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Heating a water bath

  1. Feb 20, 2005 #1
    "In the circuit in the figure, a 20-ohm resistor sits inside 102 g of pure water that is surrounded by insulating Styrofoam."

    "If the water is initially at temperature 10.1 deg. celsius, how long will it take for its temperature to rise to 58.9 deg. celsius?
    Use as the heat capacity of water, and express your answer in seconds using three significant figures."

    What I did was finding the total resistance to be equal to 30 ohms.

    I then use the mc(delta t) formula; the energy needed is then equal to 20856.144 J. I then equated this to P=V^2/R. I then found the time to be ~ 656 seconds. I'm pretty sure I did it right but MP said I did it wrong. Any suggestions?

    Thanks in advance,
     

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  3. Feb 20, 2005 #2
    I found P to be 30 J/s; V is given = 30 volt and total R was found to be 30 ohms. I thought I got it wrong cos of using 30 ohms so I tried using 20 ohms (resistance of the resistor in the water bath) instead of the resistance of the total systemk, but I still got it wrong. Any suggestions? Please help. :cry:
     
  4. Feb 20, 2005 #3

    Curious3141

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    You're calculating the required heat energy to raise the temperature of the water correctly.

    The mistake you are making is in taking the total resistance of the circuit. Remember only the 20 ohm resistor is in the water, so you only want the power output from that one. Try calculating the current through that resistor and applying [tex]P = I^2R[/tex]. Here R should be the 20 ohm resistance only.
     
    Last edited: Feb 20, 2005
  5. Feb 20, 2005 #4
    Thanks for the tip. I got the answer now. It didn't make sense to me though on why it'd take longer to heat when there're less resistors.
     
  6. Feb 20, 2005 #5

    Curious3141

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    Because you're only considering the power output from the 20 ohm resistor rather than from the whole 30 ohm resistance in the circuit. The current through the single 20 ohm resistance is equal to that through the "effective" 10 ohm resistance formed by the remaining resistors (since they're in series). When the current is the same, the power output is directly proportional to the resistance. Do you see now ?
     
  7. Feb 20, 2005 #6
    Thanks, I do now. :)
     
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