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Physics 7b
Fall 2003
Midterm 1
Section 1
September 29, 6:00-8:00pm
Answer all problems. Write neatly and clearly, explaining your work.
Partial credit for incomplete solutions will be granted provided your logic
is reasonable and clear. Indicate any parts what should not be counted
toward your grade with an “X”.
Enclose all answers in boxes. All numerical answers must be expressed
in SI units. Answers with no explanations, or disconnected comments
will not receive credit. If you obtain an answer that is questionable,
explain why you think it is wrong.
NAME:__________________________________________________
SID#:___________________________________________________
1:________________/5 8:________________/
2:________________/5 9:________________/
3:________________/10 10:_______________/
4:________________/
5:________________/5 TOTAL SCORE:___________
6:________________/
7:________________/
CONSTANTS
Avogadro’s Number: NA= 6.022 × 10
23 / mole
Boltzmann’s Constant: k = 1.381 × 10
Stefan-Boltzmann’s Constant:
8 2 4 σ 5.67*10 Wm K
− − −
Universal Gas Constant: R = 8.314 J / mol·K = 8.206 × 10
Gravity: g = 9.8 m/s
2
1 cal=4.18J
1 atm=1.013*
5 N-m
1 liter = 10
3
Material properties
cwater = 4190 J/kg-K kcopper = 2.0 W/m-C
o kair = 0.023 W/m-C
o
kwater = 0.56 W/m-C
o Lvapor(water)=2260 J/kg Atomic Weights
: N 2 :28gm/mol He:4gm/mol
FORMULAS
Thermal Linear Expansion: ∆ L =αL∆T
Work: W = pdV
Specific heat for solids: Q=mc∆T
Maxwell Distribution
Molar specific heat for gases
Q=nC∆T, where C= CV for constant volume, C=CP for constant pressure
Adiabatic expansion: =constant
Heat Transfer
Stefan’s Law:
4 P = σεAT
Thermal conductivity:
dQ dT kA dt dx
= −
No advanced calculator features may be used at any time.
3/2^2 (^2 )
8
4 2
mv v (^) kT
kT v m
dN m N v e dv kT
π
π π
−
=
= (^)
- (10pt)You place a 1.5 kg stone in a large lake of water at a temperature of 20°C.
The heat capacity of the stone is 760 J/kg-K and its initial temperature is 5°C.
When this system reaches equilibrium how much has the entropy of the water
changed?
( )
(1.5 )(760 / )(20 5 )
17,100 ( )
17100 58.96 /
293
stone stone stone fstone istone
o
stone reservoir
reservoir
Q m c T T
kg J kg K C
J heat flowed in
Q J S J K
T K
= −
= − −
= +
−∆ − ∆ = = = −
- (5 pt) A raindrop falls vertically a distance of 9000 m. Assuming that the change
in potential energy goes entirely into internal energy of the droplet, determine the
change in temperature of the droplet due to this descent.
2
: 0
(9.8 / )(9000 ) 21.
4190 J/kg-K
water
water
water
GIVEN W
PE mgh
U Q mc T
gh c T
gh m s m T K
c
=
∆ =
∆ = = ∆
= ∆
⇒ ∆ = = =
(5 pt) An aluminum cup (linear expansion coefficient=25x
/°C), capable of
holding a fluid volume of 100 cm
3 , is filled to the brim with a fluid whose volume
expansion coefficient is 4x
- /°C. How much, if any, fluid will spill out of the
cup when the temperature is raised from 300 K to 330 K?
5
4
3
4 3
3
3 7.510 /*
410 /*
100
(4 0.75)10 / 30 100*
0.
o Al Al
o
fluid
ifluid iAl
ffluid fAl
fluid Al
o
C
C
V T V V V cm
V
Amount spilled V V
T V
C K cm
cm
β α
β
β
β β
−
−
−
= =
=
∆ = ∆ = = =
= ∆ − ∆
= − ∆
= ^ −
=
g
- (5pt) The ancient Egyptians hung grass mats in doorways and windows and kept them wet to cool the dry summertime air coming into their homes. Explain the
method used here.
The principle here is evaporative cooling, where the
cooling “medium” is the mat. The hot air flowing
through the mat has its temperature lowered because
heat from the “medium” is used to evaporate the water
and cool the mat. The rate at which water is absorbed
into the air passing through the cooling media is the
evaporation rate. Can be used here because have dry
air, so low relative humidity in environment. We are
able to add vapor to the air without coming close to the
saturation pressure.
- (10pt) A solid cylindrical copper rod 10cm long has one end maintained at a
temperature of 20K. The other end is blackened and exposed to thermal radiation
at 400K. Assume no heat is radiated or conducted out of the curved surface of the cylinder. When equlibrium is reached, what is the temperature difference between
the two ends?
( ) ( )
4
8 2 4
2
1 1
20 -1 1
5
rad rad
rad
conduct rad rad end
conduct rad
rad end
dQ
P AT
dt
Wm K
for black T K
dQ
Wm A
dt
dQ kA WK cm
T T A T K
dt l m
dQ dQ
dt dt
K K T
− − −
−
− −
−
− −
Note: k(copper)=100W/Kcm
1 1
1
1 1 1
: ,
:
: ,
b a o
a a a
b b b
a o b a b
b
o c c a a o o c
c
ISOTHERMAL T constant along isotherm
T T T
Ideal gas Law PV nRT
P V nRT
P V nRT
V P P P P
V B
Adiabatic PV
regrouping PV V nRT V by Ideal gas Law
TV constant
T
T V T V T V V
T
γ
γ γ
γ
γ γ γ
− −
−
− − −
=
⇒ = =
=
=
= ⇒ =
=
⇒ =
= = ⇒ =
1
1
3 / 2
3 / 2 1
5 / 2 5 / 2
2 2
1 1 1
: 3
:
o
p p v
v v
o c
c o
c c c c
o o o c o c
o
d^ d
V
C C C (^) R
C C R
T
Given T Also d for monatomic gas by equipartition
A
so V A V
To calculate P use IGL P V nRT
T nRT P P nR A V P
A A V A
γ
γ
−
− −
−
− = − = = =
= =
=
=
= ⇒ = =
b) (10 pt) Complete the following table, expressing heat, work, and internal energy in
terms of the factors A and B, Po and Vo only. Include full and complete calculations for
each entry based on 1
st and 2
nd Law formulae, and process properties.
Process ∆U Q
a-b (^0) PoVoln(B)
b-c
3 1 1
2
o o
P V
A
− 5 / 2
o o
o o
P V
A
P V
B A
c-a
3 1 1
2
o o
P V
A
− −
Process W ∆S
a-b PoVoln(B) nRln(B)
b-c
5 / 2
3 1
2 2
o o
P V
B A
− −
-nRln(B)
c-a 3 1 1
2
o o
P V
A
−
:
:
2
, 0.
0
ln
ln( ) ln( )
:
0, ln( )
a b
a b
a b
b b b a b a a a a a
a b o a b o o
a b a b
a b o o
ISOTHERMAL
a b
d U nR T by equipartition
For an isothermal process T
U
dV V
W PdV nRT nRT
V V
W nRT B W P V B
FLT U Q W
But U so Q P V B
Entropy calc
−
−
−
−
− −
− −
−
→
∆ = ∆
∆ =
⇒ ∆ =
= = = (^)
= ⇒ =
∆ = −
∆ = =
∫ ∫
( )
( )
( , !!) :
:
2
ln ln
2
,
ln
a b
a b
a b
a b
a b
a b a b
a b a b
b b b
a a a
b a o
b
a
ulation
full derivation not required just understanding
FLT dQ dU dW
dQ (^) dU dW d dT dV
nR nR
T T T T V
dQ (^) d T V
S nR nR
T T V
For isothermal process T T T
V S nR
V
−
−
−
−
−
− −
− −
= +
= + = +
∆ = = +
= =
∆ =
∫
= nR ln( B ) = nR ln( B )
( (^ )^ (^ )(^ ))
( ) ( )
1 2
3 / 2 1 3 / 2 (^2) 5 / 2
b c o
b c o o
c
b c b
b c b b c b c b c
o o o b c o o
o
b c
d U nR T by equipartition
From table in part a
U nRT A
U P V
A
W PdV area under curve
W P V V P P V V
P P P
W V B A V B A
B B A
P V
−
−
−
−
−
^
∫
( ) ( )
3 / 2 1 3 / 2 (^2) 5 / 2
5 / 2
5 / 2
o o o
o o
b c b c b c
b c o o o o
B A P V B A
B B A
P V
B A
FLT U Q W
Q P V P V
A B A
− − −
−
−^ +^ −^ −
^
^
c a
o
c a o o
c a c a o o
a
c a c
c a
Definition o f adiabatic process Q
d U nR T by equipartition
d U nR T
nRT from table in part a A
U P V
A
W U P V by FLT withQ
A
Also W PdV
Can write adiabat
−
−
− −
−
∫
( )
1 1 1 1
i i
a
c a (^) c i i
c c a c c c a c c c c a
c a c c a a o o
b c a c a (^) a
equationas PV PV
dV W PV
V
P V V V P V V P V V
W from adiabat eqn
d W P V P V P V A
Entropy calculation no need to repeat derivation
dQ d S nR T
γ γ
γ γ
γ γ γ γ γ γ γ
γ γ
−
− − − −
−
−
− −
∫
∫ (^ )
( )
( )
0 3 / 2
3 / 2
ln ln
ln ln 2 ( / )
ln( ) ln 0 2
a a
c c
o c a o o
c a
T V
nR T V
For adiabatic process
d T V S nR nR T A A V
nR A nR A
Could also just say S
for adiabatic process by definition since Q
−
−
−
c)
(10 pt) Calculate the efficiency of the engine operating on this cycle. Is it a Carnot engine
and why?
5 / 2
5 / 2
ln( )
ln( )
L
H H
L o o o o
H o o
L
H
W Q
e
Q Q
Q PV P V
A B A
Q P V B
B
e
A B A
For a Carnot engine
T
e
T A
This clearly is not a Carnot engine
Why Have a non adiabatic isothermal porti
L
L
on
Efficiency is lower than a Carnot because now have
Q exhausted at varying T on portion b c rather
than to single T Reversible engines are defined in course
as represented on P V diagram the engine is reversible
b) (5 pt) What is the ratio of vrms for molecules in Boxes A and B before thermal
contact? After thermal contact?
Important that they mention how vrms derived.
- Internal energy is wholly kinetic energy
for gases.
- Translational kinetic energy=(1/2)mv
- By equipartition, have translational
KE=(3/2)kT
2 2 2 2 2 2 2
3
,
1
4 / 28
,
(3/7) 4 / 28 (3/49)
rms
N He
N
rms
He (^) N
rms
He
N He
N
N (^) He
rms
He N rms
He
kT v
m
After thermal equilibrium T T
v
v (^) m
m
BEFORE thermal equilibrium T T
T
v (^) T
v m
m
=
=
= =
≠
= = =
c) (5 pt) How much heat entered or left the nitrogen? The helium?
There’s no contact with the universe, AND no work
done. So change in internal energy=heat.
Heat is added to nitrogen, removed from helium, which
is calculated in part a
( )
A f o o
B f o o
Q R T T RT
Q R T T RT
d) (8 pt)What is the entropy change of the system and the entropy change of the universe?
In this problem, which is thermally isolated, there
is NO difference between the universe and the
system.
ln ln
ln ln
o A o o B o
universe A B
T
S R R
T
T
S R R
T
S S S R
^
^
