- #1
kkurutz
- 5
- 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
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