PHYS 2303: HW 1 Solutions
January 2017
1
Chapter 1
1.47 (a) The formula for the conversion to Celcius to Fahrenheit is
F =
9
C + 32
5
When temperatures in both scales are the same, C = F = T . Then,
9T
+ 32
5
5T = 9T + 160
T =
T = −40◦ C
(b) Conversion from Celcius to Kelvin: K = C + 273. Then,
F =
9
(K − 273) + 32
5
Like before, when the temperatures in both scales are the same, F = K =
T.
9
(T − 273) + 32
5
5T = 9T − 2297
T =
T ≈ 574 K
1.48 We use the same conversion formula and set F = x and C = −x.
9x
+ 32
5
14x = 160
x=−
x = 11.4◦ F
1.64 The work done during rubbing hands will be converted to heat. In general,
rubbing an arbitrary number of times
∆W = nF ∆x
1
For n = 20,
∆W = 20 × 40 × 7.5 × 10−2
= 60 J
The average heat capacity of the human body is around 3500 J/K/kg.
Then,
∆W
mc
60
=
= 0.17◦ C
0.1 × 3500
∆T =
1.65
∆Q = mc∆T
1 ∆Q
m ∆T
4.35 × 103
=
0.25 × 45
= 386.67 J/kg/K
c=
The substance is most likely composed of copper.
1.77 Assume the pond has a mass M of water. For a temperature increase of
1.5◦ C,
∆Q = M × 4186 × 1.5
To prevent this temperature increase, a certain amount, m, of water must
therefore evaporate since we know that phase change is not accompanied
by any temperature change. The latent heat of vaporization is mLv , where
Lv = 2264.76 kJ/kg.
∆Q = M × 4186 × 1.5 = mLv
m
4186 × 1.5
=
M
2264.76 × 103
= 2.77 × 10−3
Hence, 0.277% of the water needs to evaporate.
1.79 (a)The mass is given bym = ρV
m = 917 × (160 × 40) × 103 × 250
= 1.47 × 1015 kg
2
(b) Total energy required to melt the iceberg will be equal to the latent
heat of fusion of the entire mass.
∆Q = mLf
= 1.47 × 1015 × 334000
= 4.91 × 1020 J
(c) Assuming only the top surface of the iceberg receives sunlight at a
rate of P = 100 W/m2 , the total energy absorbed during one day would
be (also assuming 12 hr of sunlight)
E =P ×A×t
= 100 × (160 × 40) × 103 × 12 × 3600
= 17789 days ≈ 49 years
1.113 Consider the energy transfers that take place- the iron cylinder loses heat,
being transfered first to the latent heat used to melt the ice and then raise
the temperature to T . Let c1 and c2 be the specific heat capacities to iron
and water respectively. For the total process, ∆Q1 + ∆Q2 + ∆Q3 = 0
M c1 (T − 1000) + mLf + mc2 (T − 0) = 0
0.8 × 452(T − 1000) + 334000 × 1 + 4186T = 0
After doing the calculations, we arrive at T = 6.1◦ C All the ice has melted
since the heat transfered by the iron cylinder is more than the latent heat
of 1 kg of ice.
3
2
Chapter 2
2.25 Using the ideal gas law at constant temperature
PV
V0
1.72 × 107 × 43.8
=
= 1.8834 × 108 Pa
4
P0 =
1.8834 × 108
1.01 × 105
= 1864
no. of balloons =
2.27 From the ideal gas law
PV
kB T
10−7 × 10−6
=
1.38 × 10−23 × 293
= 2.47 × 107
N=
2.29 Volume of 100 cm3 air under tire pressure
100 × 1.01 × 105
7 × 107
= 14.4 cm3
V =
Initially,
P V = 2000 × 7 × 105 = 1.4 × 109
After the air has been released,
P 0 V 0 = 1.4 × 109 − 1.01 × 107 = 1.3899 × 109
P0 =
1.3899 × 109
≈ 6.949 × 105 Pa
2000
2.43 (a)
n=
PV
RT
80
8.31 × 273
= 0.0352 mol
=
4
(b)
3
kB T
2
3
= × 1.38 × 10−23 × 273
2
= 5.65 × 10−21 J
Eave =
(c) 200◦ C = 473 K
P V = 0.0352 × 8.31 × 473
= 138.3 J
2.46 The root-mean-square speed is given by the following formula
r
3RT
vrms =
M
where M is the molar mass. From this equation, we can form an expression
for the temperature. In order for oxygen molecules to be able to escape
the Earth,
2
mvrms
3R
0.032 × (11.1 × 1000)2
=
3 × 8.31
= 1580 K
T =
2.74 Most probable velocity
r
vp =
2RT
M
2RT
vp2
2 × 8.31 × 296
=
2652
= 0.0711 kg/mol
M=
2.79 Let (P0 , V0 , T0 and (P1 , V1 , T1 ) be thermodynamics quantities at initial
and final positions. Also V1 = 1.8V0 . At the surface, the pressure is equal
to the atmospheric pressure.
P1 V1 T0
V0 T1
1.01 × 105 × 1.8V0 × 277
=
v0 × 283
= 1.78 × 105 Pa
P0 =
5
Recall the variation of pressure with fluid depth, ∆P = ρg∆h.
∆P
ρg
0.78 × 105
=
1000 × 9.8
= 7.9 m
∆h =
2.97 We have the Maxwell-Boltzmann distribution
3/2
2
4
m
f (v) = √
v 2 e−mv /2kB T
π 2kB T
R∞
which is normalized to unity, i.e 0 f (v)dv = 1. To prove this
∞
Z
0
4
f (v)dv = √
π
Using the subsitution u =
Z
0
∞
4
f (v)dv = √
π
q
m
2kB T
m
2kB T
m
2kB T
3/2 Z
=1
6
v 2 e−mv
2
/2kB T
dv
0
v, dv =
3/2 ∞
q
2kB T
m
2kB T
m
du
3/2 Z
|0
∞
2
u2 e−u du
{z
}
√
π/4