PHGN341: Thermal Physics
Final Exam - May 13, 2010
NAME: KEY
1.
[15] A refrigerator maintains an inner temperature of -15 C. The heat rejection coils operate at +50 C. On average,
the insulation of the refrigerator allows 20 Watts of heat to leak in. Estimate the minimum annual cost of running the
refrigerator if electricity costs $0.10 per kilowatt-hour?
Solution: The coefficient of performance is limited by the second law to be
COP =
Qc
≤
W
Th
Tc
1
;
−1
so in this instance with Th = 323 and Tc = 258, we have COP ≤ 3.969. Since the refrigerator leaks 20 W, to maintain
the temperature inside, it must remove that amount of heat at the same rate, 20 W. In one year (3.15 × 107 s) the
8
Qc
J
1.59×108 J
minimum amount of work needed to do this is: W ≥ COP
= 6.30×10
= 1.59 × 108 J = 3.6×106
3.969
J/kw·hr = 44.1 kw-hr.
Electricity must provide this work. At $0.10 per kw-hr, this refrigerator will cost at least $4.41 to operate for a year.
2.
[15] Estimate the temperature of the melting point of ice at 200 bar pressure. (Data: Latent heat: L = 0.333 MJ/kg,
upon freezing 1000 cm3 of water (1 kg) expands to 1091 cm3 .)
∆P
L
∆T ' T (Vwater −Vice ) .
−5
Pa·273 K·9.1×10
m3 /kg
=
0.333×106 J/kg
Solution: The Clausius-Clapeyron relation gives the relation for the phase boundary:
200×10
5
Thus the
change in temperature needed to stay on the phase boundary is ∆T = −
−1.49 K.
Note the minus sign which is due to the fact that water expands when it freezes. Thus, the freezing temperature at 200
Pa is 271.5 K.
3.
[20]
2
P
1
3
V
Consider an engine cycle that operates in the shape of a right triangle as shown. The engine operates in the clockwise
direction and uses Helium gas as the medium. Analyze this engine by filling out the table below assuming the following
conditions: P1 = P3 = 1.0 bar, V1 = 1.0 m3 , P2 = 4.0 bar, V2 = V3 = 5.0 m3 . From your table find the efficiency
of the triangle engine under these operating conditions. Show all work and use the back if you need more space.
Step
Work (kJ)
∆U (kJ)
Q (kJ)
1→2
-600
+2850
+3450
2→3
0
-2250
-2250
3→1
+400
-600
-1000
1
R
Solution: The work done on the system is given by: W = − P dV which is nothing morethan the area under the
P-V curve for that step. For the 1 → 2 step: W1→2 = − 21 (P2 − P1 )(V2 − V1 ) + (P3 − P1 )V1 = −600 kJ. There is no
work done in step 2 → 3 since the volume is constant. For the 3 → 1 step: W3→1 = −(P1 − P3 )V1 = 400 kJ.
From the equipartition theorem for an ideal gas, we have U = f2 N kT where here f = 3 because Helium is a monoatomic gas. Using the ideal gas law, P V = N kT , we get: U = 23 P V . Thus the change in internal energy is given by:
∆Ui→f = 32 ∆(P V ) = 32 (Pf Vf − Pi Vi ). From this we find: ∆U1→2 = 23 (P2 V2 − P1 V1 ) = 2850 kJ. Likewise, we find:
∆U2→3 = 23 (P3 V3 − P2 V2 ) = −2250 kJ and ∆U3→1 = 23 (P1 V1 − P3 V3 ) = −600 kJ. Note that ∆Ucycle = 0.
Finally, for each step we can use the first law of thermodynamics to find Q = ∆U − W which gives the values shown
in the table above.
The efficiency is defined as the net work done by the system (+600 kJ) divided by the heat put into the system.
200
= 0.058. Equivalently, one can calculate the efficiency as
Only step 1 → 2 puts heat into the system; so e = 3450
|Qc |
3250
e = 1 − Qh = 1 − 3450 = 0.058
4.
[20] Satellites have measured that on average the earth is absorbing about 0.85 W/m2 more power per area than it is
radiating. The goal of this problem is to estimate an upper bound to the annual increase in temperature by assuming
this excess heat is absorbed entirely in the atmosphere.
(a) How much excess heat must the atmosphere absorb in one year? (DATA: The radius of the earth is RE = 6.4 × 106
m.)
Solution: The excess heat that must be absorbed is Qabs = 0.85 W/m2 ·4π(6.4×106 m)2 ·3.15×107 s/year = 1.38×1022
J/year.
(b) The heat capacity of an ideal diatomic gas near standard temperature and pressure is cv = 5/2N kB . To estimate
the heat capacity of the earth’s atmosphere we need to know N , i.e. how many molecules it contains. Recall the
isothermal number density versus altitude relation: n(z) = n0 e−z/z0 , where z0 = kB T /(mg) where m is the mass of a
gas molecule and g = 9.8 m/s2 is the acceleration of gravity. Calculate z0 (in m). (DATA: m̄ ' 4.8 × 10−26 kg.)
Solution: Direct substitution gives: z0 =
kT
mg
=
1.381×10−23 J/K·298 K
4.8×10−26 kg·9.8 m/s
= 8.75 × 103 m.
(c) Use the ideal gas law to evaluate n0 , where the sea-level (z = 0) atmospheric pressure is 1 bar, 105 Pa.
Solution: The number density at standard conditions is: n0 =
N
V
=
P
kT
=
105 Pa
1.381×10−23 J/K·9.8 m/s
= 2.43 × 1025 m−3 .
(d) The atmosphere is quite thin compared to the radius of the earth, so the volume element in a spherical shell at
2
dz. Integrate the number density times the volume element
height z (measuring from the earth’s surface) is dV = 4πRE
to find the total number of gas molecules in the earth’s atmosphere.
Solution: Integrating the number density gives:
Z ∞
Z
2
2
N=
n0 e−z/z0 (4πRE
)dz = 4πn0 RE
z0
0
44
Evaluation gives: N = 1.09 × 10
∞
2
dx e−x = 4πn0 RE
z0 .
0
molecules.
(e) Use your results from above to find the heat capacity of the atmosphere (in J/K) and then estimate the temperature
rise of the earth’s atmosphere in one year.
Solution: The heat capacity of the atmosphere is Cv = 52 N k = 3.76×1021 J/K. Thus the rise in atmospheric temperature
in one year is ∆T = Qabs /Cv = 3.66 K.
5.
[20] When a common star burns up its nuclear fuel, it cools and contracts until it becomes a white dwarf which is
prevented from collapsing further by the degenerate (T = 0) electron pressure.
2
(a) Assume that the nuclei making up the star have an equal number of protons and neutrons. Let mN the average
mass a nucleon and let me be the mass of an electron (mN >> me ). Derive an expression for the Fermi momentum,
pF = h̄kF , of the electrons for a white dwarf star with radius, R? , and mass, M? . (Since the star is electrically neutral,
the number of electrons, Ne , equals the number of protons.) Give your answer in terms of R? and M? and physical
contants.
Solution: The number of electrons is
M?
=
Ne =
2mN
kF
Z
2V
0
4πk 2
V kF3
dk
=
(2π)3
3π 2
The Fermi momentum is given by
h̄
R?
pF = h̄kF =
9πM?
8mN
1/3
.
3GM 2
(b) The gravitational potential energy of a white dwarf star of uniform density is UG = − 5R?? . Treating only the
kinetic energy of the degenerate electrons, the total energy of the star is Etotal = 35 Ne EF + UG , where EF is the
nonrelativistic Fermi energy (p2F /(2me )). From the total energy, find an expression for the equilibrium radius of the
white dwarf star (i.e. the radius which minimizes the total energy).
∂ETotal
∂R?
= 0, or
"
#
2/3 5/3 ∂
M?
3
3GM?2
9π
3 h̄2
2
GM?2
N e EF −
.
=
− 3 +
∂R? 5
5R?
5 4me 8
mN
R?
R?2
Solution: The equilibrium condition is
Solving for R? gives:
R? =
h̄2
2me GM?2
M?
mN
5/3 9π
8
2/3
.
(c) Evaluate the equilibrium radius for a white dwarf of one solar mass, M? = 2.0 × 1030 kg.
Solution: Substituting values for all the constants gives: R? = 7.22 × 106 m.
6.
[30] Consider a thin sheet of material of area A with N atoms which can be approximated as a two-dimensional
membrane with an area density of nA = N/A. Let cs be the speed of sound for the membrane. Treat only one phonon
d2 k
polarization; so the density of states is given by A (2π)
2 , where A is the area of the membrane.
(a) Find an expression for the 2-d cut-off wave number, kmax , that gives the correct area density, nA .
Solution: Integrating the density of states in 2-dimensions gives:
Z
Z 2π Z km ax
2
A
d2 k
Akmax
θ(k
−
k)
=
.
N =A
dθ
dk k =
max
2
2
(2π)
(2π) 0
4π
0
where θ(x) is the Heaviside step function (θ(+) = 1, θ(−) = 0). Solving gives kmax = (4πnA )1/2 , where nA = N/A.
(b) Find an expression for the 2-d Debye temperature (i.e. the temperature such that kB TD = Ephonon = h̄cs kmax ).
Solution: The Debye temperature is defined by kTD = h̄cs kmax which gives TD =
h̄cs (4πnA )1/2
.
k
(c) Write down the integral expression for the 2-d phonon energy at temperature, T (do not evaluate it).
Solution: The average 2-d phonon energy at temperature, T, is
Z kmax
A
1
U=
(2π)
dk k(h̄cs k) βh̄cs k
.
(2π)2
e
−1
0
3
(d) Calculate the phonon contribution to the internal energy in the high temperature limit.
Solution: In the high temperature limit one can expand the denominator term to find: eβh̄cs k −1 ' 1+βh̄cs k +. . .−1 =
βh̄cs k. Substituting this into the integral gives:
Z
A
h̄cs
U'
2π
kmax
dkk 2
0
2
1
A kmax
=
.
βh̄cs k
2πβ 2
(e) Calculate the phonon contribution to the internal energy in the low temperature limit.
Solution: In the low-temperature limit, β is large; so eβh̄cs k >> 1. Thus,
U'
A
h̄cs
2π
Z
kmax
dkk 2 e−βh̄cs k =
0
A k3
T 3.
π (h̄cs )3
(f) (Extra credit) Now lets do the numbers. Suppose the membrane is graphine. The area density of graphine is
nA = 42.2 atoms/nm2 and its speed of sound is cs = 12, 450 m/s. Evaluate your expressions to find the following values
in the indicated units:
Solution: Substituting the values from the previous page, find:
p
i) kmax (in nm−1 ): kmax = 4π(42.2/m3 ) = 23.03 nm−1 .
ii) Debye temperature (in K): TD = 2.189 × 103 K.
iii) Phonon contribution to the heat capacity (in eV/K) per atom in the high temperature limit evaluated at T=5000
cv
K: cv = ∂U
∂T . Using the high temperature limit of U calculated calculated in part (d), we have N = k, independent of
T.
iii) Phonon contribution to the heat capacity (in eV/K) in the low temperature limit evaluated at T= 5 K. Instead, if
we use the low-energy limit in part (e) we find: cNv = 6.3 × 10−5 k.
4