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
...
 
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.
 
\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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top