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Heat from resistance in a circuit

  1. Aug 1, 2007 #1
    So, I've been working on a problem mostly to figure out how heating a cup of coffee works and am kind of lost now.

    I set up my problem like this. I figured 0.5 L of water would be an okay amount to use and made a fictional simple circuit of a 12 V battery and a resistor with a resistance of R = 0.2 ohms. I'm trying to figure out how long it would take to heat that water a change of 70 K. I'm keeping it simple by not having it go through any phase shifts.

    The first thing I did was find out how many J of energy it would actually take to raise water 70k. After I did the # crunching (.05 kg * (4190j/kg K)*70K) and came up with a whopping 146,650 J of energy required.

    This is where I start to get lost though, as I am not that great with circuits. How do I figure out how many joules of heat energy that resistor in that circuit will give off?

    I know current = Potential V / Resistance. but I don't really know if that applies here....Any ideas on how I can figure out how to find out how many J's of energy the resistor in this circuit gives off a minute if it were placed in the water?
     
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  3. Aug 1, 2007 #2

    russ_watters

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    You added a zero to your computation somehow - it's 14,650 J.

    For the heat dissipation of the circuit, you need watts (J/s) and watts is volts times amps.
     
  4. Aug 1, 2007 #3

    Integral

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    You need the power relationships:

    P = IE = I 2 R

    Current in Amps and Voltage in volts, Resistance in Ohms give power in watts.
     
  5. Aug 1, 2007 #4

    mgb_phys

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    Or in a more useful form P = V^2/R

    Remember that although R of the resitor is only 0.2 Ohm, you also have the R of the power cables and the connectors, it's going to be difficult to make connections with a resistance much lower than 0.2 Ohm.

    If you are planning to actually build this, V=IR so I=V/R with a 12V battery and R=0.2 thats 60 Amps which is going to require some fairly hefty cables and will probably damage the battery if you run it for more than a few minutes.
     
  6. Aug 2, 2007 #5

    ranger

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  7. Oct 12, 2011 #6
    hey..try this fromula...Q=iirt...(i square rt)....it says the amount of heat required
     
  8. Oct 12, 2011 #7
    I think the OP was correct. I have (roughly) 4.2 J / (K*g) * 70K * 500g = 147000 Joules.

    EDIT: If you use P = I^2 * R, and you use a 12V battery which somehow can supply 60 Amps of current indefinitely, then it will take about 200 seconds to heat your water (if its insulated perfectly) by 70 degrees. 200 seconds seems like a valid answer.
     
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