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Temperature question

  1. Sep 8, 2005 #1
    This is not a homework question. Just some basic logic. Suppose that we have two identical pieces of solid metal. One is heated to a temperature of 100 degrees Celsius and one is cooled to zero degrees Celsius. Now we put both of them in a perfectly isolated container with negigible heat capacity. At what temperature would both pieces end up with?

    Is this a first grade question and what is the correct answer? 50 degrees Celsius? I think not. Somehow, I believe the answer to be 53,8 degrees Celsius. Am I right?
     
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  3. Sep 8, 2005 #2

    G01

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    The two pieces of metal wouldn't necessarily end up at the same tempurature but they would contain the same amount of heat at the end. If they were both the same type of metal I would think they would end up at the same temp. but I could easily be wrong on this.
     
  4. Sep 8, 2005 #3
    I agree They would both end up at the same amount of heat being identical that would be 53,8 degrees celsius and not at 50 degrees Celsius? No?
     
  5. Sep 8, 2005 #4

    G01

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    [tex]\delta[/tex]Q = m[tex]\delta [/tex]TC right?

    Use this equation to find out how much heat both start out with. C is the specific heat of the object. Then find an average of the heats and work out what each metal' temperature would be at those heats. I think that should work.
     
  6. Sep 8, 2005 #5

    G01

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    sry i wanted uercase delta there, woops
     
  7. Sep 8, 2005 #6
    I was thinking of using [tex]T=\sqrt{\frac{(t1^2+t2^2)}{2}}[/tex], with t’s in kelvin.
    Again this is no homework question. Just some brain exercise. Is this correct?
     
  8. Sep 8, 2005 #7

    Doc Al

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    Where did you get this formula from? Common sense and symmetry should tell you that the final temperature will be right in the middle between the two initial temperatures.

    To figure it out, realize that the net heat flow must be zero, so:
    [tex]m_1 c_1 \Delta t_1 + m_2 c_2 \Delta t_2 = 0[/tex]

    But since the two objects are identical:
    [tex]m c \Delta t_1 + m c \Delta t_2 = 0[/tex]
    [tex]\Delta t_1 = -\Delta t_2[/tex]

    Since the objects reach thermal equilibrium, they have the same final temperature. Which, per the above, must be right in the middle.
     
    Last edited: Sep 8, 2005
  9. Sep 8, 2005 #8
    But in the middle of what? The middle of the temperature or the middle of the energy?

    It's excruciating obvious that 50C is the middle of the temperature but perhaps that that 53.8 degrees Celsius is the average (middle) of the energy. no? After all it's the energy that we have a conservation law of, it's not about about conservation of temperature.

    We do not measure the temperature but we measure the lenght of a fluid column that depends on the volume of that fluid in a bulb at the underside of the thermometer. We define the temperature to be proportional to the volume of that fluid. What is temperature again? Wasn't it the same as average speed of the individual molecules.

    Now what I'm not sure of, but assume that the volume of that fluid is directly proportional to the average speed of the molecules. Everything is linear up to now.

    But the energy the heat is proportional with the square of the speed of the molecules. So double temperature is four times more energy No?

    I'm quite puzzled here. Did we overlook something so obvious in the most plain basics of physics or am I way of track?
     
  10. Sep 8, 2005 #9

    Doc Al

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    Both!

    No. The average (translational) KE of the molecules is proportional to the temperature, not the temperature squared.

    The amount of energy transferred as heat is proportional to the change in temperature.
     
  11. Sep 8, 2005 #10
    If the KE would be proportional to the temperature and you add a fixed amount of energy for each degree C to the bulb of the thermometer, the speed of the molecules of the fluid would increase with the square root of the delta energy. We assume that the speed of those molecules is proportional to the expansion of the volume of the fluid (it's probaly more complex), hence the column of the thermometer would grow with the square root of the added energy.

    But the ticks on the thermometer are on a lineair scale.
     
  12. Sep 8, 2005 #11

    Doc Al

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    Sounds good to me.
    Why would you assume that? You could make a handwaving argument that the average amplitude of the molecular vibrations in a solid is proportional to the energy (and thus the temperature), but not the speed. Think of a mass on a spring: Double the energy and you double the amplitude.


    Right. And they are used to measure temperature which is proportional to energy.
     
  13. Sep 8, 2005 #12
    Good one. But this is true for harmonic oscilations. Just thinking. I can imagine that molecar motion is anti harmonic. Molecules speeding up when nearing others due to interacting gravity. So the speed could be highest on the end of the trajectory before the collision. If energy and temperature and thermal expansion are truly proportional then I could also image that the decrease in density is a factor for the square root relation to the average speed.
     
  14. Sep 8, 2005 #13

    Bystander

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    V is proportional to T for an ideal gas. For solids and liquids,

    V ~ V0(1 + 3εΔT) ,​

    where V0 is volume at any reference T you pick, ΔT is the difference between temp of interest and ref., and ε is the coefficient of linear expansion.

    The capillary cross-sectional area in liquid in glass thermometers is a function of εglass, much smaller than for liquids used in thermometry, therefore, the expansion of the liquid in the t-mom bulb, proportional to T, is manifested by a nearly linear increase in the length of the column of liquid in the capillary (Vliq = Vbulb + Lcapillary Acapillary).

    Any other questions?
     
  15. Sep 9, 2005 #14
    Yes, physics is a long time ago. But I'm still puzzled, what am I doing wrong in the relationship between energy and temperature? Let's look indeed at the ideal gas law:

    [tex]pV=nRT[/tex]

    or

    [tex]\frac{p_1V}{T_1}=\frac{p_2V}{T_2}[/tex]

    Now if T2 is double T1 and the volume V remains constant in a fixed container, then the pressure P2 is double P1 as well.

    Pressure is force per area. So double pressure is double force as the area is constant.

    the force is the sum of the momentums of the individual molecules (mv) per second:

    [tex]F=ma=\sum \frac{m_iv}{s}[/tex]

    So doubling the force means doubling the individual speeds of the molecules as m and time remains constant, which brings me to square one. Energy is proportional to the square of the speed. So, am I forced to conclude that doubling the temperature means four times more energy? What am I doing wrong.
     
  16. Sep 9, 2005 #15

    russ_watters

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    I thought you had it earlier, but temperature is related to energy, not velocity. From post 10, it sounds like your question is about how thermometers work, not how temperature works.
     
  17. Sep 9, 2005 #16

    russ_watters

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    More...
    No, temperature is defined to be proportional to kinetic energy, which is why it makes for a linear scale with volume of a fluid. There is no discrepancy there.
    What you are imagining just isn't reality: those quantities - energy, temperature, and thermal expansion, really are proportional.
     
  18. Sep 9, 2005 #17

    Doc Al

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    So far, so good.

    The force exerted on the walls is the average rate of change of momentum of the particles hitting that wall. It depends the speed of the particles and on how often they hit the wall.

    What you are doing wrong is assuming that the rate that particles hit the wall remains the same. But, since the particles are moving faster, they hit the walls more often. The change in momentum of a single collision is proportional to speed, but the number of collisions per second is also proportional to speed. Thus the pressure is proportional to the square of the speed.

    So it all makes sense: Double the temperature and the pressure doubles, since both are proportional to speed squared.
     
  19. Sep 9, 2005 #18
    wat do u mean by "time remains constant".....i think u r gettin confused between velocity for K.E with temperature for heat energy......the temperature increases linearly with the increase with addition of energy until the oject phase tranformation stage occurs....at this point the temperature ceases to increase as the energy added is used for phase transformation.....looking at the P-v, P-T, and the energy graphs might help i reckon.....it should however be noted the K.E increases exponentially wit increase in velocity......hope tis helps
     
  20. Sep 9, 2005 #19
    That's it indeed. More collisions. Thanks all, I was lost for a moment.
     
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