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Homework Help: Thermal Expansion of Mercury

  1. Apr 11, 2005 #1
    I have tried this question 6 times (getting the same answer more than once) and am still not getting the correct answer. I think I need some help understanding where I'm going wrong.

    The question: A container is filled to the brim with 1.35 L of mercury at 20°C. When the temperature of container and mercury is raised to 60°C, 7.4 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the container.

    I have tried using the equation, (delta V/Vo) = 3 (alpha) (delta T), knowing that Beta=3(alpha) for the relationship between linear and volumetric expansion coefficients. I also know that the volume of mercury in the container doesn't decrease that much (1.35L --> 1.3426L) and that the temperature change is 40K (since degrees celsius and kelvin are the same scale). I am very frustrated with this problem and hope someone can help me see my error. :frown:

    Thanks.
     
  2. jcsd
  3. Apr 11, 2005 #2
    I'm not familiar with this but I want to know, why does the volume of mercury decrease with an increase in tempeture?
     
  4. Apr 11, 2005 #3

    Gokul43201

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    As far as I know, it doesn't. The volumetric thermal expansion coefficient for mercury at room temperature is about 1.8 * 10^(-4) /K. That makes the mercury in the container increase in volume by about 10 ml.
     
  5. Apr 11, 2005 #4
    thank you for pointing that out, I don't know what I was thinking in my reasoning that the volume would decrease. Perhaps I had just been used to doing something similar in previous questions I had been working on prior to trying this one. Nonetheless, wouldn't I set the problem up with delta V = (7.4e-3L)/(V0=1.35L)(delta T=40k)? Thus I get an answer that needs to be divided by 3 to convert from the volumetric coefficient to the linear coefficient. My answer is 4.567...e-5 K^-1. I tried this as one of my answers and it was wrong. Am I making an incorrect assumption still? What am I missing in my logic here?
    THank you to the two previous threads and any that may be able to help me out!!
     
  6. Apr 12, 2005 #5

    Gokul43201

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    I'm doing this very roughly. I previously calculated that the mercury will expand by about 10 mL. But only 7.4 mL spills out. This means that the vessel must increase its volume by the difference = 2.6 mL (This is the logical step you are missing). Dividing this by the initial volume = 1350 mL, the fraction is about 0.2% or 0.002. Dividing this by 40 and then by 3 should give you the linear thermal coefficient of expansion of the material of the container. I get rougly 0.000016 /K.

    Do this correctly using a calculator. You should get a number that isn't terribly far from what I got.
     
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