Q(J) Wen = 0 &UCD = QCD - WCD ~ QCD = -60 L atm. .<.-^-l- M -«K

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$Lea & Burke & 'i-ye Physics: The Nature of Things 22.5 GIveu^rT^N CD: Isochoric = - (2 aim) (20 L) = 60 L atm T c =^"410 K, T H = 750 K, Q H = 3.0 x 10 3 J Find: -WOT\<.$

1 Lea & Burke & 'i-ye Physics: The Nature of Things 22.5 GIveu^rT^N CD: Isochoric = - (2 aim) (20 L) = 60 L atm T c =^"410 K, T H = 750 K, Q H = 3.0 x 10 3 J Find: -WOT\<., W and the heat rejected, Q c, T c n 410 = -(40 L )(-! atm) = -60 L atm Wen = 0 &UCD = QCD - WCD ~ QCD = -60 L atm DA: isobaric compression QDA = Pi(Vi - V 2 ) - (1 atm)(20 L - 40 : = 50 L atm WDA - Pi(Vi - V 2 ) = 1 atm(-20 L ) = -20 L -atm = 30 L atm Table: Since 1 L atm = 101 J, we have: Process: AB BC CD DA Q(J) W(J) A /(J) As a check, note that the sum of the third column gives: A[7 n et J J J J = 0 As required for a cyclic set of processes, b) The efficiency is denned by: _ net work obtained heat input from the table in 3a), W Q- in W aei = 4040 J-2020 J =+2020J Q in = 3030 J J = J therefore, 22.4 e = 44% W nei 2020 J Q- m J The work done per cycle is: W =^&oqn ~ (0.453)(3.0 x 10 3 J) ('= 1.4 x lo 3^; <^. The waste heat exhausted is: \Q C \ =_ [1._- e)qg = (0.547)(3.0 x 10 3 J) fl~= 1.6xl0 3 J\ ^' - H ^ As a check on the answers, note that W + \Q C \ jr. 1.4 x 10 3 J x 10 3 J = 3.0 x 10 3 J = Q H / as required by the 1st Law of Thermodynamics No. For example, an irreversible processes carried out inside an insulated container involves no exchange of energy with the environment. Therefore the only entropy change possible is that of the system. When an ice cube melts, heat is transferred to the cube from the environment. The entropy change of the environment is negative, while the entropy change of the cube is positive and sufficiently large that the total,- entropy change (of cube plus environment) is also positive J/K 22.9 The Carnot cycle is the most.efficient heat engine, and has the maximum efficiency possible, e < e c = I ~ 7j=r for T c = 300 K, T H = 600 K.<.-^-l- M -«K The engine can be no more than 50% efficient K E l = -5.4 x HT 19 J; <? = -2.2 x 10~ 18 J; 9-2 = 2 T = 10 4 K

$22.14 The number of states is 52!. Thus the entropy is: 5 = Jfeln52! = Jfelne 156 = 156fc = 156 (1.38 x ID" 23 J/K) = 2. 15 x 10~ 21 J/K 2 /<w 1 " I 0 i c- \ i-* ^>A >lcv- ^ ] ^- 22.$

2 22.14 The number of states is 52!. Thus the entropy is: 5 = Jfeln52! = Jfelne 156 = 156fc = 156 (1.38 x ID" 23 J/K) = x 10~ 21 J/K 2 /<w 1 " I 0 i c- \ i-* ^>A >lcv- ^ ] ^ See solutions manual In both cases the gas undergoes a complete cycle of processes and then returns to its initial state. Since internal energy is a state variable, its value is the same at the end of the cycle as at the beginning - the change in internal energy is zero independent of whether the processes are reversible. Of course the irreversible cycle does produce a. net change of entropy while the reversible cycle does not ^vvfe consider each process in detail to find the heat input and net work done. "22?: isobaric compression. Point B is on a lower isotherm than point.a, so heat transfer is out of the system during this leg. The work done by the system is negative: W AB = P A (V s - V A ] = P A (\V A - V A ] = -\P A V A \& J 4 BC: Constant volume process. Point C is on a higher isotherm than point B, so heat transfer is into the system: Then, using the ideal gas law: QBC = \(Pc- QBC = PB Since the volume does not change, no work is done: = XR (T c - T B ) = (1P B - P B ) \V A = \P B V A = \ \J W BC = 0 CA: Along the isotherm, there is no change in internal energy, so Q = W. Both are positive. Using equation 19.13: QCA = W CA = P G V C In?± = In In2 Thus the efficiency of the cycle is: _ net work done W AB + W 3C + W CA _ ~\ heat input QBC + QCA %P A V A + P A V A }n2 ~l + 2hi n2 ^ 2122 We consider each leg of the cycle to compute the heat input and the net work done. Since Q = 0 along both adiabats, the calculations are most easily done in terms of heat transfers. For a complete cycle, there is no change of internal energy. Thus from the first law of thermodynamics, = 0 = QMt - W net => Q» Qin

$Lea &: Burke Physics: The Nature of Thingi from Figure 22.25, 8JL 22.27 a) The coefficient of performance for a Carnol "T"i\ "**^ refrigerator is: ^ ^?$

3 Lea &: Burke Physics: The Nature of Thingi from Figure 22.25, 8JL a) The coefficient of performance for a Carnol "T"i\ "**^ refrigerator is: ^ ^? > ediesel r 2 = 8_L 2L _ 1 = ^(36 F-32 F)+273K = K - ^(82 F-32 F)+273K for r = T2- For diatomic gas, 7 and so 7 1 =. Then e Diese! = eotto = (For the numbers in the Figure, 7 =.) A. large volume of water may approximate an ideal heat reservoir through its very great heat capacity (C^o ~ 4184 J/K kg) However, the thermal conductivity of water is extremely low, being only about 0.6 ^5- While the thermal conductivity of copper is extraordinarily high (fa 400 TJ-rg), it's usefulness as an ideal heat reservoir is limited by the small value of its specific heat capacity (393 J/K kg) By increasing the hot temperature while keeping VA constant, the entire cycle moves upward in the P-V diagram. P 4 A\\ b) a) 260 K = K K (300.8 K) - (275.2 K) -11 Q dq = Kc\W\ = (10.8)(220 W) = 2400W b) The efficiency increases by 1% c) decreases by 1% According to Newton's law of cooling, the rat< at which heat leaks into the freezer is proportional to the temperature difference between inside and outside. The freezer uses net power P to remove thif heat: p _ Wnet per cycle At per cycle For the Carnot cycle ( 22.3) W net - Nk(T H D to remove heat PA increases, and since AD is an adiabat PjyV^ also increases. The product PD^D is fixed by the value of the temperature TC of the reservoir, thus VD must increase. The pressure and volume throughout the cycle increase. Similarly if TC is decreased, pressure and volume throughout the cycle decrease K each cycle. To remove heat faster, the refriger ator must cycle more rapidly. AJ Qcoid T c H Thus the power used is (Tn- P- oc oc (T Ai H ~ T c f

4 3MUUIC3 both the first and second laws. After sufficient water is absorbed, the bird becomes mechanically unbalanced. When the torque arising from the weight of absorbed water exceeds the torques due to friction in the axle and due to our inventor's generating equipment, the bird will move to its "wet" state. In this process, ordered gravitational energy is transformed into ordered electrical energy plus some thermal energy due to friction. Only the latter leads to entropy increase. NUW, exposed to ory air, water molecules evaporate, their entropy increasing as they become still less well known in position. Evaporation reducesthe water temperature, resulting in heat flow from the air into the bird and further production of entropy. Eventually the bird becomes unbalanced again and its head drops to the water for another drink. Again ordered gravitational energy is transformed to electrical energy via the inventor's generator. Energy transfer to the system during the cycle occurs as heat transfer during the wet state, accounting for frictional dissipation and any electrical energy produced. Net entropy is produced in each phase of the cycle. So both laws are satisifed overall. \Vater in contact with dry air is not in equilibrium, and evaporation results in energy flow and entropy production. The "dippy bird" is an example of how devices can exploit nonequilibrium to carry on organized activity. Life on earth is another example, which takes advantage of entropy production as the earth converts sunlight to infrared radiation. ( ^22.38 First the water is heated from room temperature to the boiling point, at constant ^StiOMSBS* 551 pressure: = (4.0 x 10-3 m 3 ) (1.0 x 10 3 kg/m 3 ) (4186 J/kg - K) In =*' 4.0 x 10 3 J/K^ The change in entropy due to conversion of water to steam at 100 C = 373 K is given by: = 9. = H^i = (4-0 * 10" 3 m 3 ) (1-0 x 1(}3 k 8fa 3 ) ( 2272 x 1Q3 J/k g) =JI^ 373 K So the total entropy change is me sum of the two amounts we have calculated: A5 tota i = 2.8 x 10 4 J/K The rate of heat transfer through the roof is: where A is the roof's area, R/ is the thermal resistance of the roof material, TH is the (hot) temperature inside the building and T c is the (cold) temperature outside. From Example 22.5, we have : ds dqc/ dq H / ~~ T c T H. Heat flow is into the cold exterior and out of the hot interior, so: ds H H A. I l\ A(T H -T c f R f T C T S Putting in the numbers: T H = 69 F = - (69-32) K = K T c = 44 F = (44-32) K = K ds 510 m 2 ( K K) 2 ~0.20m 2 -K/W ( KU K]

5 22. Entropy and the Second Law of Thermodynamics 409 Then ~- x w T 6300 K (See front endpapers of text) ~ = -6.2 x W/K At least as much entropy is generated elsewhere in the universe by the absorption of solar radi ation AS = 2.8 x 10 4 J/K Given m = 1 kg Ti = 15 C = 288 K = 80 C = 353 K T 2 The entropy increase of the water is AS = ^ = f rp - mcln T Ti mcdt = me in -22, JThe measured efficiency, e, may be expressed = " """ln terms of the Carnot efficiency, &c'- therefore e = (0.55)e c = 0.17 e c = 0.31 But we know ec = 1 Tc/Tn with so K T c = ( ) K = 278 K TH = v_ The required work is: 278 K = K = 4.0 x 10 2 K T C = (1.0 x 10 3 g)(4.2 = 850 J/K W/K 1j2.4rShow: -IM f 353 " H288, differentiating with respect to time yields W_ = L_Tc\ dqn \ TH I with T H in Kelvin. Thus t for an ideal gas., Solution: for an ideal gas 1st Law: dq = dq T r^ _ PV = NkT du = - du + PdV ^> du dv T ~f~ dv mc v ±P-mcv^ + N. fc~ PV \Nk, dv T fdq CD (2) Tc = 150 K (to 2 sig fig.) 35 W (268 K) = K Only the fast molecules near the surface of a liquid can overcome the attraction of the ' molecules in the surface and escape by evaporation. The average kinetic energy of the remaining molecules drops correspondingly, leaving the liquid at a lower temperature The Maxwellian velocity distribution (Eqn ) m,27r/btv exp{ mv2 ^ 2kT J gives the fraction of molecules in a speed range v to v + dv. For a fixed value of v and dv, two V things happen as T increases by a factor of 2: as reauired. i) The normalization factor by a factor 2 = 2.83 and [2irkT decreases

Handout 12: Thermodynamics. Zeroth law of thermodynamics

Handout 12: Thermodynamics. Zeroth law of thermodynamics 1 Handout 12: Thermodynamics Zeroth law of thermodynamics When two objects with different temperature are brought into contact, heat flows from the hotter body to a cooler one Heat flows until the temperatures