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Temperature Change and Volume Expansion

by yellowgators
Tags: expansion, temperature, volume
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yellowgators
#1
Jul27-08, 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
(delta-V)/(V-initial)=(beta)(delta-T)
(delta-L)/(L-initial)=(alpha)(delta-T)
(delta-A)/(A-initial)=2(alpha)(delta-T)

3. The attempt at a solution
The change in the volume of water:
(delta-V)=(beta-water)(delta-T)(V-initial)=(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 cross-sectional area of the cylinder.
(delta-L)=(alpha-Pyrex)(delta-T)(L-initial)=(3.25*10^6/K)(70K)(L-intial)
final length=.9997725(L-initial)
(delta-A)=2(alpha)(delta-T)(A-initial)=2(3.25*10^6/K)(70K)(A-initial)
final area=.999545(A-initial)
final volume of cylinder=final length*final area=.9993176(V-initial)=.9993176(40L)=39.9727 L

final volume of cyliner-final volume of water=0.5523L

Does my method make sense?
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alecsing
#2
Jul28-08, 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:-

(V-final)=(V-initial)[1 + (beta)(delta-T)]
(beta)=3*(alpha)

For water, (V-final)=(V-initial)[1 + 3*(alphaw)(delta-T)]
where alphaw is coefficient of linear expansion of water

Similarly for pyrex, (V-final)=(V-initial)[1 + 3*(alphap)(delta-T)]
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 = (V-initial)[3*(alphap - alphaw)(delta-T)]
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-

(final-L)=(L-initial)[1 + (alpha-Pyrex)(delta-T)]
(final-A)=(A-initial)[1 + 2*(alpha-Pyrex)(delta-T)]

Now multiply the two equations. You will get:-
(final-V)=(L-initial)(A-initial)[1 + (alpha-Pyrex)(delta-T)][1 + 2*(alpha-Pyrex)(delta-T)]
=(V-initial)[1 + 3*(alpha-Pyrex)(delta-T) + 2*((alpha-Pyrex)(delta-T))^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.
yellowgators
#3
Jul28-08, 08:33 PM
P: 15
Quote Quote by alecsing View Post
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.
I double-checked my method versus the method using beta to find the final volume of the pyrex container. The final volumes are not equal using both methods; the volume of the one I used is larger. This makes sense because beta would be used, in this case, for a solid cylinder of Pyrex. A solid cylinder would not expand at the same rate as an open container simplified into 1- and 2-dimensional parts.

alecsing
#4
Jul29-08, 10:35 AM
P: 5
Temperature Change and Volume Expansion

Quote Quote by yellowgators View Post
I double-checked my method versus the method using beta to find the final volume of the pyrex container. The final volumes are not equal using both methods; the volume of the one I used is larger. This makes sense because beta would be used, in this case, for a solid cylinder of Pyrex. A solid cylinder would not expand at the same rate as an open container simplified into 1- and 2-dimensional parts.
You are right in saying that the volume calculated by you would be larger and also more accurate. But the point I tried to make was that although it would be larger, but it would not be significantly larger (in fact very small relative to the rest of the terms). Therefore, it is simpler and less tedious to calculate it directly.


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