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Thermal expansion question

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    When the temperature of liquid mecury increases by one degree celsuis , its volume increases by one part in 5500. The fractional increase in colume per unit change in temperature(when the pressure is held fixed ) is called the thermal expanision coefficient , [tex]\beta[/tex]:


    For mercury. [tex]\beta[/tex]=1.81 x 10^-4 K^-1(The exact value varies with temperature, but between 0 degrees C and 200 degrees C the variation is less than 1 percent. )

    a) Get a mercury thermometer, estimate the size of the bulb at the bottom and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

    b)The thermal expansion coefficient of water varies significantly with the temperature: It is 7.5 x 10^-4 K^-1 at 100 degrees C, but decreases as the temperature is lowered until it becomes zero at 4 degrees C. Below 4 degrees C it is slightly negative, reaching a value of -.68 x 10^-4 K^-1 at zero degrees C. (this behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal coefficient of water were always positive.

    2. Relevant equations

    3. The attempt at a solution

    for part a, I did not really have the laboratory equipment to measure the radius and diameter of the bulb of a mercury thermometer, so I google the estimates and I found that the radius of a bulb was r=.012 cm and D=2r =.024 cm. Think does are good estimates?

    b) wouldn't the lake not be freezing over if [tex]\beta[/tex] since the temperature change and the volume change would have to be positive. The lake would need to be melting and the size of the lake would need to be expanding since the final volume would have the be larger than the initial volume and the final temperature would have to be larger than the initial temperature.
  2. jcsd
  3. Jan 18, 2009 #2


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    The point of the question is to demonstrate that a large bulb and a thin tube give you a lot more accuracy. It doesn't really matter what bulb size is - but I would say 1 cm long by 0.5cm diameter is probably more reasonable.

    For B consider what happens to denser and less dense water - which rises, which sinks.
    Now at what temperature is water denser / less dense?
  4. Jan 18, 2009 #3
    water is more dense than ice, at higher temperatures liquid water is more likely to be dense than at lower temperatures. And more dense object are more likely to sink than less dense objects. I want to say because more dense object are more likely to be smaller in size and there not as heavy as less dense objects , assuming they have the same mass.
  5. Jan 18, 2009 #4


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    Ignore the ice for now.
    As the water cools to 4deg C it becomes denser and sinks.
    So initially warm water is on the top and cooler water at the bottom.
    Now what happens to the 4deg C water at the bottom if it cools below 4deg C - does it get lighter or heavier - and so does it float or sink?
    What does this tell you about the temperature at the bottom of the water.
  6. Jan 18, 2009 #5
    as the 4deg C water cools below C , it become denser, and therefore sinks , but as it approaches zero degrees it turns to ice and ice is generally less dense than water , so the water will turn to ice will float once it reach a temperature of zero degrees celsuis. the temperature of the bottom of the water is cooler than the surface of the lake . Doesn't the temperature of water stay around 4 degrees celsuis at the bottom of the water, since it will float to the top if you continue to cool the water to zero degrees since ice is typically zero degree and if you warm the water the water will rise?
  7. Jan 18, 2009 #6


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    No it has it's maximum density at 4deg C, below this it becomes less dense and rises.
    This is a weird feature of water-but is very important for life. It means that lakes freese at the top first and that 4deg C water lays on the bottom.
    Normally warm water rises and cool water sinks - which transfers heat from bottom to top, but once it gets below 4degC colder water goes up and 4deg C water goes down.
    This means that it is very rare for a lake to freeze solid
  8. Jan 18, 2009 #7
    I was trying to say that as you cool water it becomes less dense until you reach 4 degrees celsuis. Once it surpasses 4 degrees celsuis , this trend is reverse and water rises and become less dense . Sorry I didn't not make my explanation clear.
  9. Jan 18, 2009 #8
    at zero degrees , temperature is zero and coefficient is negative. The problem asks me if the process would be any different if the coefficient were positive.I wouldn't think so because even if change in temperature is negative , and change in delta V is negative, since water is larger than ice, the coefficient is still positive.
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