Physics 101 Quiz #8 Solution
November 20, 2002
Name
20 Minutes. Box your answers.
density
water
copper
aluminum
(kg/m3 )
1000
8890
2700
specific heat
thermal
coefficient of
capacity
conductivity thermal expansion
(J/kg◦ C)
(J/m s ◦ C)
(1/◦ C)
4186
0.60
69 × 10−6
387
390.
17 × 10−6
900
240.
23 × 10−6
1. A 4 kg aluminum rod has a temperature of 100◦ C. It is placed into an insulated
bucket containing 50 kg of water at 0◦ C.
(a) [4 points] What is the equilibrium temperature of the water and
aluminum?
Qgained by water = Qlost by aluminum
mH2 0 cH2 0 (T − 0◦ C) = mAl cAl (100◦ C − T )
(50 kg) 4186 kgJ◦ C (T − 0◦ C) = (4 kg) 900 kgJ◦ C (T − 0◦ C)
2.09 × 105 (T − 0◦ C)
2.09 × 105 T
2.13 × 105 T
T
= 3600 (100◦ C − T )
◦
= 3.6 × 105 C − 3600 T
◦
= 3.6 × 105 C
= 1.7◦ C
(If you don’t get the answer to part a, use T = 50◦ C in part b.)
(b) [2 points] The original length of the aluminum rod was 0.50 m. By how
much did its length change?
∆L = αL0 ∆T = (23 × 10−6 )(0.5 m)(100.0◦ C − 1.7◦ C) = 0.0011 m = 1.1 mm
Continued on the other side....
2. In an industrial lab, a vat of liquid at 20◦ C must be heated. This is accomplished
by running a hot water pipe of length 0.50 m and radius 4.0 cm (0.04 m) through
the vat. The pipe is made of 2.0 mm (0.002 m) thick copper. The water flowing
through the pipe has an average temperature of 60◦ C.
(a) [2 points] How much heat flows through the wall of the pipe in ten seconds?
The pipe is a cylinder of length L = 0.5 m and radius r = 0.04 m, so the pipe
wall has area A = 2πLr = 0.126 m2 . It has thickness l = 0.002 m, so
390 msJ◦ C (0.126 m2 )(60◦ C − 20◦ C)(10 s)
kA∆T t
Q=
=
= 9.8 × 106 J
l
0.002 m
(b) [2 points] The water in the pipe flows at a velocity of 20.0 m/s. How much
does the temperature of the water change as it flows through the pipe?
(Hints: (1) The temperature change is small, so your answer to part a is
valid even though T is changing; (2) What mass of water flows through
the pipe in 10 seconds? How can you use this number?)
(Note: this solution uses different definitions of A and ∆T than part a. Flow
rate is written as Qf to distinguish it from heat, which is written as Q.)
2
The cross-sectional area of the pipe is A = πr2 = π (0.04 m) = 0.00503 m2 .
2
The volume flow rate is Qf = vA = (20 m/s)(0.00503 m ) = 0.101 m3 /s. In
t = 10 s, V = Qf t = (0.101 m3 )(10 s) = 1.01 m3 of water flows through the
pipe. This has mass m = ρV = (1000 kg/m3 )(1.01 m3 ) = 1010 kg. The water
loses the energy that passes through the pipe wall, so Q = −9.8 × 106 J, and
Q = mc∆T
⇒
∆T =
Q
−9.80 × 106 J
= −2.3◦ C
=
mc
(1010 kg) 4186 kgJ◦ C
Rewrite and sign the honor pledge: “I pledge my honor that I have not violated the
Honor Code during this examination.”
Signature
Physics 101 Quiz #8 Solution
November 19, 2004
Name
20 Minutes. Box your answers.
Constants needed in the quiz:
Radius of the earth: 6.38 · 106 m
Density of snow: 200 mkg3
Density of water: 1000 mkg3
Specific heat capacity of water: cH2 O = 4186 kg·J◦ C
Latent heat of fusion for water/ice: Lf = 33.5 · 104
J
Stefan-Boltzmann constant: σ = 5.67 · 10−8 K4 ·m
2 ·s
J
kg
Problem 1 Soon after the earth was formed, heat released by the decay of radioactive
elements raised the average internal temperature from 300 to 3000 K which roughly
is the value today.
(a) [2 points] Assume that the coefficient of volume expansion is β = 3 · 10−5 K−1 ,
how much did the radius of the earth increase when it heated. (Hint: 3 · α = β)
∆R
=
=
∆R
=
α · R · ∆T =
1
· β · R · ∆T
3
1
1
· 3 · 10−5
· 6.38 · 106 m · 2700 K
3
K
172 km
(b)[2 points] The heat of the inner parts of the earth
is conducted to the surface and can be measured. In
North America in a surface area of A = 1 m2 the heat
flow coming from underneath is Q/t = 54 · 10−3 J/s
(see figure on the right). If the surface temperature is
15◦ C and the thermal conductivity of the rocks in the
upper part of the earth is k = 2.5 s·m·J◦ C , what is the
temperature at a depth of d = 35 km?
A
d
earthsurface
heatflowQ/t
fromthecenter
oftheearth
A
· ∆T · t
d
d
J
35000 m
Q
·
= 54 · 10−3 ·
∆T =
= 756◦ C
t k·A
s 2.5 smJ◦ C · 1 m2
Q=k·
=⇒
The absolut temperature at 35 km is then T = 756 ◦ C + 15◦ C ≈ 770◦ C.
Continued on the other side....
(c)[2 points] If an area of 1 m2 at the earth surface emits Q/t = 54 · 10−3 J/s in
radiation, what would the temperature at the earth surface be, if there would be no
other energy sources? Assume that the earth is a perfect black body emitter (e = 1).
Q = e · σ · T 4 · A · t,
T =
!
4
Q
1
·
=
t σ·A
"
4
54 · 10−3
e=1
J
1
·
≈ 31 K
J
−8
2
s 5.67 · 10
K4 ·m2 ·s · 1 m
(d)[2 points] The amount of heat that reaches the earth from the sun is much larger
than this. A surface area of 1 m2 receives 1400 J per second from the sun. If all the
energy would be absorbed by a 0.5 m layer of snow, how long would it take to melt
the snow?
mass of the snow: msnow = ρsnow · V = 200
kg
m3
· 0.5 m3 = 100 kg
heat needed to melt the snow: Q = msnow · Lf = 100 kg · 33.5 · 104
time needed to melt the snow:
Q
1400J/s
=
33.5·106 J
1400 J/s
J
kg
= 33.5 · 106 J
= 24000 s = 6.6 h
(e)[2 points] After the snow melted, the water of 0◦ C stays on the surface and will
absorb 10.1 · 105 J during the rest of the day. What is the final temperature of the
water?
Q = cH2 O · mH2 O · ∆T
∆T =
Q
10.1 · 105 J
=
= 2.4◦ C
cH2 O · mH2 O
4186 kg·J◦ C · 100 kg
Rewrite and sign the honor pledge: “I pledge my honor that I have not
violated the Honor Code during this examination.”
Signature
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QUIZ 8
Physics 101, Fall 2005
SOLUTIONS
Problem 1: Ideal Monatomic Gas Engine
One mole of an ideal, monoatomic gas is
contained in a cylinder closed on the top
by a movable piston, and is taken through
the cycle shown on the figure (A → B →
C → D → A). The temperature at point
A is TA = 293 K. The volume at the point
A of the cycle is VA = 0.023 m3 .
2PA
B
C
PA
A
D
VA
3VA
a. (1 pt) Determine the total internal energy UA at point A. Box your answer.
UA =
3
3
nRTA = (1 mol)(8.31 J mol−1 K−1 )(293K) = 3650 J
2
2
b. (2 pts) Determine the net work performed by the gas on the surroundings (W ) in one
complete cycle (A → B → C → D → A). Box your answer.
PA =
(1 mol)(8.31 J mol−1 K−1 )(293K)
nRTA
=
= 105 Pa
VA
0.023 m3
WABCDA = (2PA − PA )(3VA − VA ) = 2PA VA = (2 × 105 Pa)(0.023 m3 ) = 4600 J
c. (2 pts) Determine the net heat exchanged with the surroundings (Q) in one complete
cycle (A → B → C → D → A). Hint: pay attention to the sign! Q is positive if heat enters
the system, negative if heat exits the system. Box your answer.
∆UABCDA = QABCDA − WABCDA = 0
QABCDA = WABCDA = 4600 J
QUIZ 8
Physics 101
d. (2 pts) Determine the temperatures at points B, C, and D (TB , TC , and TD respectively).
Box your answers.
PA V A
PB V B
PB V B
2PA VA
=
→ TB = TA
= TA
= 2TA = 2(293 K) = 586 K
TA
TB
PA V A
PA V A
In a similar way, for points C and D:
TC = TA
PC V C
2PA 3VA
= TA
= 6TA = 6(293 K) = 1758 K
PA V A
PA V A
TD = TA
PA 3VA
PD V D
= TA
= 3TA = 3(293 K) = 879 K
PA V A
PA V A
Problem 2: Liquid Helium Storage
(3 pts) Liquid helium is stored at its boiling point temperature of 4.2 K in a spherical
container with radius = 0.3 m. The container is a perfect blackbody radiator (i.e.: emissivity e = 1). The container is surrounded by a spherical shield cooled at liquid nitrogen temperature (77 K). The space between the container and the shield is under vacuum.
The latent heat of vaporization of Helium is LV = 2.1 × 104 J/kg. What mass of helium
boils away through the venting valve in one hour? Hint: the Stefan-Boltzmann constant is
σ = 5.67 × 10−8 J s−1 m−2 K−4 ). Box your answer.
4
mLV = Q = eσ4πr2 (TN4 − THe
)t
m=
4
eσ4πr2 (TN4 − THe
)t
(1)(5.67 × 10−8 J s−1 m−2 K−4 )(4π)(0.3 m)2 [(77 K)4 − (4.2 K)4 ](3600 s)
=
LV
2.1 × 104 J kg−1
m = 0.39 kg