(adsbygoogle = window.adsbygoogle || []).push({}); 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]:

[tex]\beta[/tex]=[tex]\Delta[/tex]V/V/([tex]\Delta[/tex]T)

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.

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# Homework Help: Thermal expansion question

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