Final Solutions

a) The mass is (rho_f)(t)(A)= 100 kg.

b) Heat should be added at the average rate k(A)(87-17)/10 ^{-3} =7 x 10 ^6 Watts.

c) v_{rms} is proportional to sqrt(T), with T in Kelvins. T_{in}=360 and T_{out}=290. So v_{rms,in}/v_{rms,out}=sqrt(360/290) = 1.11.

d) The mean free path is proportional to V/N which, using the ideal gas law with p=1 atm held fixed, is proportional to T. So lambda _{in}/lambda _{out}=360/290 =1.24

e) n=pV/RT= (10 ^5)(3 x 10 ^3)/(8.31)(360)= 10 ^5

f) Using instead T=290 in above, n=1.24 x 10 ^5.

g) 2.4 x 10 ^4 moles difference, times the molar mass, gives M_{displ} = 720 kg

h) Since it just floats, the ratio must be 1.

2)

a) The heat required is m c _i (10) + m L_f + m c_w(50)= (0.5)(2 x 10 ^3 x 10 + 3 x 10 ^5 + 4 x 10 ^3 x 50) =2.6 x 10 ^5 J.

b) The 2.6 x 10 ^5 J must have been lost by the water in cooling the 50 degrees, so M x (4 x 10 ^3)(50)=2.6 x 10 ^5, which gives M = 1.3 kg.

c) The entropy of the water which was ice increases, since it is heated.

d) The entropy of the water which was hot decreases, since it was cooled.

e) The entropy of the entire system increases, since this is not a reversible process.

3)

a) f_{beat}=|f -f'|. Since she is coming toward you, her frequency is bigger, i.e. you hear her horn at f'= 1000 Hz.

b) v = v_s/10 =34 m/s.

c) After passing, you hear her horn at frequency (10/11)(900)= 818 Hz. The beat frequency is thus 82 Hz.

4)

a) With both ends closed, the fundamental harmonic has wavelength 2L=2 m. This gives frequency f_1=v/lambda = 171 Hz.

b) With one end uncovered, the wavelength is doubled, so the frequency is halved, i.e. f_2= 86 Hz.

c) Since the intensity is proportional to f^2, the new intensity is 1/4 that of the old one. This gives beta _2 = beta _1 + 10 dB log (1/4)= 74 dB.

5) f=v/lambda and both have the same lambda, so v_2=2v_1. Since v=sqrt{tau/mu}, The tension of string 2 must be 80N.

6)

a) Using Av=const, the new velocity is 40 m/s. b) Plugging into Bernoulli's eqn gives p_{upper}=2.7 x 10 ^5 Pa.

7)

a) For the first lens, p=infinity gives 1/i_1=1/f_C. Putting this in the second lens, -1/f_C + 1/i_2=1/f_E. Setting i_2=f gives 1/f=1/f_E+1/f_C. Solving gives f_E=0.02 meters.

b) 1/f_C=(n-1)(1/R_1-1/R_2) gives 1/R_1=1/R_2+1/(n-1)f_C, which gives R_1=0.04 meters.

8)

a) At the end of step 2, V=2 m^3.

b) The work of step 1 is zero, step 2 is 4 x 10 ^5 J, and step 3 is Nk_BT ln (1/2)=-8 x 10 ^5 ln (2). Adding gives total work Delta W=-1.5 x 10 ^5 J It is negative.

c) The total heat change is also -1.5 x 10 ^5 J.

d) In step 1, V is constant, so dW=0 and dQ=dE=(f/2) N k_B dT = 3 V dp. So Delta Q = 3 (1) (-4 x 10 ^5) = -1.2 x 10 ^6 J.

e) dS=dQ/T=3 N k_B dp/p, so Delta S = 3 N k_B ln (1/2) =-3(8 x 10 ^5)(1)(1/500) ln (2)= - 4.8 x 10 ^3 ln (2) =-3.3 x 10 ^3 J/K.

9)

The two paths have a pi RPS difference, so there is destructive interference when 2L=m lambda /n_2, so lambda = 2L n_2/m. So 6 x 10 ^{-7} = 2Ln_2/m and 4.5 x 10 ^{-7}=2Ln_2/(m+1) for some m. The ratio of these two is 6/4.5 = 4/3 = m+1/m, so m=3. Thus 2L n_2= 18 x 10 ^{-7} meters. Since n_2=3, we get

a) L= 3 x 10 ^{-7} meters.

b) For fully constructive interference, we must have lambda = 18 x 10 ^{-7} meters/(m+1/2). The only solution in the visible range is for m=3, which gives lambda = 5.14 x 10 ^{-7} meters.