Use the density of water as 210*10^-6
so
(210*10^-6)(0.5*10^-3)(90-19)
= 7.455*10^-6
or 0.000007455
Please tell me where i went wrong.
During an all-night cram session, a student heats up a one-half liter (0.50 10-3 m3) glass (Pyrex) beaker of cold coffee. Initially, the temperature is 19°C, and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to 90°C. The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?
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Physics - bobpursley, Friday, April 13, 2007 at 10:08am
Wouldn't you use the volume coefficent of water, and compute the new volume? I will be happy to critique your work.
delta V = (coeffiecient of volume expansion of water)(volume of coffee)(change in temp)
delta V = (207e-6)(0.5e-3m^3)(90deg C-19 deg C)
delta V = 0.000007349 m^3
delta V = (0.5e-3) - 0.000007349
delta V = 0.000492651
delta V = 0.000007349 m^3. That is how much the coffee expands. That is , lets see, 7.3 cc. I think that is what the problem is asking.
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