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Temperature of thermometer and temperature of material constituents

  1. Aug 22, 2009 #1
    Hello--

    Suppose that I insert a thin-needle thermometer into a porous material (such as snow) that consists of air, ice, and water. The thermometer is used to measure a temperature. Is it reasonable to assume that the temperature measured by the thermometer is the same as the temperatures of the air, ice, and water which are the constituents of the porous material?

    Now suppose that I insert another needle into the snow at a distance d0 from the thin-needle thermometer. An electric current is passed through the metal needle, which functions as a resistor. The needle heats up, and transfers thermal energy to the snow. Note once again that the snow is a mixture of air, ice, and water.

    Once again, I measure the temperature of this mixture using the thin-needle thermometer.

    Is it still reasonable to assume that the temperature measured using the thermometer is also the temperature of the air, ice and water?

    When I am writing this, I am thinking about the zeroth law of thermodynamics. Is my assumption reasonable, given this and other laws of thermodynamics?
     
  2. jcsd
  3. Aug 23, 2009 #2
    aaaaa...might be, might be not...analysis of such non-homogeneous situations are pretty complex.

    Usually air is warmer.

    Definitely not.

    It takes time for different constituents to gain a specific temperature...the rate of heat transfer, and ultimately rise in temperature is governed by newton's laws of cooling and Fourier's law...also a function of specific heat capacity.

    If situations are made static and given time, air, water, ice...or whatever materials are in contact will reach a common temperature, although the energy stored in terms of heat in each of the components might be different.
     
  4. Aug 23, 2009 #3
    Thank you for your response, dE_logics! I think this makes things a little more clear.

    If I set up an experiment, is there a way to measure the temperature of the constituents?

    How would I know when the materials in contact have reached thermal equilibrium?
     
  5. Aug 23, 2009 #4
    If I know the total temperature of the mixture and the specific heat of the constituents (air, ice, and water), then is it possible to be able to calculate the temperature of each constituent?
     
  6. Aug 23, 2009 #5
    However, although it may not be physically reasonable to make the assumption that the temperature measured using the thermometer is also the temperature of the air, ice and water, isn't this the assumption that is being made when a measurement is made with the thin-needle thermometer?

    Is it being assumed is that the air, ice and water are in thermal equilibrium with the thermometer?
     
  7. Aug 23, 2009 #6
    Pretty difficult job, as said before analyzing such random heterogeneous mixture is pretty difficult.

    By doing the thing that you wanna do, as stated above.

    Total temperature?...there's nothing like that.

    Do you mean algebraic addition of the temperature of each component?

    The temperature measured depends more on what the thermometer is buried into...for instance if it's buried in ice, it'll measure the temperature of ice.

    If it's submerged in water, it will give you the temperature of water.

    Depends highly on the situation. But given time and in an isolated system, they will reach a common temperature.
     
  8. Aug 23, 2009 #7
    Okay, thanks dE_logics, this makes sense. Yes, when I say total temperature, I mean the algebraic addition of each component.

    It is only when the system is in thermal equilibrium that all of the components (air, ice and water) will have the same temperature. This may take time, and it also depends on the situation.
     
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