Transcript Lecture 1: Rotation of Rigid Body

PHY 126-07 Mid-term Exam II
Chapters 15,16,17,18
Problem 1 (10+10 points)
A scientist is making a mockup model of a human ear using a small can with the top surface
open and the bottom closed. A human ear is especially sensitive to sound whose frequency
is 3,440 Hz. If this is the lowest frequency that the ear is especially sensitive to, (a) what the
height of the mockup model h should be? Assume that the speed of sound is 344 m/s.
(b) What is the next lowest frequency the ear especially sensitive to?
Solution:
(a) This is an example of standing waves in a stopped pipe. When
the incoming sound wave’s frequency coincides with or is very
close to the frequencies of standing waves the ear is especially
sensitive to the sound as resonance occurs (4 points). Therefore
the fundamental frequency f1=v/(4h) where v is the speed of
sound (3 points). From 3440 s-1 = (344 m/s)/(4h), h=1.00/40.0=
0.025 m (3 points).
(b) The n-th harmonic frequency is described by:
fn=nv/(4h) where n=1,3,5,…(odd integer) (5 points). A harmonics
with an even integer n does not exist. Therefore the next lowest
frequency is when n=3. f3=3v/(4h)=3(344 m/s)/(4 x 0.025 m) =
10,300 Hz (5 points).
open
h
closed
Problem 2 (3+3+3+3+3+5 points)
6.00 kg of ice at -5.00oC is added to a cooler holding 30 liters of water at 20.0oC. The
density of water is 1000 kg/m3. The latent heat of fusion of water Lf is 3.33 x 105 J/kg ,
the specific heat of water cwater and ice cice are 4190 and 2090 J/(kg K), respectively.
(a) What is the mass of 30-liter water mwater ?
(b) To melt 6.00 kg of ice how much heat Qmelt is needed?
(c) To change the temperature of the ice to 0.00oC how much of heat Qice is needed?
(d) Once all the ice melts how much of heat Qice-water is needed to raise the temperature of
the water from the ice to T?
(e) To lower the temperature of 30 liter water from 20.0oC to T, how much heat Qwater is
needed?
(f) What temperature of the water T when it comes to equilibrium?
Solution:
(a) mwater=(30.0 L) x (1.00 kg/L)=30.0 kg (3 points).
(b) Qmelt=(6.00 kg) x (3.33 x 105 J/kg) = 2.00 x 106 J (3 points).
(c) Qice=(6.00 kg) x cice x [0.00oC –(-5.00oC)]=(6.00 kg) x [2090 J/(kg K)](5.00 K)
=6.27 x 104 J (3 points).
(d) Qice-water=(6.00 kg) x cwater x [T-(0.00oC)]=2.51 x 104 x T J (3 pints).
(e) Qwater=(30.0 kg) x cwater x [T-(20.0oC)]=(12.57 x 104 J)x(T-20.0oC) (3 points).
(f) From Qmelt+Qice+Qice-water+Qwater=0 (2 points),
(200 x 104 J) + (6.27 x 104 J) + [(2.51 x 104 J/K) x T] + (12.57 x 104 J/K) x (T-20.0oC) = 0,
So (15.1 J/K)T=45.7 J, and therefore T=3.03oC (5 points).
Problem 3 (2+4+4+2+4+4 points)
A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than
cooler air at the same pressure. If the volume of the balloon is V=500 m3 and the surrounding
air is at T=15.0oC, you will find the temperature of the air in the balloon be for it to lift a total
load m=290 kg (in addition to the mass of the hot air). The density of air at T=15.0oC and
atmospheric pressure is r = 1.23 kg/m3. Ignore the weight of balloon’s skin. Answer the
following questions:
(a) Find the buoyant force B by the air in the atmosphere in terms of V, r and g?
(b) What must the density of the hot air r’ be for the balloon and the load to float in the
atmosphere in terms V, r, and m?
(c) From the ideal gas equation pV=nRT, derive the relation rT= constant at a constant
pressure.
(d) If the temperature of the hot air is T’, express T’ by T, r, and r’.
(e) Express T’ in terms of T, r, m and V.
(f) Plug-in all the numerical values needed and find the temperature of the hot air.
Solution:
(a) B = rVg (2 points).
(b) Fnet = mg + r’Vg –B = 0. Therefore r’ = r – m/V (4 points).
(c) pV=nRT=(mtot/M)RT where p, n, mtot, M and R are the pressure, number of moles,
the mass of air, the molecular mass of air and universal gas constant. Knowing that
p, M, and R are constant and mtot=rV, rT=r’T’ =constant (4 points).
(d) From the result of Part (c) T’=Tr/r’ (2 points).
(e) Using the results of Part (b) and (d) T’=Tr/(r-m/V)=T/[1-m/(Vr)] (4 points).
(f) Using the result of Part (e), T’=(288.15 K)/{1-(290 kg)/[(500 m3)(1.23 kg/m3)]}=545 K=272oC
(4 points).