PHY 1361

**First Hour Exam**

February 6, 2002

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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.

From Table 19.2, p 588, we find that the coefficient of thermal expansion for aluminum is = 24 x 10

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

L_{o} = 5.000 0 cm (or L_{o} =
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 L_{o}
T

T
= [L ] / [ a L_{o} ]

T = [ 5 x 10^{
- 4 }m ] / [ ( 24 x 10^{ - 6} (C^{o})^{ - 1 })
( 5 x 10 ^{- 2 }m )]

T = 417 C^{o}

T_{f} = 437^{o}C

T = [ 5 x 10

T = 417 C

T

**2. (20.***) **A calorimeter contains 500 mL of water at 30^{o}C and 300 g of ice at 0^{o}C.

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 0^{o}C, 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.

Q_{water} + Q_{ice} = 0

Q_{water} = c_{w} m_{w} T

Q_{water} = (4186 J/kg-C^{o}) (0.500 kg) (T_{f} - 30^{o}C)

Q_{water} = 2093 T_{f} (J/C^{o}) - 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 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} = 25 g = 0.025
kg. m_{2} is the mass of the ice after it has melted and is water at
0^{o}C.

Q_{ice} = Q_{melt} + Q_{raise}

Q_{ice} = m_{ice} L_{f} + c m_{2} T

Q_{ice} = (0.025 kg)(3.33 x 10^{5} J/kg) + (4186 J/kg-C^{o})
(0.025 kg) (T_{f} - 0^{o}C)

Q_{ice} = (0.025)(3.33 x 10^{5} J) + (4186 J/C^{o})
(0.025 kg) T_{f}

Q_{ice} = 8,325 J + 105 T_{f} (J/C^{o})

Q_{water} + Q_{ice} = 0

[2093 T_{f} (J/C^{o}) - 62,790 J] + [ 8,325 J + 10^{5}
T_{f} (J/C^{o})] = 0

(2093 + 10^{5}) Tf (J/C^{o}) - (62,790 - 8,325) J= 0

2198 T_{f} (J/C^{o}) - 54,465 J= 0

2198 T_{f} (J/C^{o}) = 54,465 J

T_{f} = (54,465 / 2198) ^{o}C

T_{f} = 24.8^{o}C

**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 P_{1} and
P_{2} and the volumes V_{1} and V_{2} and other constants
like n, R, k. Do **not** use T in your answers.

In terms of the pressures P_{1} and P_{2} and the volumes V_{1}
and V_{2};

Find the two temperatures of the two isotherms, T_{1} and T_{2},
and

Find the work done along the path from A to B, W_{AB}.

Look at state A, where V = V_{A} = V_{1} and P = P_{1}
and T = T_{1}.

There, the ideal gas law

P V = n R T

becomes

P_{1} V_{1} = n R T_{1}

or

T_{AB} = T_{1} = P_{1} V_{1} /
n R

Likewise, T_{2}, the temperature along the isotherm from C to D, is

T_{CD} = T_{2} = P_{2} V_{2} /
n R

The work done along an isotherm is

W_{isotherm} = n R T ln[V_{f}/V_{i}]

For our path from A to B, this means

W_{AB} = n R T ln[V_{B}/V_{A}]

We know V_{A} = V_{1}. But what is V_{B} in terms of
the given state variables? For that, we will go back to the Ideal Gas Law,

P V = n R T

P_{B }V_{B} = n R T_{B}

P_{2 }V_{B} = n R T_{1}

P_{2 }V_{B} = n R ( P_{1} V_{1} / n R ) = P_{1}
V_{1}

V_{B} = (P_{1} / P_{2}) V_{1}

Now we can complete the expression for W_{AB},

W_{AB} = n R T_{1} ln[V_{B}/V_{A}]

W_{AB} = n R T_{1} ln[{(P_{1} / P_{2}) V_{1}}/V_{1}]

W_{AB} = n R T_{1} ln[P_{1} / P_{2}]

But we know T_{1} = P_{1} V_{1} / n R

W_{AB} = n R (P_{1} V_{1} / n R) ln[P_{1} /
P_{2}]

W_{AB} = (P_{1} V_{1} ) ln[P_{1} / P_{2}]

**
**

**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

DU = (3/2) n R DT

DU = (3/2) (3 moles) (8.31 J/mole-K) (2.0 K)

DU = 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

Q_{h} = W/e

Q_{h} = 200 J/0.30

Q_{h} = 667 J

(b) expelled?

1 - [Q_{c}/Q_{h}] = e = 0.30

Q_{c}/Q_{h} = 1 - 0.30

Q_{c}/Q_{h} = 0.70

Q_{c} = Q_{h} (0.70)

Q_{c} = (667 J) (0.70)

Q_{c} = 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.

At the same temperature, the average value of the translational kinetic energy [(1/2) m <v

D.

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.