Download Name: SOLUTIONS Physics 240, Exam #1 Sept. 24 2015 (4:15

Name: SOLUTIONS
Physics 240, Exam #1
Sept. 24 2015 (4:15-5:35)
Instructions: Complete all questions as best you can and show all of your work. If you just
write down an answer without explaining how you got it, you will not receive any points.
Do not use any device to retrieve text, equations, or algorithms during the exam. You can
use a scientific calculator but no tablets nor smartphones.
You may also use one handwritten index card (3” x 5”) but do not consult with or discuss
this exam with anybody else. Good luck, and budget enough time (8 problems)
Use the space below if you run out of room to answer the questions on any upcoming sheet.
-------------------------------------------------------------------------------------------------------------------TOTAL POSSIBLE POINTS = 67 / 63 [because of bonus points]
1. CONCEPTUAL QUESTIONS (instead of math “work” just briefly justify your answer.)
6 + 4 POINTS TOTAL
(a.) An inventor has allegedly created a material called “starlite” that when painted on an
egg can prevent it from cooking even when exposed to 1,000-degree Celsius heat. What
material property you learned this semester can be responsible for this feat? Would it be
high or low?
(b.) In the example of the Japanese bees killing the hornet by surrounding it, from the
chapter on the first law of thermodynamics, how did they accomplish that? Which equation
was primarily used to demonstrate the fact? Would the property in (a.) prolong its life?
(c.) In the kinetic theory chapter, why did the bottle of champagne create fog upon opening?
(d.) BONUS QUESTION: What is the ‘cute’ way of describing all 4 thermodynamic laws?
(2 points)
(a.) specific heat or heat capacity OR thermal conductivity; high, high, low, respectively
(3 points)
(b.) raised their temperature and thus hornet’s too without harming themselves, but raising
temperature above maximum for hornet, and heat not transmitted/emitted away fast
enough. Equation used was P=sigma*e*A*T^4 (blackbody radiation). Yes, higher specific
heat would mean the hornet’s temperature would go up a smaller delta-T for a given Q in.
(1 point)
(c.) a sudden increase in volume leading to drop in temperature to below the boiling point
or sudden drop in pressure below the vapor pressure, causing condensation (both accepted)
(4 BONUS points)
(d.) You must play the game (OR, there is a game); You can’t win; You can’t break even
(OR, You must lose); You can’t quit (OR, you can’t get out of the game)
(in this order: 0-3)
2. Imagine that you have a piece of ice with mass m = 720 g at −10°C.
(a.) How much thermal energy must be absorbed by it to take it to a liquid state at 15°C?
(b.) If we supply it with a total energy of only 210 kJ (as heat), what then are the final state
and temperature of this system (how much water and how much ice)?
10 POINTS TOTAL
Use 4,190 J/(kg-K) as the specific heat of water and 2,220 for ice. Heat of fusion = 333 kJ/kg.
(5 points)
(a.) SOLUTION The first Key Idea is that the heating process is accomplished in three steps.
Step 1 The Key Idea here is that the ice cannot melt at a temperature below the freezing point—
so initially, any thermal energy transferred to the ice can only increase the temperature of the ice.
The energy Q1 needed to increase that temperature from the initial value Ti = − 10°C to a final
value Tf = 0°C (so that the ice can then melt) is given by Q = mc ∆T. Using the specific heat of
ice cice as provided above gives us
Q1 =c_ice m(Tf -Ti)
= (2220 J/(kg × °C))(0.720 kg)[0°C - (-10° C)] =15,984 J ~ 15.98 kJ.
Step 2. The next Key Idea is that the temperature cannot increase from 0°C until all the ice
melts—so any energy transferred to the ice due to heating now can only change ice to liquid
water. The thermal energy Q2 needed to melt all the ice is given by Q = L m. Here L is the heat of
fusion LF, with the value given above. We find
Q2 = LFm = (333 kJ/kg)(0.720 kg) = 239.8 kJ.
Step 3. Now we have liquid water at 0°C. The next Key Idea is that the energy transferred to the
liquid water during heating now can only increase the temperature of the liquid water. The Q3
needed to increase the temperature of the water from the initial value Ti = 0°C to the final value
Tf = 15°C is given by the following (with the specific heat of liquid water cliq):
Q3 =cliqm(Tf -Ti) = (4190 J/(kg × °C))(0.720 kg)(15°C - 0°C) = 45,252 J ~ 14.25 kJ.
The total required thermal energy transfer Q
steps:
Q
tot
tot
is the sum of the amounts required in the three
=Q1+Q2+Q3
= 15.98 kJ + 239.8 kJ + 45.25 kJ ~ 300 kJ. (Answer)
Note that the energy transfer required to melt the ice is much greater than the energy transfer
required to raise the temperature of either the ice or the liquid water.
(5 points)
(b.) SOLUTION From step 1 in part (a.), we know that 15.98 kJ is needed to raise the
rem
temperature of the ice to the melting point. The remaining energy required Q
is then 210 kJ –
15.98 kJ, or about 194 kJ. From step 2, we can see that this amount of energy is insufficient to
melt all the ice. Then this Key Idea becomes important: Because the melting of the ice is
incomplete, we must end up with a mixture of ice and liquid; the temperature of the mixture must
be the freezing point, 0°C.
We can find the mass m of ice that is melted by the available energy Q
LF:
m=
Q
rem
/ LF =
194kJ /
rem
by using Q = L m with
333 kJ/kg = 0.583 kg ~ 580 g.
Thus, the mass of the ice that remains is 720 g – 580 g, or 140 g, and we have 580 g water and
140 g ice at 0°C. So the final state is mixed, with 81% water and 19% ice by mass, with final
temperature freezing.
3. On a hot day in Las Vegas, an oil trucker loaded 37,000 L of diesel fuel. She encountered
cold weather on the way to Payson, Utah, where the temperature was 23.0 K lower than in
Las Vegas, and where she delivered her entire load. The coefficient of volume expansion for
diesel fuel is 9.5 x 10-4/°C. (Note that the thermal expansion of the steel tank is not relevant.)
5 POINTS TOTAL
(a.) How many liters did she deliver?
(b.) Who paid for the “missing” diesel fuel?
(3 points)
(a.) SOLUTION The Key Idea here is that the volume of the diesel fuel depends directly on the
temperature. Thus, because the temperature decreased, the volume of the fuel did also. From the
first reading assignment and lecture notes, the volume change is
ΔV =V β ΔT = (37,000 L)(9.50 x 10
-4
/°C)(-23.0°C) = -808 L.
Thus, the amount delivered was Vdel = V + ΔV = 37,000 L - 808 L = 36,192 L. (Answer)
(2 points)
(b.) No one, this is a trick question, because it is the same number of gas atoms before and after
thermal contraction. However, the reality is that gas is sold by unit volume not mass or energy
content, so unless the temperature changes back, the customer in Utah paid for more liters than
they got and thus they paid for the “missing” fuel.
4. A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the
water (and thus the helium) undergoes a temperature increase ΔT of 20.0 C° at constant
pressure. As a result, the bubble expands. The helium is monatomic and ideal.
(a.) How much thermal energy is added to the helium during the increase and expansion?
(b.) What is the change ΔEint in internal energy of the helium during temperature increase?
(c.) How much work W is done by the helium as it expands against the pressure of the
surrounding water during the temperature increase? What if the water cannot expand out?
Solve part (c.) two different ways: with and without using the first law of thermodynamics.
10 POINTS TOTAL
Ideal gas constant R = 8.31 J/(mol-K).
(3 points)
(a.) SOLUTION One Key Idea here is that the thermal energy transferred Q is related to the
temperature change ΔT by the molar specific heat of the gas. Because the pressure P is held
constant during the addition of energy, we use the molar specific heat at constant pressure CP and
equation
Q = n CP ΔT
to find Q. To evaluate CP we recall that for any ideal gas, CP = CV + R. Also, we know that for
any monatomic gas (like helium) with only 3 degrees of freedom in motion,
CV = 3/2 R . Thus, that equation above gives us
Q = n (CV +R) ΔT = n (1.5 R + R) ΔT = n (2.5 R) ΔT
= (5.00 mol)(2.5)(8.31 J/(mol × K))(20.0 C°) = 2077.5 J ~ 2080 J. (Answer)
(3 points)
(b.) SOLUTION Because the bubble expands, this is not a constant-volume process. However,
int
the helium is nonetheless confined (to the bubble). Thus, a Key Idea here is that the change ΔE
is the same as would occur in a constant-volume process with the same temperature change ΔT.
int
We can easily find the constant-volume change ΔE with:
int
ΔE = n CV ΔT = n (1.5 R) ΔT
= (5.00mol)(1.5)(8.31J/(mol×K))(20.0 C°) = 1246.5 J ~ 1250 J. (Answer) OR skip to c. then
come back and use 1st law
(4 points)
(c.) SOLUTION One Key Idea here is that the work done by any gas expanding against the
pressure from its environment is given by integrating P dV. When the pressure is constant (as
here), we can simplify that to W = P ΔV. When the gas is ideal (as here) and the number of moles
is unchanging, we can use the ideal gas law (PV = nRT) to write P ΔV = nR ΔT. We end up with
W = nRΔT = (5.00 mol)(8.31 J/(mol × K))(20.0 C°) = 831 J. (Answer)
Because we happen to know Q and ΔE
int
, we can work this problem another way. The Key Idea
now is that we can account for the energy changes of the gas with the first law of
thermodynamics, writing
W by helium on water = Q - ΔE
int
= 2077.5 J - 1246.5 J = 831 J. (Answer)
Note that during the temperature increase, only a portion (1250 J) of the thermal energy (2080 J)
that is transferred to the helium goes to increasing the internal energy of the helium and thus the
temperature of the helium. The rest (831 J) is transferred out of the helium as work that the
helium does during the expansion. If the water were frozen, it would not allow that expansion.
Then the same temperature increase of 20.0 C° would require only 1250 J of energy, because no
work would be done by the helium.
If the surrounding water has nowhere to expand (like ice example above, or very rigid, closed
container) then its pressure must increase as the bubble tries to push the liquid out of the way but
there is nowhere to go (pressure is no longer constant, so can’t be pulled out of work integral any
more.) As long as the bubble is changing size however, that means the gas inside of it is doing
work due to change in volume, so W is integral of P x dV.
5. One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature T
fixed at 310 K from an initial volume Vi of 12 L to a final volume Vf of 19 L.
(a.) How much work is done by the gas during the expansion?
(b.) What would be the final temperature if the gas had expanded adiabatically to this same
final volume? Oxygen (O2) is diatomic and at this temperature has rotation but not
oscillation.
(c.) What would be the final temperature and pressure if, instead, the gas had expanded
freely to the new volume, from an initial pressure of 2.0 Pa [N/m2]?
(d.) The cylinder containing 12 L of oxygen is instead at 20°C and 15 atm. The temperature
is raised to 35°C, and the volume reduced to 8.5 L. What is the final pressure (in unit atm)?
10 POINTS TOTAL
0 degrees Celsius = 273.15 Kelvin, and 1 Celsius degree is the same “size” as one Kelvin unit.
(3 points)
(a.) SOLUTION The Key Idea is this: Generally we find the work by integrating the gas pressure
with respect to the gas volume. However, because the gas here is ideal (P V = n R T) and the
expansion is isothermal (T not in integral), that integration leads to
W = n R T ln (Vf / Vi)
= (1 mol)(8.31 J/(mol × K))(310 K) ln (19 L / 12 L) = 1180 J. (Answer)
(3 points)
(b.) SOLUTION The Key Ideas here are as follows:
1. When a gas expands against the pressure of its environment, it must do work.
2. When the process is adiabatic (no thermal energy is transferred as heat), then the energy
required for the work can come only from the internal energy of the gas.
3. Because the internal energy decreases, the temperature T must also decrease. We can relate the
initial and final temperatures and volumes for an adiabatic process with:
Ti Vi
γ-1
=Tf Vf
γ-1
. (Equation 19-62)
Because the molecules are diatomic and have rotation but not oscillation, we can take the molar
specific heats from Table 19-3. Thus,
γ (gamma)= Cp / Cv = (7/2)R / (5/2)R = 1.40
Solving Eq. 19-62 for Tf then inserting known data yields
Tf = TiVi gamma-1 / Vf gamma-1 = (310 K)(12 L)^(1.40-1) / (19 L)^(1.40-1)
= (310 K) ( 12 / 19 ) ^ 0.40 = 258 K. (Answer)
(2 points)
(c.) SOLUTION Here the Key Idea is that the temperature does not change in a free expansion:
Tf = Ti = 310 K. (Answer)
We find the new pressure using Eq. 19-61 (Pi Vi = Pf Vf, for constant T), which gives us
Pf = Pi (Vi / Vf) = (2.0 Pa) (12 L / 19 L) = 1.3 Pa. (Answer)
(2 points)
(d.) SOLUTION The Key Idea here is that, because the gas is ideal, its pressure, volume,
temperature, and number of moles are related by the ideal gas law, both in the initial state i and in
the final state f (after the changes). Thus, we can write PiVi = nRTi and PfVf = nRTf. Dividing
the second equation by the first equation and solving for Pf yields
Pf =Pi Tf Vi / TiV f (19-9)
Note here that if we converted the given initial and final volumes from liters to SI units of cubic
meters, the multiplying conversion factors would cancel out of Eq. 19-9. The same would be true
for conversion factors that convert the pressures from atmospheres to the more accepted SI unit of
pascals. However, to convert the given temperatures to kelvins requires the addition of an amount
that would not cancel and thus must be included. Hence, we must write
Ti = (273 + 20) K = 293 K
and
Tf = (273 + 35) K = 308 K.
Inserting the given data into Eq. 19-9 then yields
Pf = (15 atm)(308 K)(12 L) / [(293 K)(8.5 L)] = 22 atm. (Answer)
6. Imagine an ideal Carnot engine that operates between the temperatures TH = 850 K and
TL = 300 K. The engine performs 1,200 J of work each cycle, with the duration of each cycle
being 0.25 seconds.
7 POINTS TOTAL
(a.) What is the efficiency of this engine?
(b.) What is the average power of this engine?
(c.) How much thermal energy QH is extracted from the high-temperature reservoir per
cycle?
(d.) How much thermal energy QL is delivered to the low-temperature reservoir per cycle?
(e.) What entropy change is associated with the energy transfer to the working substance
from the high-temperature reservoir? From the working substance to the low-temperature
reservoir?
(f.) An inventor claims to have constructed a heat engine that has an efficiency of 75% when
operated between the boiling and freezing points of H2O. Is it possible compared to Carnot?
(1 point)
(a.) SOLUTION The Key Idea here is that the efficiency ε of an ideal Carnot engine depends
only on the ratio TL/TH of the temperatures (in kelvins) of the thermal reservoirs to which it is
connected. Thus, from Eq. 20-11, we have
ε = 1- TL / TH = 1 - (300 K) / (850 K) = 0.647 ~ 65% (Answer)
(1 point)
(b.) SOLUTION Here the Key Idea is that the average power P of an engine is the ratio of the
work W it does per cycle to the time Δt that each cycle takes. For this Carnot engine, we find
P = W / t = 1200 J / 0.25 s = 4,800 W = 4.8 kW. (Answer)
(1 point)
(c.) SOLUTION Now the Key Idea is that, for any engine, including a Carnot engine, the
efficiency ε is the ratio of the work W that is done per cycle to the thermal energy QH that is
extracted from the high-temperature reservoir per cycle. This relation, ε = |W| / |QH| (Eq. 20-9),
gives us
QH = W / ε = 1200 J / 0.647 = 1855 J. (Answer)
(1 point)
(d.) SOLUTION The Key Idea here is that for a Carnot engine, the work W done per cycle is
equal to the difference in energy transfers |QH| − |QL|. (See Eq. 20-6.) Thus, we have
|QL| = |QH| − W = 1855 J − 1200 J = 655 J. (Answer)
(2 points)
(e.) SOLUTION The Key Idea here is that the entropy change ΔS during a transfer of thermal
energy Q at constant temperature T is given by Eq. 20-2 (ΔS = Q/T). Thus, for the transfer of
energy QH from the high-temperature reservoir at TH, we have
ΔSH = QH / TH = 1855 J / 850 K = +2.18 J/K (Answer)
For the transfer of energy QL to the low-temperature reservoir at TL, we have
ΔSL = QL / TL = -655 J / 300 K = -2.18 J/K. (Answer)
Note that the algebraic signs of the two thermal energy transfers are different. Note also that, as
Eq. 20-8 requires, the net entropy change of the working substance for one cycle (which is the
algebraic sum of the two quantities calculated above) is zero, the minimum allowed by the second
law of thermodynamics.
(1 point)
(f.) SOLUTION The Key Idea here is that the efficiency of a real engine (with its irreversible
processes and wasteful energy transfers) must be less than the efficiency of an ideal Carnot
engine operating between the same two temperatures. From Eq. 20-11, we find that the efficiency
of an ideal Carnot engine operating between the boiling and freezing points of water is
ε = 1- TL / TH = 1 - (0 + 273) K / (100 +273) K = 0.268 ~ 27%
Thus, the claimed efficiency of 75% for a real heat engine operating between the given
temperatures is impossible. (So, answer to question posed is no.)
7. One mole of nitrogen gas is confined to the left side of a container. You open a stopcock
in the middle and the volume of gas doubles. Calculate the change in entropy of the gas for
this irreversible process. (As always, treat the gas as ideal. Hint: Find a reversible process.)
5 POINTS TOTAL
(5 points)
SOLUTION We need two Key Ideas here. One is that we can determine the entropy change for
the irreversible process by calculating it for a reversible process that provides the same change in
volume. The other is that the temperature of the gas does not change in the free expansion. Thus,
the reversible process should be an isothermal expansion.
int
Since the internal energy of an ideal gas depends only on temperature, ΔE = 0 here and so Q =
W done by gas on environment from the first law. Combining this result with Eq. 20-14 gives
Q = W = n R T ln (Vf / Vi),
in which n is the number of moles of gas present. From Eq. 20-2 the entropy change for this
isothermal and reversible process is
ΔSrev = Q / T = n R T ln ( Vf / Vi ) / T = n R ln ( Vf / Vi ).
Substituting n =1.00 mol and Vf / Vi = 2, we find
ΔS
rev
= n R ln ( Vf / Vi ) = (1.00 mol) (8.31J/(mol×K)) (ln 2)
= +5.76 J/K.
Thus, the entropy change for the free expansion (and for all other processes that connect the
initial and final states being considered here) is
ΔS
irrev
= ΔS
rev
= + 5.76 J / K. (Answer)
ΔS is positive, so the entropy increases, in accordance with the entropy postulate of Section 20-1.
10 POINTS TOTAL
8. The figure below (left) shows two identical copper blocks of mass m = 1.5 kg: block L at
temperature TiL = 60°C and block R at temperature TiR = 20°C. The blocks are in a
thermally insulated box and are separated by an insulating shutter. When we lift the
shutter, the blocks eventually come to the equilibrium temperature Tf = 40°C (right). What
is the net entropy change of the two-block system during this irreversible process? (Note the
specific heat of copper is 386 J per kg per Kelvin.)
Use the next page as necessary for additional work on this, or any of the other exam problems.
(10 points)
SOLUTION The Key Idea here is that to calculate the entropy change, we must find a reversible
process that takes the system from the same initial state at the left to the same final state at the
rev
right. We can calculate the net entropy change ΔS
of the reversible process using Eq. 20-1, and
rev
then the entropy change for the irreversible process is equal to ΔS . For such a reversible
process we need a thermal reservoir whose temperature can be changed slowly (say, by turning a
knob). We then take the blocks through the following 2 steps, illustrated in the figure below.
FIGURE 20-6 The blocks of the figure up top can proceed
from their initial state to their final state in a reversible way if
we use a reservoir with a controllable temperature (a) to
transfer thermal energy reversibly from block L and (b) to
transfer thermal energy reversibly to block R.
Step 1. With the reservoir’s temperature set at 60°C, put block L on the reservoir. (Since block
and reservoir are at the same temperature, they are already in thermal equilibrium.) Then slowly
lower the temperature of the reservoir and the block to 40°C. As the block’s temperature changes
by each increment dT during this process, thermal energy dQ is transferred from the block to the
reservoir. Using Eq. 19-5, we can write this transferred energy as dQ = mc dT, where c is the
specific heat of copper. According to Eq. 20-1, the entropy change ΔSL of block L during the full
temperature change from initial temperature TiL (= 60 °C = 333 K) to final temperature Tf (= 40
°C = 313 K) is
ΔSL=integral(dQ/T from i to f)=integral(mc dT/T, TiL, Tf) = mc*integral(dT/T,TiL,Tf)
= m c ln ( Tf / TiL )
Inserting the given data yields
ΔSL = (1.5 kg) (386 J/[kg×K]) ln ( 313 K / 333 K )
= - 35.86 J / K.
Step 2: With the reservoir’s temperature now set at 20°C, put block R on the reservoir. Then
slowly raise the temperature of the reservoir and the block to 40°C. With the same reasoning used
to find ΔSL, you can show that the entropy change ΔSR of block R during this process is
ΔSR = (1.5 kg) (386 J/[kg×K]) ln ( 313 K / 293 K )
= + 38.23 J / K.
The net entropy change ΔS
process is then
ΔS
rev
rev
of the two-block system undergoing this two-step reversible
= ΔSL + ΔSR = -35.85 J/K + 38.23 J/K = 2.4 J/K. (Answer)
irrev
Thus, the net entropy change ΔS
for the two-block system undergoing the actual irreversible
irrev
rev
process is ΔS
=ΔS
= PLUS 2.4 J/K. This result is positive, in accordance with the entropy
postulate of Section 20-1.