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Temperature Change and Volume Expansion 
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#1
Jul2708, 10:40 AM

P: 15

1. The problem statement, all variables and given/known data
A Pyrex container is filled to the very top with 40.0L of water. Both the container and the water are at a temperature of 90.0 degrees C. When the temperature has cooled to 20.0 degrees C how much additional water can be added to the container? 2. Relevant equations (deltaV)/(Vinitial)=(beta)(deltaT) (deltaL)/(Linitial)=(alpha)(deltaT) (deltaA)/(Ainitial)=2(alpha)(deltaT) 3. The attempt at a solution The change in the volume of water: (deltaV)=(betawater)(deltaT)(Vinitial)=(207*10^6/K)(70K)(40L)=0.5796L the final volume of water 39.4204 L change in holding capacity of Pyrex container: I used the equation for change in linear direction for the height of the cylinder and change in the area for the crosssectional area of the cylinder. (deltaL)=(alphaPyrex)(deltaT)(Linitial)=(3.25*10^6/K)(70K)(Lintial) final length=.9997725(Linitial) (deltaA)=2(alpha)(deltaT)(Ainitial)=2(3.25*10^6/K)(70K)(Ainitial) final area=.999545(Ainitial) final volume of cylinder=final length*final area=.9993176(Vinitial)=.9993176(40L)=39.9727 L final volume of cylinerfinal volume of water=0.5523L Does my method make sense? 


#2
Jul2808, 10:33 AM

P: 5

The approach is quite correct. But if you are aware of the fact that the coefficient of volume expansion (i.e. beta) is equal to 3*(alpha), then you might have used a simpler method. You could have used the equations:
(Vfinal)=(Vinitial)[1 + (beta)(deltaT)] (beta)=3*(alpha) For water, (Vfinal)=(Vinitial)[1 + 3*(alphaw)(deltaT)] where alphaw is coefficient of linear expansion of water Similarly for pyrex, (Vfinal)=(Vinitial)[1 + 3*(alphap)(deltaT)] where alphap is coefficient of linear expansion of pyrex Now to calculate the final answer i.e. Volume of additional water, you just have to subtract the final volume of water from the final volume of pyrex container. This would result in the simple equation: Volume of additional water = (Vinitial)[3*(alphap  alphaw)(deltaT)] This approach simplifies calculations, and also less no. of equations are formed. P.S. You might have some problem visualizing that the final volume of the pyrex container can be calculated directly without using the initial length and the initial base area. To understand this, you can analyze your approach (finalL)=(Linitial)[1 + (alphaPyrex)(deltaT)] (finalA)=(Ainitial)[1 + 2*(alphaPyrex)(deltaT)] Now multiply the two equations. You will get: (finalV)=(Linitial)(Ainitial)[1 + (alphaPyrex)(deltaT)][1 + 2*(alphaPyrex)(deltaT)] =(Vinitial)[1 + 3*(alphaPyrex)(deltaT) + 2*((alphaPyrex)(deltaT))^2] If you neglect the term containing the square of the coefficient of linear expansion (as that term would be much less than the other terms), then you can obtain the equation that I used in my solution. 


#3
Jul2808, 08:33 PM

P: 15




#4
Jul2908, 10:35 AM

P: 5

Temperature Change and Volume Expansion



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