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Homework Help: Joule Heating, Calculating Filament Resistance

  1. May 1, 2008 #1
    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 ?
    Last edited: May 1, 2008
  2. jcsd
  3. May 1, 2008 #2
    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
  4. May 2, 2008 #3


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    Homework Helper

    Hi reaxn,

    Please post some details of what you have done (the numbers you used in the equations and your final answer).
  5. May 2, 2008 #4
    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 :(
  6. May 2, 2008 #5


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    Homework Helper

    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?
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