# Guide for Giancoli Chapter 19

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After two chapters that were "sparse" in new fundamental laws and problems, we now encounter a chapter that is jam-packed with them. We will cover all sections, and will probably spend as much time on this chapter as on the two previous chapters combined.

Section 1 establishes "heat" as a transfer of energy. Thus heat becomes very similar to work, except that it is a different type of transfer: it is a transfer of energy "because of a difference in temperature." (p. 497) [Work, you should remember, involves a force and a movement over a macroscopic distance.] Thus, the basic unit of heat is the JOULE, just as with all types of energy quantities. The conversion from Joules to "calories" or "kilocalories," which are older (but still used) units for heat, is discussed. By the way, a "food" Calorie is a "kilocalorie" or kcal = 4186 J.

Section 2 defines "internal energy" which will be an extremely important concept for us. Be sure you clearly understand "heat," "temperature," and "internal energy." This is more than just vocabulary. Section 3 defines specific heat (Eqn. 19-2), which also gives the quantitative relation between heat and temperature change. This equation only works so long as the substance does not change states (i.e., it remains a solid, or liquid, or gas throughout the change in T). Section 4 then illustrates how to solve problems when heat is exchanged between objects (while NO work is done). The basic law is simply the conservation of energy: if no work is done and no changes of state take place, the sum of all heat transfers must balance. Thus Eqn. 19-2 can be used for each object and the sum of heat gains for all objects must total zero: $\Sigma \$Qj = mjcj($\Delta \$T)j = 0 (where a negative Q means a heat LOSS, since in that case $\Delta \$T < 0 for that particular object).

What, then, is different when we take up cases where there are changes of state? Section 5 addresses this, and defines "latent" heats of fusion and vaporization. Armed with these ideas and the data in Tables 19-1 and 19-2, we can work most any problems that involve only heat transfer ("only" means no work is done). The Examples in Sections 3, 4, and 5 are worth examining.

Sections 6, 7, and 8 take up cases where both heat transfer and work are involved. For the first time we write a conservation of energy equation that includes heat: it is called the "First Law of Thermodynamics:" $\Delta \$U = Q - W. The sign convention is crucial. The vocabulary is also important ["isothermal" => T is constant and etc.]. You must be able to interpret work as the area under the curve on a PV-diagram (see p. 509). The "molar" specific heats which are defined in Section 8 differ from "ordinary" specific heats only in that C is defined as specific heat "per mole" rather than "per kilogram" as previously. In the formula for Q, then, we use n, the number of moles, rather than m, the mass in kilograms. I will not ask you to convert on exams, just read carefully to see which specific heat is given in a particular problem. I will spend a good bit of time on Section 8, and will expect you to know what is meant by the "equipartition of energy" and "degrees of freedom."

Section 9 treats one especially important case: the adiabatic (Q = 0) expansion (or contraction) of a gas, and derives an equation (Eqn. 19-15, p. 514) that will occasionally be extremely useful. I will not expect you to know the derivation, but will expect you to be able to use the equation.

Section 10 classifies the three ways that heat transfer takes place: conduction, convection, and radiation. I'll expect you to know what each means. There are some problems involving conduction (Eqn. 19-16a on p. 515), and I'll do one or two in class. I want you to know what the "R-factor" is for heat insulation(see p. 517; it has nothing to do with the gas constant!). The Stefan-Boltzmann law for radiation (Eqn. 19-17) is useful for some easy plug-in type problems. Mainly, note the very strong dependence of the total radiated energy (per time, per surface area) as the temperature of an object increases! Doubling T causes the radiated energy to increase by a factor of 24 = 16!

## Recommended Questions/Problems for Chapter 19

Questions: 4, 8, 10, 13, 19, 31, 35

Section 1 Problems: 3, 6

Sections 2 & 3 Problems: 8, 11, 15, 16, 17

Section 5 Problems: 18, 19, 20, 24, 26

Sections 6 & 7 Problems: 31, 32, 33, 34, 35, 38

Section 8 Problems: 42, 44, 49

Section 9 Problems: 51, 53,

Section 10 Problems: 60, 62, 63 (see p. 519 for Solar Constant)

General Problems: None

## Selected Hints and Solutions

Q8 : In a liquid each molecule is "attached" to its neighbors by attractive forces. In order to escape the liquid and become "free" (i.e., a gas molecule) work must be done: a force must act on the molecule to "pull" it away from its attracting neighbors. This work requires energy; that is, a portion of the internal energy of the liquid must be "expended" as work. Thus the internal energy of the remaining liquid (and thus the temperature, which is a measure of the average kinetic energy per particle) decreases. Another factor is that it is the FAST molecules (whose KE is above the average for the liquid) that are lost, thus lowering the average of those left behind.

Q13 : Yes. If the process is isothermal (T does not change) then the internal energy does not change. If, therefore, 3700 J of energy leave as the gas does work, then 3700 J of heat have been added. If less were added, T would decrease; if more were added, T would increase.

Q19 : See Problem II(b) of the 2008 Test I

P6 : Energy is supplied by the heater at the rate of 350 Joules/second so the total supplied in a time $\Delta \$t will be 350$\Delta \$t Joules. The total energy required is Q = mc$\Delta \$T. The mass is 0.25 kg (because the density of water is 1.00 kg/liter and we have 1/4 liter of water), c is 4187 Joules/(kg-C) and $\Delta \$T is 50 C. So set Q = 350$\Delta \$t and find $\Delta \$t. The answer is 180 seconds.

P15 : See Problem I of the 2008 Test I which is quite similar. This is a standard "heat exchange" problem in which no work is done, various objects exchange heat, some warming and some cooling. If we add all the "Q's" (which are positive when heat is added, negative when heat leaves) we must get zero, since energy must be conserved.

In this problem we have three "objects:" the iron, the cup, and the glycerin. For each one the heat added will be given by Q = mc$\Delta \$T. The sum of the three must be zero. (Note that $\Delta \$T can be positive or negative; clearly if the sum of three numbers is zero, ONE of them - and possible two, but not all three - must be negative!) We know the masses, we can look up (Table 19-1, p. 499) the specific heats for iron and aluminum, and we are given the initial and final temperatures of all three. Thus the only unknown is the specific heat of glycerin.

P20 : Another heat exchange problem, but now some of the heat is used to change the STATE of some nitrogen from liquid to gas. This heat is given by QNitrogen = MVLV where MV is the amount of nitrogen vaporized and LV is the "latent heat of vaporization" of nitrogen (Table 19-2, p. 503). This value of heat is positive if the change-of-state is from liquid to gas, since heat must be "gained" by the nitrogen that escapes from the liquid and forms vapor. We STILL add up all the Q's and equate the total to zero (just as in the previous problem).

So: the ice cools from 0 C = 273 K to a final temperature of 77 K; we can get the specific heat of ice from Table 1, so we can compute the QIce (which will be negative) for the ice: QIce = MIcecIce$\Delta \$TIce. Our equation is simply QNitrogen + QIce = 0. We can compute the MV. The answer is 72 grams.

P31 : Note that the curved line is an "isotherm" along which T is constant. Since Tc = Ta, the change in internal energy between a and c is zero, and the first law then gives that: Qac = Wac.

Clearly Wac = Wab + Wbc. Now the work Wab = 0 since V is constant (no area underneath the vertical line ab). The work Wbc is just the area of the rectangle, which is easily computed (height x width) HOWEVER we need to express P in Pascals and V in cubic meters (1.0 m3 = 103 liters) so that the work computed is in Joules. Thus we have the work Wac, and this is also the heat Qac for this isothermal process.

So picture what happens: as the system moves from a to c, heat is added (internal energy rises) but work is done (internal energy falls) the net result being that the internal energy doesn't change. So clearly Q = W.

P38 : Pay close attention to the Figure (Fig. 19-32). In the past, versions of this problem have shown up on my tests frequently.

The system starts at a and ends up at c via two paths on the P-V diagram. Call these paths ac (the curved path), abc (a to c via b). We are given:

Wac = - 35 J

Qac = - 63 J

Wabc = - 54J

(a) We want Qabc. We know that $\Delta \$Uac must be the same for both paths, and $\Delta \$U = Q - W. Thus (Q - W)ac = (Q - W)abc which we can write as Qac - Wac = Qabc - Wabc. We know three of these, so we solve for Qabc = - 82 J.

(b) Work is area under the curve for any process. We know that Wab = Wabc (since NO work is done from b to c) and equals = - 54 J. If the pressure at c (and d) is half that at a (and b), the area under the curve cda (really just cd since there is NO area under da) is just half that under abc, but is POSITIVE. Thus Wcda = Wcd= + 27 J.

(c) From the given information and the first law: $\Delta \$Uac = - 28 J. So $\Delta \$Uca = + 28 J for ANY path. Thus Qcda - Wcda = +28 J. We just found Wcda in part (b), so Qcda = + 55 J.

(d) We have already found that $\Delta \$Uac = - 28 J. The task is to find Ua - Uc = $\Delta \$Uca = + 28 J.

(e) We are given that $\Delta \$Ucd = + 12 J and we know that $\Delta \$Ucda = + 28 J. Thus $\Delta \$Uda = + 16 J. Also, $\Delta \$U = Q - W for any path, and Wda = 0. Thus Qda = $\Delta \$Uda = + 16 J.

Comment: The high number of subscripts make this problem tedious. It is absolutely vital to WRITE DOWN things that you know, and to refer FREQUENTLY to the figure. If you can get a "feel" for which corner is which (a, b, c, d) then the connection between the visual and the symbolic date begins to make sense; otherwise it is very, very easy to get confused by simply mis-copying subscripts! Go slow: do PARTS of this problem one-at-a-time and take short breaks between parts. The test-problem version of this is simpler and has fewer parts.

P49 : Nitrogen is diatomic, so f = 5 degrees of freedom. Since we know that U is (1/2)kT per particle per degree of freedom, we have U = (f/2)NkT = (5/2)nRT. Any CHANGE in U comes from a change is temperature: $\Delta \$U = (5/2)nR$\Delta \$T.

(a) A straightforward computation using the above equation: we know n, f = 5, and we know $\Delta \$T.

(b) Work at constant pressure is just P$\Delta \$V. (This is true because work is the integral of P over dV and if P is constant it factors out of the integrand leaving the integral of dV from initial to final values, which is just $\Delta \$V.) We have to find V1 from the initial state (we are given n, P, and T initially). Then we use the "comparison" form of the ideal gas law to get V2 and thus we know $\Delta \$V.

(c) From the first law: Q = $\Delta \$U + W.

P53 : See Problem III on the 2008 Test I, which is very similar.

We are given n, f (= 5 because gas is diatomic), Po and Vo. We are also given Vf and told that the process is adiabatic (Q = 0). Since f = 5, the parameter $\gamma \$ in Equation 19-15 (p. 514) will be 1.4.

(a) We can immediately compute To from the ideal gas law, obtaining 404 K. To find Tf we will also use the ideal gas law, but first we must first compute Pf using Equation 19-15:

PoVo$\gamma \$ = PfVf$\gamma \$

and the result is 0.078 atm.

Once we have Pf we can compute Tf from the ideal gas law: it is 195 K.

(b) Simple: $\Delta \$U = (f/2)nR$\Delta \$T = - 16,000 J

(c) Q = 0 since the process is adiabatic.

(d) From the first law: the work done BY the gas is W = Q - $\Delta \$U. The work done ON the gas is the negative of this: $\Delta \$U - Q; in this case just $\Delta \$U. So the answer is the same as part (b): -16,000 Joules. Conceptually: since the gas expands, it does work. Since it expands adiabatically, Q = 0. Thus the internal energy of the gas must decrease by the amount of work it does.

P62 : When equilibrium is established, the heat/sec arriving at the junction must equal the heat/sec leaving it, otherwise the temperature at the junction will continue to change. This means that the heat/sec conducted through the copper rod must equal the heat/sec conducted through the aluminum rod. So we use Equation 19-16a for each rod and equate them (we can use either centigrade or K):

kCuACu(225 - Tx)/LCu = kAlAAl(Tx - 0)/LAl

But the cross-sectional areas are equal: ACu = AAl and also the lengths are equal: LCu = LAl. So these drop out and we can solve for Tx. The answer is 147 degrees C.