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Homework Help: The conversion of electrical energy to heat energy.

  1. Oct 23, 2009 #1
    Question: Using a Joule's calorimeter to investigate the efficiency of the conversion of electrical energy to heat energy.

    Joule's calorimeter
    thermometer (0-100degrees celsius)
    power supply (0-12V)
    ammeter (0-5A)
    electrical leads and distilled water

    We carried out the prac to find the efficiency of the calorimeter with a copper coil, But we had a slight hick-up, we kept getting over 100% efficiency instead of under 100%. This isnt possible because we're not turning heat energy into electrical energy. Here are our results after we did the prac:

    Initial temperature = 23degrees
    Final temperature = 44.5degrees
    Temperature equilibrium = 21.5degrees
    Mass of distilled water = 0.094kgs
    Mass of calorimeter = 0.122kgs
    Current (I) = 2.2A
    Potential difference (V) = 3.8V
    Time = 880seconds
    Specific heat capacity of water = 4180
    Specific heat capacity of copper = 390
    Relevant equations
    P = IVt
    Q=mc (delta) t

    My attempts at the solution

    P = 3.8*2.2*880
    P = 7356.8W or 7356.8Js

    Q(copper+water) = (0.0943*4180*21.5) + (0.122*390*21.5)
    Q(copper+water) = 9497.7


    9497.7/7356.8 x 100% = 129.1% efficiency. How is this so? shouldn't it be lower.

    Does anyone know what could be the problem/s. How to fix this issue? we tried 5 times and cept getting over 100% efficiency, Is there anything we can change to correct the efficiency?

    Thank you, Sirsh.
  2. jcsd
  3. Oct 23, 2009 #2

    Andrew Mason

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    There is a measurement error of some kind. How are you measuring the water volume? Try changing your ammeter and voltmeter.

  4. Oct 23, 2009 #3
    We assumed that it had something to do with the Temperature equilibrium being too high. We tried it 5 times, with a 500mA ammeter and another Voltmetre of the same type (Cant remember the dial value) The instructions told us to raise the temperature by 20-30degrees from the original temp of the water which was around 23degrees.

    The water was measured on a measuring scale, it was supposed to be 100ml's exactly but we only got 94.3ml's.

    I've tried this investigation through theory work; like changing around variables, but the only way I have found it too work is if i change the water to ethylene glycol and the copper to aluminium, and that's not going to happen lol.
  5. Oct 23, 2009 #4

    Andrew Mason

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    Measure the resistance of the calorimeter. Use P = I^2R and see what you get. It should be the same if you are measuring the voltage at the right place (across the calorimeter) and using a voltmeter that does not affect the voltage.

    Accuracy in measuring the mass of the water is critical. Are you measuring volume of water? How are you doing this?

  6. Oct 23, 2009 #5
    How would i go about getting the P value to work out the resistance of the calorimeter? would putting more volts into the circuit work? Um, the beaker was accurately measured by taking the mass of the pyrex beaker from the mass of the beaker + water in it.
  7. Oct 23, 2009 #6

    Andrew Mason

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    According to your data, the resistance is 1.73 ohms. If you put an ohmeter on it, that is what it should read. If not, there is something wrong with where you are taking your readings.

    If the electrical measurements are correct and your water and temperature measurements are correct and if you are using a copper calorimeter, then you are losing heat somewhere.

  8. Oct 23, 2009 #7
    Okay, so 1.73Ohms. The wires are insulated, the calorimeter is heavily insulated (might be the issue, possibly if it wasnt, enough heat would escape to make the Q value in Q=mcT+mcT lower and the efficiency lower.) The only reason for whats going on from what i can think of is that the water is heating to quickly.

    Because the Q(copper+water) is 9497.7J, I thought that i could try 9497.7 = 3.8*2.2*t, to find the specific time that it would needto make it close to 100% efficiency. but if i did that it would make the time greater, therefore making the temperature rise even more then it is, and then having the exact same problem..
  9. Oct 23, 2009 #8

    Andrew Mason

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    No. This cannot be the reason. You are not going to discover that the specific heat of water is not 4180 J/kg-K. over this temperature range. You are doing something wrong in your measurements. Use an ohmmeter to verify that the resistance is 1.73 ohms. You also have to leave the calorimeter for a time in order for the heat in the heater coil to reach equilibrium with the calorimeter/water. Are you taking into account the mass of the heater coil?

    You have to figure out what you are doing wrong. If you are using a copper calorimeter, either your electrical measurements are wrong, your time measurements are wrong, your mass measurements are wrong or your temperature measurements are wrong, or the heat is not transferring to or staying in the water/calorimeter.

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