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March 3, 2015

March 3, 2015

Posted by **Maggie** on Sunday, April 6, 2008 at 9:46am.

m3

I know the volume of the pipe is

pi R^2 L = 1.93*10^-2 m^2 = 19.3 liters

I calculated how much that volume of water increases when being heated from 20 to 75 C. I used the thermal expansion coefficient of water. It varies from 20 to 75 C.

delta V = V*4*10^-4*(75 - 20)

= 0.43 liters

My finnally answe has to be in m^3 so I divided .43l liters by 10^-3 m^3 but I can't come up with right answer for this problem.

- Physics -
**Damon**, Sunday, April 6, 2008 at 10:50amchange in V/V = beta *(Tf-Ti)

= 4*10^-4 (55) = 220*10^-4 = 2.2 * 10^-2

pi r^2 L = pi * (9.1*10^-3)^2 m^2 * 75m

= 1.95*10^-2 METERS CUBED NOT liters

so

delta V = 1.95*10^-2 * 2.2*10^-2 = 4.29 * 10^-4 meters cubed

delta v/V can be liters per liter or meters^3 per meter^3 or thimbles per thimble but in this case everything is in meters.

- Physics -
**Happy**, Sunday, April 6, 2008 at 12:20pmmy computer home work is tell me .000429 is not correct I have reworked the problem but it keep tell me the answer is not correct

- Physics -
- Physics -
**Damon**, Sunday, April 6, 2008 at 2:13pmWell, I assumed your thermal expansion coef for water was correct. Otherwise I did it all independently from your work. However water is non-linear and even changes sign just above freezing (why ice freezes at the surface of the pond) so if it is a little off it is because it is hard to define that coef. If it is a lot off, either we are both crazy or the computer homework is wrong.

- Physics -
**Tom**, Friday, April 18, 2008 at 10:04pmTry this one.

you seem to have forgotten about the coeficient for copper pipe. Remember volume gained (reservoir) equals volume lost (pipe)

deltaV =beta_water x V_i x deltaT

this gives the change in V for water

then

deltaV = beta_copper x V_i x deltaT

this gives the change in V for Copper

delta_T and V_i are the same for each equation. Then SUBRACT the lower number from the higher number. I had a similar problem and got the correct answer.

hope it helps.

- Physics -
**Anonymous**, Monday, March 2, 2009 at 8:37amwhat?

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