Problem Set 13: | 9 | 17 | 19 | 23 | 31 | 33 | 35 | 37 | 10:1 | Go up
- by David Harms, 2002

9. A concrete highway is built of slabs 14 m long (20Cº).  How wide should the expansion cracks be (at 20Cº) between the slabs to prevent buckling if the temperature range is -30Cº to +50Cº?

I know I use the formula:
DL  = aLoDT
DL  = ? (Change in length of the slabs. We are solving for this.)
a    =  12e-6 (Coefficient of expansion. I looked it up on page 388.)
Lo   =  14 m  (Initial length of slabs.)
DT  =  30 Cº (50Cº - 20Cº)  (Change in temperaure.  You only care about the hottest number since you are dealing with expansion.)

DL = 12e-6 *14 m * 30Cº
DL = .00504 m
DL = .50 cm

The space between the slabs must be the same as the expansion of the the slabs.  Therefore .50 cm is the correct answer.

17. A quartz sphere is 14.5 cm in diameter.  What will its change in volume if it is heated from 30Cº to 200Cº?

DV = gVoDT
DV = ? (Volumetric change in the sphere.  We are solving for this.)
g     =  1e -6 (Cº)-1 (Coefficient of volumetric expansion.  It is on page 388 listed as b.)
Vo   =  (4/3) * p(.145 m/2)3 (Initial volume.  You must take the diameter and divide by to to get the radius.  Then you must stick it in the formula for volume of a sphere: (4/3)pr3
Vo   =  .001569 m3
DT  =  170Cº (Change in temperature.  200Cº -  30Cº.)

DV = 1e-6*.001569m3*170Cº
DV = 2.7 e-7 m3 *1e6 (Conversion from m3 to cm3.)
DV = .27 cm3

The answer is therefore .27 cm3.

19. If the fluid is contained in a long, narrow vessel so it can expand in essentially one direction only, show that the effective coefficient of linear expansion a is approximately equal to the coefficient of volume expansion b (g in the IB packet.)?

Since it can only expand in one direction, is is really linear expansion, not volumetric expansion. Therefore the two coefficients would be the same.

23. A 23.4 kg solid aluminum wheel of radius 0.52 m is rotating about its axel in frictionless bearings with an angular velocity w = 32.8 rad/s.  If its temperature is now raised from 20.0Cº to 75.0Cº, what is the fractional change in w?

To solve this problem we must find I (Moment of Inertia) before it heats up and after it heats up.

I have the formula:
I1 = Smiri2
I1 = ? (Moment of Inertia.  We are solving for this.)
= 1/2 (looked it up on page 22).
m = 23.4 kg (mass)

I1 = (1/2)*23.4kg * .522
I1 = 3.16368 kg m2

DL  = aLoDT
DL  = ?
Lo   = .52 m
a    =  25e-6Cº-1
DT  = 55Cº

DL = 25e-6Cº-1 * .52 m * 55Cº
DL = 7.14 e -4 m

I then had to find I2:
I2 = Smiri2
I2 = ?
S  = 1/2
m = 23.4 kg
r = .520715

I2 = .5 * 23.4 *.5207152
I2 = 3.172386 kg m2

I also know I have the formula:
I1w1 = I2w2
w1 = 32.8 rad/s (Angular velocity.)
I
1   = 3.16368 (Found at beginning of problem.)
w2 = ? (We are solving for this.)
I
2  =  3.172886 (Just solved for.)

w2 =  32.7048 r/s

To find the percentage difference I took
(w1-w2)/w1
(
32.8-32.7028)/32.8

.3%

31. If 3.0 m3 of a gas initially as STP is placed under a pressure of 4 atm, the temperature of the gas rises to 38.0Cº.  What is the volume? ?

I know I had to use the equation:
(P1V1)/T1 = (P2V2)/T2

P1 = 1.013e5 Pa (1 atm* 1.013e5 Pa/atm)
V1 = 3 m3
T1 = 273 k (Number of Kelvins at STP)
P2 = 405200 Pa (4 atm * 1.013e5 Pa/atm)
V2 =  ? (We are solving for this)
T3 = 311 k (273 k +38)

I then solved this as got:
V2 =  .85 cm3

33. <Question goes here>?

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35. <Question goes here>?

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37. <Question goes here>?

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10:1. <Question goes here>?

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