Joule Heating, Calculating Filament Resistance

[reaxn]

The experiment is joule heating and I'm trying to calculate the filament resistance.1.) mass of the calorimeter cup 49.5g
specific heat of the aluminum cup: 0.22 cal/kg*K
mass of calorimeter cup w/ water 232.5g
initial water temp. 20.5 C
voltage across heater 7.3V
current through heater 5A
final water temp 32.7 C
time interval 10mins
calculated filament resistance = ?ΔQ = mcΔT ; have to use this twice, mass and heat capacity of water + mass and heat capacity of the aluminum calorimeter cupU = I squared * Resistance * time
3.) I have no clue where to go :| I've attempted numerous times but I haven't come out with a resistance anywhere close to what I should be getting.

can anybody help me out ?

 

[reaxn]

if any1 can log onto aol instant messenger or mirc, post your screename or an irc server/channel name, so that we can chat :) thanks

 

[alphysicist]

Hi reaxn,

Please post some details of what you have done (the numbers you used in the equations and your final answer).

 

[reaxn]

ok i subsituted in all the values and came out with a Q of 9897.1

i interchanged Q and U and plugged in U = I^2*R*t i used 5 for I and time 600s? is that correct or should i have used 10 mins?

the R came out to be .660 once i solved for R


seems very wrong for some reason :(

 

[alphysicist]

Yes, it should be 600 seconds since you're using 5 amps( and an amp is a coulomb/second).

However, I was more interested in the numbers you used to get the Q value, because some of your numbers seem like they might have problems. For example, you have the specific heat of aluminum as 0.22 cal/(kg*K), but I think that is not right; I think it needs to be 0.22 cal / (g*K).

By the way, do you know your answer is wrong? What do you think it should be?