PHY 1361
First Hour Exam

February 6, 2002

Statistics:

High: 92%

Mean: 76%

Low: 46%

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Entropy is important. In picking and choosing topics for questions, I did not choose an entropy question for this hour exam. I expect ideas of entropy may reappear on the final exam.


1. (19.27) At 20.0oC, an aluminum ring has an inner diameter of 5.000 0 cm and a brass rod has a diameter of 5.050 0 cm.
From Table 19.2, p 588, we find that the coefficient of thermal expansion for aluminum is = 24 x 10 6(Co) - 1 and the coefficient of thermal expansion for brass is = 19 x 10 - 6 (Co) - 1
To what temperature must the ring be heated so that it will just slip over the rod?


We need to increase the diameter of the aluminum ring from

Lo = 5.000 0 cm (or Lo = 5 x 10 - 2 m) to 5.050 0 cm.

That is an increase of

L = 0.050 cm = 5 x 10 - 4 m.
L = a Lo T

T = [L ] / [ a Lo ]
T = [ 5 x 10 - 4 m ] / [ ( 24 x 10 - 6 (Co) - 1 ) ( 5 x 10 - 2 m )]
T = 417 Co
Tf = 437oC

2. (20.***) A calorimeter contains 500 mL of water at 30oC and 300 g of ice at 0oC.
Determine thefinal temperature of the system or how much ice melts.
That is, . . .
If all the ice melts, determine the final temperature of the system.
If the final temperature is 0oC, determine how much ice melts.

Be sure to take care of a change of state -- due to the melting of the ice -- as well as a temperature change.

Qwater + Qice = 0
Qwater = cw mw T
Qwater = (4186 J/kg-Co) (0.500 kg) (Tf - 30oC)
Qwater = 2093 Tf (J/Co) - 62,790 J

Now, how much heat is gained by the ice? Does all of the ice melt? We do not know. So we must make an assumption -- or a guess. We then have to see if our result is consistent with that assumption.

Assumption: All the ice melts and Tf > 0oC
Be sure to check for consistency with our final answer and this assumption.

This ice is initially at 0oC so heat that it gains immediatly starts to melt the ice.  Remember, mice = m2 = 300 g = 0.300 kg. m2 is the mass of the ice after it has melted and is water at 0oC. After it melts, it is just ordinary water so the specific heat c is that of water, c = 4186 J/kg-Co.

Qice = Qmelt + Qraise
Qice = - [ mice Lf + c m2 Tf ]

The minus sign just indicates that this is heat that is absorbed by the 0.300 kg of ice or water

Qice = - [ (0.300 kg)(3.33 x 105 J/kg) + (4186 J/kg-Co) (0.300 kg) (Tf - 0oC) ]
Qice = - [(0.300)(3.33 x 105 J) + (4186 J/Co) (0.300 kg) Tf]
Qice = - [ 99,900 J + 1,256 Tf (J/Co) ]
Qwater + Qice = 0
[2093 Tf (J/Co) - 62,790 J] - [ 99,900 J + 1,256 Tf (J/Co)] = 0
(2093 - 1,256) Tf (J/Co) - (62,790 + 99,900) J= 0
837 Tf (J/Co) - 162,690 J= 0

837 Tf (J/Co) = 162,690 J
Tf = (162,690 / 837) oC
Tf = 194oC

While this is consistent with our assumption, as I stated it, it is still nonesense! We can not have a final temperature that is greater than the highest initial temperature. So our assumption is wrong and we must try a different assumption:

Assumption 2: Tf = 0 and only part of the ice melts.

Then we will find how much of the ice melts

Qwater = - 62,790 J
Qice = mmelt Lf
Qice + Qwater = 0
mmelt Lf + ( - 62,790 J) = 0
mmelt Lf = 62,790 J
mmelt (3.33 x 105 J/kg) = 62,790 J
mmelt = (62,790 J)/(3.33 x 105 J/kg)
mmelt = [62,790 /(3.33 x 105)] kg
mmelt = 0.189 kg
(Tf = 0oC)

And this value is, indeed, consistent with our assumption and with the system. 0.189 kg of ice melts. 0.111 kg of ice remains as ice in the water at zero degrees Celcius!


3. (20.39) An ideal gas undergoes a thermodynamic process that consists of two isobaric and two isothermal steps as shown here:


Give all your answers in terms only of the pressures P1 and P2 and the volumes V1 and V2 and other constants like n, R, k. Do not use T in your answers.
In terms of the pressures P1 and P2 and the volumes V1 and V2;
Find the two temperatures of the two isotherms, T1 and T2, and
Find the work done along the path from A to B, WAB.

Look at state A, where V = VA = V1 and P = P1 and T = T1.
There, the ideal gas law

P V = n R T

becomes

P1 V1 = n R T1

or

TAB = T1 = P1 V1 / n R

Likewise, T2, the temperature along the isotherm from C to D, is

TCD = T2 = P2 V2 / n R


The work done along an isotherm is

Wisotherm = n R T ln[Vf/Vi]

For our path from A to B, this means

WAB = n R T ln[VB/VA]

We know VA = V1. But what is VB in terms of the given state variables? For that, we will go back to the Ideal Gas Law,

P V = n R T
PB VB = n R TB
P2 VB = n R T1
P2 VB = n R ( P1 V1 / n R ) = P1 V1
VB = (P1 / P2) V1

Now we can complete the expression for WAB,

WAB = n R T1 ln[VB/VA]
WAB = n R T1 ln[{(P1 / P2) V1}/V1]
WAB = n R T1 ln[P1 / P2]

But we know

T1 = P1 V1 / n R
WAB = n R (P1 V1 / n R) ln[P1 / P2]
WAB = (P1 V1 ) ln[P1 / P2]


4. (21.13) Calculate the change in internal energy of 3.0 moles of an ideal gas when its temperature is increased by 2.0 K.

For an ideal gas, we have

U = (3/2) n R T
U = (3/2) n R T
U = (3/2) (3 moles) (8.31 J/mole-K) (2.0 K)
U = 74.8 J


5. (22.2) A heat engine performs 200 J of work in each cycle and has an efficiency of 30%. For each cycle, how much thermal energy is

(a) absorbed and

Qh = W/e
Qh = 200 J/0.30
Qh = 667 J



(b) expelled? 

1 - [Qc/Qh] = e = 0.30
Qc/Qh = 1 - 0.30
Qc/Qh = 0.70
Qc = Qh (0.70)
Qc = (667 J) (0.70)
Qc = 467 J

An easier, more direct method to find the heat expelled would be to use the First Law of Thermodynamics or the Work-Energy Theorem,

W = Qh - Qc

Qc = Qh - W

Qc = 667 J - 200 J

Qc = 467 J


Conceptual Questions —
A.
Metal lids on glass jars can often be loosened by running them under hot water. How is this possible?

19.Q17: Metal lids on glass jars can often be loosened by running them under hot water. How is this possible?

The coefficient of thermal expansion is much greater for metals than for glass.Heating a metal lid on a glass jar will cause the metal lid to expand far more than the glass. This should allow you to unscrew the metal lid from the glass jar.


B.
Why can you get "burned" by biting into the hot CHEESE on a steaming pizza far easier than by biting into the CRUST of the pizza?

Now we can also answer the "Puzzler" question about the pizza.

Cheese is primarily water with a large heat capacity while the crust has very little water and a very small heat capacity.

C.
One container is filled with helium gas and another with argon gas. If both containers are at the same temperature, which molecules have the higher rms speed?

Q21.2 One container is filled with helium gas and another with argon gas. If both containers areat the same temperature, which molecules have the higher rms speed?

At the same temperature, the average value of the translational kinetic energy [(1/2) m <v2>] will be the same for both helium and argon. Argon has a larger mass so it will have a smaller value of <v2> or vrms. Helium has a smaller mass so it will have a larger value of <v2> or vrms.

D.
Use the first law of thermodynamics to explain why the total energy of an isolated system is always constant.

Q22.3 Use the first law of thermodynamics to explain why the total energy of an isolated system is always constant.

The first law of thermodynamics is just a restatement of Energy Conservation. If a system is isolated no work is done on it and no heat is transferred to or from it. Energy may be transferred from one part of the system to another but the total energy remains constant.

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