SOLUTION SET 4 - COMPLETE

Chapter 15: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 530-534.

1. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.

So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.~ 8.1 x 102 N.

3. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.

9. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.

13. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.

19. A circular wine barrel is shown in figure 15-26 on page 530. The barrel has a diameter of 75 cm. If a net upward force exceeds 643 Newtons it will burst. A tube of 1.0 cm in diameter extends into the barrel through a hole in the top. Initially the barrel is filled to the top and the tube is empty above that level. What weight of water must be poured into the tube in order to burst the barrel? And why would anyone want to do that anyway?
Solution:
Pressure = force/area
Density = mass/volume
Force = m g = 643 N (in barrel)
W = weight of water
Force density = r g
PE = mgh
PEdensity = r g h
Pbarrel = Ptube
F/Abarrel = W/Atube
The weight of the water must be poured into the tube, such that the above holds true:

W = [(Atube)/(Abarrel)] F

Abarrel = p (0.375)2 = 0.4415625 m2 . (Barrel’s area)
Atube = p (0.005)2 = 0.0000785 m2 . (Tube’s area ). So,

W = [(Atube)/(Abarrel)] F = [(0.0000785)/(0.4415625)] (643 N) = [0.000178](643) = 0.114 N.

Or, W = 0.11 N.~ 1.1 x 10-1 N.


25. Referring to example 15-4 on page 507 in the book, suppose that some vegetable oil has been added to both sides of the U-tube. Not YouTube. On the right side of the tube the depth of oil is 5.0 cm as before. On the left side of the tube the depth is 3.0 cm. Find the difference in fluid level between the two sides of the tube.
Solution:
We know the density of the oil and the water, respectively:

roil = 9.2 x 102 kg/m3.
rwater = 103 kg/m3.

The pressure point at “C” is given by,

PC = Pat + rogh1 + rwgx

and the pressure point at “D” is given by,

PD = Pat + rogh2 + rwgy.

Now we must find the difference in fluid level between the two sides of the tube:

(y + h2) – (x + h1)

At equilibrium, PC = PD, or, Pat + rogh1 + rwgx = Pat + rogh2 + rwgy..This leads us to:

rogh1 + rwgx = rogh2 + rwgy. Because the atmospheric pressure, Pat, is on both sides.

An ordinary orangutan can also see that the acceleration of gravity, g, is in each term, so we can divide that out, and now end up with:

roh1 + rwx = roh2 + rwy. We can rewrite this as rw (x - y) = ro(h2 – h1).

(h2 – h1) = [(rw)/(ro)](x – y). Or we can invert it to be: (x – y) = [(ro)/(rw)](h2 – h1)].

Now, multiply both sides by negative one ( - 1), then add (h2 – h1) to both sides: Then we will have:

[(h2 – h1) - (x – y)] = (h2 – h1) - [(ro)/(rw)](h2 – h1)..And that can be factored to look like this:

(y + h2) - (x + h1) = (h2 – h1)[1 - (ro)/(rw)]. And this equals (5 cm – 3 cm)[1 - (9.2 x 102 kg/m3)/(103 kg/m3)] = (2 cm)[1 – 0.92] = (2 cm)(0.08) = 0.16 cm.~ 1.6 x 10-2 m.


29. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III

31. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.

39. A hydrometer is a device for measuring fluid density. It's constructed as shown in Figure 15-30 on page 532. If the hydrometer samples fluid 1, the small float inside the tube is submerged to level 1. When fluid 2 is sampled the float is submerged to level 2. Is the density of fluid1 greater, less, or the same as fluid 2's density? [This is how auto mechanics test the antifreeze level in your car engine. Since antifreeze (ethylene glycol) is denser than water, the higher the density of coolant in your radiator the more antifreeze protection the engine has.]
Solution:
How would I know? I'm not an auto mechanic. He he he.

A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.

Chapter 16: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 566-567

1. The official record for lowest temperature ever recorded on Earth was set at Vostok, Antarctica, on July 21, 1983. The temperature on that day fell to T = ( – 89.2° C), which is well below the temperature of dry ice (frozen CO2). Give this temperature in degrees Fahrenheit.
Solution: We use the “famous” equation that °F = (9/5) °C + 32°. So, let's “plug and chug” and that will be it: °F = (9/5) °C + 32° = (9/5)(-89.2°C) + 32° = (1.8)(-89.2°C) + 32° = (- 160.56°) + 32° =
- 128.56 °F.~ - 129 °F.

3.Normal body temperature for humans is T = 98.6° F. Convert this temperature to degrees:
a. Celsius (aka Centrigrade)
b. Kelvin
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°).
a. °C = (5/9)(°F – 32°) = (5/9)(98.6° – 32°). = (0.556)(98.6° – 32°). = (0.556)(66.6) = 37°C .
b. Using °K = °C + 273, then °C + 273 = 37 + 273 = 310°K.

9. A world record for the greatest change in temperature was set in Spearfish, North Dakota on January 22, 1943. At 7:30 AM, the temperature was T = ( - 4° F); two minutes later, at 7:32 AM, the temperature was T =+ 45° F. Calculate the mean rate of temperature change, per second, using Celsius or Kelvin scales (they use the same units); Hint: find DT.
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°). At the lowest temperature, °C = (5/9)(- 4° – 32°). = (0.5555)( - 36) = - 20°C. At the highest temperature, °C = (5/9)(45° – 32°). = (0.5555)(13) = 7.2°C. Then, DT. = [7.2 - ( -20)] = 27.2 °C. in two minutes. Or, 27.2 / (120 sec) =
+ 0.227 °C/s.

13. Referring to Table 16 - 1 on page 545, which would be more accurate for an all-season, outdoor use: a tape measure made of steel or one made of aluminum?
Solution: Steel (iron) because it will shrink and expand less.

19. It is desired to slip an Aluminum ring over a steel bar as shown in Figure 16-14 on page 567. At T = 10.00° C, the inside diameter of the Aluminum ring is da = 4.000 cm, and the diameter of the steel rod is ds = 4.040 cm. The steel bar maintains a constant temperature of Ts = 10.0° C.
a. In order for the ring to slip over the bar, should the ring be heated, or cooled? Explain.
b. Find the temperature, Ta, of the ring at which it fits over the over the bar.
c. Who is the Lord of the Rings?
Solution: In this case, we must consult the coefficients of linear expansion, or, a. If the aluminum expands at a faster rate than the steel as we heat them, then we should heat the aluminum. Otherwise, cool it.
a. a(Al) = 24 x 10- 6 per degree K; and a(Steel) = 12 x 10- 6 per degree K. This means that Aluminum expands faster, so we heat the aluminum.
b. The Aluminum ring will NOT fit over the Steel bar at this temperature (10° C), so we need it to expand to exceed the diameter of the steel, i.e. Da > ds. So, Da > 4.040 cm, such that Da = (4.000 cm )+ Dd. According to the linear expansion formula, DL = aL0DT. So, for this problem, Dd = a(da DT). And, of course, Dd = ds – da = 0.040 cm; DT = Tf – Ti = Tf - 10.00° C. Where We are looking for Tf. So, let's re-write this as:

(Dd)/ (ada ) = DT = (Tf – Ti). And now, we can re-arrange that to get Tf.

Tf = Ti + (Dd)/ (ada ) = 10.00° C + (0.040 cm) / [(24 x 10- 6)(4.000 cm)] =

10.00° C + (0.040 cm) / [(96 x 10- 6)] = 10.00° C + (4.17 x 102 °C) = 427°C.

c. Exactly.

25. An aluminum saucepan with a diameter of d = 25.0 cm and a height of h = 6.0 cm, is filled “to the brim” with H2O. The initial temperature of the pan and water is T = 19° C. But then the pan with the water in it is placed on a range where it is heated to T = 88° C.
a. Will the water overflow from the pan, or will it decrease? Explain.
b. Calculate the volume of water that either overflows or decreases.
c. Go make some popcorn. Turn off the stove.
Solution: In this case, we must consult the coefficients of linear expansion for Aluminum, or, a. and volumetric expansion for water, or b. If the aluminum expands at a faster rate than the water as we heat them, then the water level will go down. Otherwise, it will overflow. We will use both DL = aL0DT for Aluminum, and DV = bV0DT for water.
a. a(Al) = 24 x 10- 6 per degree K; and b(water) = 0.21 x 10- 3 per degree K. This means that the water will expand faster in volume than the Aluminum will in linearity. The pot will overflow. By 24 cm3 .
b. The volume of the pan (and thus, the water) is the area of the pot, which is the area of the circle of the pan, multiplied by its height, or, (h)(p r2) where r = ½ d. Thus, the volume is:

V = (h)(p r2) = (6.0 cm)(p)(12.5)2 = (6.0 cm)(3.14)(156.25) = (6.0 cm)(3.14)(156.25) = 2945 cm3. So, this is the “original” volume, or V0. DT = 88° C – 19° C = 69° C.

Thus, DV(water) = bV0DT = (0.21 x 10- 3 )(2945)(69) = 42.7 cm3.

But the Aluminum pot will expand, too, by:

DV(Al) = 3aV0DT = (3)(2.4 x 10- 5)(2.945 x 103)(6.9 x 101) =

(3)(2.4)(2.945)(6.9)(10- 5)(103)(101) = 146.31 x 10- 1. = 14.63 cm3.

Finally, overall change of volume is [DV(water) - DV(Al)] = (42.7) - (14.63) = 28 cm3.

c. Okay.



29. Consider the apparatus the Joule used in his experiments on the mechanical equivalent of heat, shown in Figure 16-8 on page 550. Suppose both blocks have a mass of m1 = m2 = 0.95 kg and that they fall through a distance of h = 0.48 meter.
a. Find the expected rise in temperature, DT, of the water, given that 6200 Joules are needed for every increase of DT = 1.0° C. Give your answer in degrees Celsius (°C).
b. Would the temperature rise in degrees Fahrenheit (°F) be greater than, or less than, the result in part (a)? Explain.
c. Find the rise in temperature, +DT, in Fahrenheit degrees (°F).
Solution: In each case, the stored gravitational energy is PE = m g h = (0.95 kg)(9.8 m/s2)(0.48 m) = 4.4688 J. Since there are two of them, then the PE = 2(4.4688) = 8.9376 J.
a. If all the PE were converted to heat (which it is not), then 8.9376 J of heat went into the water. And if 6,200 J are needed for +1.0° C, then the rise is +[(8.9376) / (6200)](1.0° C) = + 0.00144° C which is negligible.
b. Greater.
c. Using the “famous” relationship: we find that each degree rise in Celsius is the same as each 1.8° rise in Fahrenheit. So, +DT(Fahrenheit) = (1.8)(0.00144) = + 0.002595° F.

31. Two objects are made of the same material, but have different temperatures, T1 and T2. Object 1 has a mass of “m” while object two has a mass of “2m.” If the objects are brought into thermal contact,
a. Is the temperature change of object 1, DT1, greater, lesser, or equal to the temperature change of object 2, DT2?
b. Choose the best explanation from among the following three choices:
I The larger object gives up more heat and therefore its temperature change is greatest
II The heat given up by one object is taken up by the other. Since the objects have the same heat capacity the temperature changes are the same
III One object loses heat of magnitude Q the other gains heat of magnitude Q. With the same magnitude of heat involved the smaller object has the greater temperature change.
Solution: DQ = m C DT, this is the change of heat energy relative to Temperature. In this case, Q is heat energy; m is the mass; C is the specific heat capacity, and you know T.
a. The smaller one.
b. III

39. If 2200 Joules of heat are added to a 190-gram object, its temperature increases by +DT = 12° C.
a. Find the heat capacity of the object.
b. Find the specific heat of the object.
Solution: We already know that DQ = m C DT, so...
a. DQ = m C DT, and we have DQ, m, and DT. We need only to find C: Let's re-write to isolate C:

C = (DQ) / (m)(DT) = (2200 J) / [(0.190 kg)(12)] = (2200 J) / (2.28) ~ 965 Joules / kg-deg = 9.65 x 102 kg-deg.

b. The same, 9.65 x 102 kg-deg..

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