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Thermal condution? little confusing can u please help to remove my doubt.

  1. Jan 1, 2012 #1
    thermal condution?????? little confusing can u please help to remove my doubt.

    According to law of thermal conduction dQ/dt=-kAdθ/x. variable have their usual meaning.
    first of all i want to know what is meaning of dQ/dt in this law??

    If we see it in a cylindrical object with it's curved part fully insulated. then one end will have higher temperature and other will have lower temperature So one end is getting more energy (which is more hot) but other end is cooler which means it radiate less energy so where does some energy goes??????????

    I think i m wrong somewhere in understanding this law can u please help me in find out my error;
     
  2. jcsd
  3. Jan 1, 2012 #2
    Re: thermal condution?????? little confusing can u please help to remove my doubt.

    [itex]\frac{dQ}{dt}[/itex] is the rate of heat flow in the system.
    Your second paragraph seems off to me since the cylinder has symmetry between the ends so how one end could be chosen over the other to be the hotter one is beyond me.
     
  4. Jan 1, 2012 #3
    Re: thermal condution?????? little confusing can u please help to remove my doubt.

    Still in doubt

    If you doesn't understand second para then; think of this condition that there is a cylindrical metal rod of certain length you put one of it's end in a stove and other end in your hand; after a steady u will observe that temperature the end in ur hand is not increasing. why it stopped increasing.(Consider rod is insulated in curved region)
    Temperature of rod end in your hand is lower than that of stove; It should same that of stove since the amount of heat enter in the from stove should equal to heat released from other end.
    where my explanation is wrong. I think it depend on the more deeper meaning of dQ/dt ; that's why my first question is dQ/dt.
     
  5. Jan 1, 2012 #4
    Re: thermal condution?????? little confusing can u please help to remove my doubt.

    Ah now I understand the situation, the rate of cooling is proportional to the temperature so as the end gets up to a high enough temperature its rate of cooling becomes equal to the rate of heating coming from the stove.
     
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