PHY 1361

**First Hour Exam**

February 7, 2001

**Statistics:**

High: 98%Mean: 75%

Low: 55%

Pace yourself! Do ** not** spend too much
time on any one particular problem or question. There are

Read through the entire exam first. If you find one of the questions or problems is taking too long,

Problem 3 is clearly the longest and most involved; allot a little more time for Problem 3.

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 to reappear on the final exam.

How much will the 25.0 m section expand as the temperature changes from 10.0

For concrete, the coefficient of thermal expansion is

a = 12 x 10

The temperature change, T,
is 40 C^{o}; T
= 40 C^{o}

Now we can put all the pieces together and evaluate the expansion,
L
;

Since each concrete section will expand by 1.2 cm, there must be 1.2 cm between each section so they do not run into each other and cause compressive stresses.

If all the ice melts, determine the final
temperature of the system.

If the final temperature is 0^{o}C, determine how much ice
melts.

Qwater = c

Q

Q

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 T_{f} >
0^{o}C

Be sure to check for consistency with our final answer and this
assumption.

This ice is initially at 0^{o}C so heat that it gains
immediatly starts to melt the ice. Remember, m_{ice} =
m_{2} = 300 g = 0.300 kg. m_{2} is the mass of the
ice after it has melted and is water at 0^{o}C.

Qice = m

Q

+ (4186 J/kg-C

Q

Q

Q

[2,093 T

(2,093 + 1,256) T

3,349 T

3,349 T

T

Is that "believeable" or "reasonable"? It is
between the lowest and highest initial temperature so that means it
is, at least, "possible". No, this is **colder** than the ice! Our
**assumption** was **wrong!** We must make another
assumption.

**Assumption:** T_{f} = 0 and only **part** of the ice
melts. Now we must determine **how much** of the ice
**melts.**

Qwater = c

Q

Q

Q

This ice is initially at 0^{o}C so heat
that it gains immediatly starts to melt the ice. Remember, this
time when some of the ice melts it does not change its temperature
since T_{f} = 0^{o}C. We are now looking
for M_{melt} .

Qice = M

Q

Q

- 62,790 J + (M

(M

M

M

M

As always, ask if this is "believeable" or "reasonable" -- or consistent with our assumptions. Yes, this is consistent with our assumption of only part of the ice melting. This amount of ice that melts is, indeed, less than our total, initial amount of ice. So this answer seems plausible.

First, its pressure is tripled under constant volume.

It then expands isothermally to its original pressure and

finally is compressed isobarically to its original volume.

(a) Draw a PV diagram of this cycle.

Here is a sketch:

(b) What is the temperature of the isotherm?

P V / T = constant

P

V

P

T

T

T

(c) What is the volume at the end of this isothermal expansion?

P V / T = constant

P

T

P

P

V

V

**V _{C} = 12 L**

(d) How much work is done in each process of this cycle?

W

W

W

W

P V = n R T

n R = P V / T

n R = (12 atm) (1 L) / 900 K

n R = 0.0133 L-atm/K

W

W

W

W

W

W

W

W

W

(e) What is the net work done during this cycle?

W

**4. (22.7)** An engine absorbs 1 600 J from a
hot reservoir and expels 1 000 J to a cold reservoir in each
cycle.

(a) What is the efficiency of the engine?

e = 1 - [ 1 000 J/1 600 J]

e = 1 - 0.625

e = 0.375

e = 37.5%

(b) What is the power output of the engine if each
cycle lasts for 0.30 s?

For each cycle,

W = 1 600 J - 1 000 J

W = 600 J

Then the power is

P = 600 J / 0.30 s

P = 2 000 W = 2 kW

**A. (19.Q14) **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. (20 “Opener")** 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?

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

**
C. (Q21.2)** 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?

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

**
D. (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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(C) 2001; Doug Davis; all rights reserved