COLLEGE PHYSICS II TEST NO. 1 SPRING 2008
Problems
1. Divers use lift bags to lift objects off the sea floor. If a lift
bag has a volume of 50 L (liters) and a maximum gauge pressure of
150 kPa when fully inflated, what maximum volume of air can be put in
the bag at a depth of 20 m so as not to exceed the maximum pressure
when it reaches the surface? The density of sea water is 1030 kg/m^3
and atmospheric pressure is 101 kPa. Assume the temperature of the
air is the same at depth and at the surface.
You use the ideal gas equation of state for this problem, which means
your pressure has to be absolute. Since the initial and final
temperatures are the same, you don't have to worry about changing
to kelvin. (The temperature is not given anyway.) The initial
absolute pressure at the sea floor is
Pi = 1 atm + dgh = 101 kPa + (1030 kg/m3)(9.81 N/kg)(20 m)(1 kPa/1000 Pa) = 303 kPa.
The final absolute pressure is the gauge pressure of 150 kPa in the
bag plus atmospheric pressure.
Pf = 251 kPa.
With a final volume of 50 L, the initial volume must be
Vi = Vf(Pf / Pi).
Vi = (50 L)(251 kPa / 303 kPa) = 41 L
2. You have 100 g of ice in a calorimeter at -20 deg C. How much water
at 100 deg C would you have to add to completely melt the ice and
achieve a final temperature of 0 deg C in the calorimeter? Assume no
heat is lost to or gained from the environment during the process.
Take the specific heat of ice to be 0.50 cal/g/K and the heat of
fusion of ice to be 80 cal/g. The specific heat of water is exactly
1 cal/g/K. Express your answer in grams of hot water.
The heat needed to raise the temperature of the ice and to melt it
is given by
Q = micecice(Tf - Ti) + Lfmice
Q = (100 g)(0.5 cal/g·K)(20 K) + (80 cal/g)(100 g) = 9000 cal
The heat given up by the water is
Q = mwatercwater(Ti - Tf).
(Note that the initial and final temperatures for the ice and water are
different.) This Q has the be the same as that found for the ice to
satisfy the conditions of the problem. So,
9000 cal = mwater(1 cal/g·K)(100 K).
Solving,
mwater = 90 g.
3. You want to construct a box-shaped Styrofoam cooler with an area of
0.85 square meter (including all six walls) with a wall thickness
sufficient such that it would take 24 hours for a ten-pound bag
(about 4.5 kg) of ice to completely melt. (a) What rate of heat flow
in watts through the walls would melt 4.5 kg of ice at 0 deg C to
4.5 kg of meltwater at 0 deg C in 24 hours? The heat of fusion for
water ice is 3.34E+5 J/kg. (b) How thick would the Styrofoam walls
have to be to achieve this rate of heat flow if the outside tempera-
ture is 27 deg C? The thermal conductivity of Styrofoam is 0.010
W/m-K.
(a) The heat needed to melt 4.5 kg of ice is
Q = mLf = (4.5 kg)(3.34 × 105 J/kg) = 1.5 × 106 J.
This heat is to be supplied over 24 h, so
Q/t = (1.5 × 106 J) / [(24 h)(3600 s/h)] = 17 W
(b) The heat flow equation for conduction is
Q/t = kTA(ΔT) / d,
where kT = 0.01 W/m·K, A = 0.85 m2, ΔT = 27 K,
and Q/t = 17 W. Solving for d and putting in the numbers,
d = (0.01 W/m·K)(0.85 m2)(27 K) / 17 W = 0.013 m = 1.3 cm.
4. A charge of +20 nC is placed on the x axis at x = 20.0 cm, and a
charge of -10 nC is placed on the y axis at y = 20.0 cm. What is
the magnitude and direction of the electrostatic (Coulomb) force on
a third charge of +10 nC placed at (x,y) = (20.0 cm, 20.0 cm)?
The diagram, as best I can do it using a text editor, is
y
| ^
| |
| | F1
| F2 |
-10 nC (-) <----- (+)
| +10 nC
|
|
|
|_________________(+)________ x
+20 nC
The magnitudes of F1 and F2 are
found by the usual point charge formula.
F = ko q1q2 / r2.
Substitute for the charges (in coulombs!) and r (in meters!) to get
the force in newtons. You get
F1 = 4.5 × 10-5 N = 45 μN
and
F2 = 2.25 × 10-5 N = 22.5 μN.
These forces are perpendicular to each other. Hence you can find the
resultant magnitude from the Pythagorean theorem.
F = [(45 μN)2 + (22.5 μN)2]1/2 = 50 μN.
From the diagram above, the resultant force has to be in the second
quadrant. The angle is
θ = tan-1(45 μN / 22.5 μN) = 63° above the
negative x direction.