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 cup


U = 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?