Assignment 9, solutions
Problem 1:
A polar bear partially
supports herself by pulling part of her body out of the water onto a rectangular
slab of ice of volume 10 m3. The ice sinks down so that only half of
what was once exposed now is exposed, and the bear has 60% of her volume (and
weight) out of the water. Assume that the bear and the water have mass density
1 g/cm3 and ice has mass density 0.9 g/cm3. Estimate the
bear’s mass.
Solution:
Problem 2:
Pure water can be super-cooled at standard atmospheric
pressure to below its normal freezing point of 0 °C. Assume that a mass of
water has been cooled as a liquid to –5 °C, and a small (negligible mass)
crystal of ice is introduced to act as a “seed” or starting point of
crystallization. If the subsequent change of state occurs adiabatically and at
constant pressure, what fraction of the system solidifies? Assume the latent
heat of fusion of the water is 80 kcal/kg and that the
specific heat of water is 1 kcal/(kg oC).
Solution:
 | Concepts, principles, relations
that apply to the problem:
Specific heat, latent heat |
| Why do they apply?
The latent heat released as some of the water freezes heats
the system. |
| How do they apply?
The freezing stops when the ice and the water are at 0
oC. Let M be the mass of the water, M’ the amount that turns to ice.
Energy conservation: M(1 kcal/(kg oC))*(5 oC)
= M’(80 kcal/kg).
M’/M = 5/80 = 1/16. 6.25% of the system solidifies. |
| Details of the calculation:
None |
Problem 3:
A thin vertical uniform wooden rod is pivoted at the top
and immersed in water as shown.
The pivot point is slowly lowered. At a certain moment the
rod begins to deflect from the vertical. What fraction of the rod is still in
the water at that moment if the density of the rod is one-half of the density of
water?
Solution:
| Concepts, principles, relations
that apply to the problem:
The buoyant force, stable versus unstable equilibrium |
| Why do they apply?
If the rod is displaced
by a small angle q gravity and the
buoyant force will exert a torque about the pivot point. The direction of the
torque determines if we have a restoring force and a stable equilibrium.
|
| How do they apply?
Let the rod have length L and cross sectional area A (ÖA
<< L), and let aL be the length of the section of the rod in the water. Assume
the rod is displaced by a small angle q.
In the small angle approximation, the torque about the pivot point is
t = mgLq/2
– B(1 - a/2)Lq = ½rwatergAL2q/2
- rwatergaAL2(1 -
a/2)q, counterclockwise.
As long as t is
positive we have a restoring force and a stable equilibrium. When
t is negative we have an unstable
equilibrium and the rod will deflect from the vertical.
At the moment the rod begins to deflect from the vertical
t = 0.
½rwatergAL2q/2
- rwatergaAL2(1 -
a/2)q = 0, ½ - a(2 - a)
= 0, a2 - 2a + ½ = 0,
a = (1 -
Ö½).
When the fraction of the rod
under water is (1 - Ö½)
@ 0.3, the rod begins to deflect from
the vertical. |
| Details of the calculation:
None |
Problem 4:
A vertical open glass tube of length h is half-submerged in
mercury. The top end of the tube is then closed and the tube is slowly pulled
out until the bottom of the tube is barely submerged in the mercury. What is
the length of the mercury column remaining in the tube? The atmospheric
pressure corresponds to the pressure of a column of mercury of height H. Assume
the temperature is constant.
Solution:
| Concepts, principles, relations
that apply to the problem:
The ideal gas law |
| Why do they apply?
At constant temperature PV = constant (Boyle’s law).
When a length h/2 of the tube is above the mercury, the
volume of the gas in the tube is Ah/2 and the pressure is atmospheric pressure P0.
When a length h of the tube is above the mercury, the
mercury rises to a height h1 and the gas occupies a volume A(h - h1).
Since PV = constant (Boyle’s law), the gas pressure is
P = ½P0h/(h-h1).
We have P0 = ½P0h/(h-h1) +
rmercurygh1. |
| How do they apply?
Given: rmercurygH
= P0, so P0 = ½P0h/(h-h1) + P0h1/H.
Therefore
½hH = h1H + h1h - h12,
h12 - h1(H + h) + ½hH = 0
h1 = (H + h)/2 – (H2 + h2)½/2 |
| Details of the calculation:
None |
Problem 5:
The figure below shows a
maximally efficient Carnot cycle in the pressure-volume plane for a heat engine
operating between two heat reservoirs to perform work.
(a) For each label 1 through 4 identify the process, say whether work is done
by the working fluid or on it and whether heat is added to the working fluid or
extracted from it.
(b) Make a diagram showing the same cycle in the temperature-entropy plane.
Label the parts of your diagram that correspond to the parts labeled in the P-V
diagram and put arrows on each segment indicating the direction it is traversed.
Solution:
| Concepts, principles, relations
that apply to the problem:
The Carnot cycle, isothermal and adiabatic processes, entropy. |
| Why do they apply?
We are asked to describe the Carnot cycle in various ways. |
| How do they apply?
(a)
(1) Isothermal expansion: The engine is in contact with a heat
reservoir at temperature T1.
The engine does work. DW =
òPdV. Heat is added to the working fluid.
(2) Adiabatic expansion: During
the expansion the temperature falls to T2.
The engine does work. DW =
òPdV. No heat is added or extracted from
the working fluid.
(3) Isothermal compression: The engine is in contact with a heat
reservoir at temperature T2.
Work is done on the engine. DW =
-òPdV. Heat is extracted from the
working fluid.
(4) Adiabatic compression: During
the compression the temperature rises to T1.
Work is done on the engine. DW =
-òPdV. No heat is added or extracted from
the working fluid. |
| Details of the calculation:
(b)
Change in entropy:
 ,
dS = dQr/T
The subscript r denotes a reversible path.
 |
Problem 6:
A nursery uses natural gas heating to keep the greenhouses at 30 oC
all year. An engineer points out that the water at the bottom of a nearby lake
is at a constant temperature of 5 oC, and that he can build an ideal
heat pump that will work at maximum possible efficiency to pump heat from this
lake water into the greenhouses. He claims that the nursery will come out ahead
with his system, even though it uses electricity instead of natural gas at three
times the cost per Joule.
Is he right? Neglecting capital and maintenance costs by what factor would
their energy bill change?
Solution:
| Concepts, principles, relations
that apply to the problem:
The coefficient of
performance for a heat pump |
| Why do they apply?
The coefficient of
performance for a heat pump is the ratio of the energy delivered at the
higher temperature to the work put into the system, COP = Qhigh/(Qhigh
- Qlow). Knowing the COP, we can determine the cost
of the delivered energy. |
| How do they apply?
The best possible coefficient of performance is
COPmax(heat pump) = (Qhigh/(Qhigh - Qlow))max
= Thigh/(Thigh - Tlow) = Troom/(Troom
- Toutside)
Here COPmax = 303/25.
The gas heating system and the heat pump have to deliver the same amount of
energy.
For the heat pump: energy paid for = 25/303 times energy delivered
For the gas heater: energy paid for = energy delivered
Cost for heat pump delivery = 3*25/303 times cost for gas heater delivery
Their new bill would be approximately ¼ their old bill. |
| Details of the calculation:
None |
Problem 7:
In a Wilson cloud chamber at a
temperature of 20 degrees C, particle tracks are made visible by causing
condensation on ions by an approximately reversible adiabatic expansion of the
volume in the ratio
final
volume/initial volume = 1.375.
The ratio of the specific
heats of the gas at constant pressure and at constant volume is CP/CV
= 1.41. Estimate the gas temperature after the expansion.
Solution:
| Concepts, principles, relations
that apply to the problem:
Specific heat |
| Why do they apply?
We are supposed to use the definitions of the specific heats of the gas at
constant pressure and at constant volume and the given ratio of the two to
estimate the temperature. |
| How do they apply?
Definitions:
at constant pressure: dQ = nCPdT
at constant volume: dQ = nCVdT
Energy conservation:
at constant pressure: dU = dQ - dW = nCPdT - PdV = nCPdT
- nRdT
at constant volume: dU = nCVdT
dU depends only on the change in temperature,
so nCPdT - nRdT = nCVdT, CP - CV
= R.
Adiabatic expansion: dU = - PdV = nCVdT, dT = -PdV/(nCV).
Ideal gas law: PdV + VdP = nRdT = -nRPdV/(nCV) = -(CP
- CV)PdV/CV.
(Cp/CV)PdV + VdP = 0, (Cp/CV)dV/V
+ dP/P = 0.
Integrate: (Cp/CV)lnV + lnP = 0. PVg
= constant, g = Cp/CV.
P1V1g = P2V2g. P1 = P2(1.275)
g = 1.57 P2.
P1V1 = nRT1, 1.57 P2V1
= nRT1, P2V2 = nRT2.
Ratio: 1.57 (V1/V2) = T1/T2.
T2 = 293 K/1.14 = 257 K. |
| Details of the calculation:
None |