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Unformatted text preview: Review Problems (1) Calculate the coeﬃcient of density expansion for a material of volume coeﬃcient of expansion β ; meaning ρ → ρ0 (1 + γ ∆T ), what is γ ? (2) (Problem 18-49) Calculate the average number of collisions a molecule in an ideal gas makes per second, called the collision frequency, f , in terms of N , V , T , m and r (the mass and radius of the molecules). (3) A rod of cross-sectional area A and length l has its left end held at constant temperature T1 and its right end held at T2 < T1 . If the conductivity varies with distance from the left end, x, according to the relationship k = x/R + k0 (R and k0 are positive), what is the steady-state heat ﬂow, H , through the rod? (4) A black-body sphere of radius r, mass m, speciﬁc heat c and initial temperature T0 is losing heat due to radiation. What is the temperature of the sphere as a function of time? Does the sphere ever reach absolute zero? (5) (Problems 20-49, 20-79) We know that thermodynamic processes can be represented on a PV diagram (when they are reversible) but they can also be represented on a TS diagram (temperature-entropy diagram). (a) Draw a TS diagram for a Carnot cycle. (b) What does the area within the curve represent? (c) Determine the slope of the tangent to the curve on a TS diagram for a constant volume process involving an deal gas of n moles with constant-volume molar speciﬁc heat CV for some value of temperature T . (6) Calculate the maximum amount of work that can be extracted from a heat engine using two identical objects of mass m, speciﬁc heat c and temperatures T1 and T2 < T1 as its heat reservoirs. [Hint: calculate the ﬁnal temperature of the two objects] (b) What is the eﬃciency of this engine? (c) Is it the Carnot eﬃciency? Explain. (7) Extra Practice: pick a PV cycle and calculate the eﬃciency. Solutions 1 2 (1) V ρ ρ0 (1 + γ ∆T ) γ ∆T = = = = V0 (1 + β ∆T ) m m 1 = = ρ0 ⇒ V V0 (1 + β ∆T ) 1 + β ∆T 1 ρ0 1 + β ∆T 1 −β ∆ T −1= ≈ −β ∆ T ⇒ γ = −β 1 + β ∆T 1 + β ∆T (2) lM , the mean free path, is the average distance a molecule travels before making a collision, hence lM /v is the average time between collisions. Thus the number ¯ of collisions in a second is one over this, which equals v /lM = f . ¯ 8kT 1 and lM = √ πm 4 2π r2 (N/V ) v= ¯ √ ⇒ f = 4 2π r2 (N/V ) 8kT πm (3) Let us look at the heat ﬂow between x and x + dx, where k is approximately constant; remember that heat ﬂow is the same everywhere in the state-state situation, hence H = constant. H dT dx T (x) T1 T2 H = = = = = = k(x)A dQ dT = T (x) − T (x + dx) = −k(x)A ⇒ dt dx dx H − ⇒ A(x/R + k0 ) HR C− ln(x/R + k0 ) A HR T (0) = C − ln k0 A HR T (l) = C − ln(l/R + k0 ) ⇒ A A (T1 − T2 ) l/R+k0 R ln k0 (4) Let us consider how much heat is lost between times t and t + dt, noting that because the temperature is going down, dT < 0. 3 −mcdT
T T0 = = = = = = dT T4 dT T4 |dQ| = σ (4π r2 )T (t)4 dt 4πσ r2 dt mc t 4πσ r2 − dt mc 0 − − 4πσ r2 t mc 11 1 −3 3 3 T0 T 1 T3 T 1 12πσ r2 + t 3 T0 mc
1 3 T0 + 12πσ r 2 t mc As you can see, T → 0 when t → ∞, but it never reaches zero. (5) The change in entropy of an ideal gas is ∆Sgas = d N k ln 2 Tf Ti + N k ln Vf Vi (a) A Carnot cycle consists of an isothermal expansion (T is constant and S increases by the above formula), followed by an adiabatic expansion (S is constant because dS = dQ/T = 0 and T decreases because T V γ −1 = constant implies that T goes down if V goes up), followed by an isothermal contraction (T is constant, S goes down), followed by an adiabatic expansion (S is constant, T goes up). This process is represented by a clockwise rectangle in the TS plane. (b) The integral of T as a function of S is represents the net heat absorbed. T dS = dQ, hence the area enclosed (c) For a constant volume process involving an ideal gas dS = dQ d dT p d dT dT = Nk + dV = N k = nCv T 2 T T 2 T T dS T , is equal to dT nCv hence the slope, which is just (6) Heat will be taken from the hotter object, used to do work, and some of it will be expelled in the form of exhaust to the colder object. This process will repeat itself until the two objects have the same temperature at which point no further work can be performed. If Tf is the ﬁnal temperature, then the heat in is |QH | = mc(T1 −Tf ), the heat out is |QL | = mc(Tf −T2 ) and the work done is the diﬀerence W = |QH | − |QL | = mc(T1 − 2Tf + T2 ). All we have to do is calculate Tf and we can ﬁgure out the work done. Can we just set heat in equal to heat out to get Tf = T1 +T2 ? Of course not, because if 2 heat out equals heat in then W = 0. So what else can we use? We know that the maximum possible work always occurs when the engine is reversible, so we can set the total change in entropy equal to zero in order to solve for Tf . 4 Tf 0 = ∆S =
T1 mc dTH + TH Tf mc
T2 dTL = mc ln TL 2 Tf T1 T2 To get zero the argument of the natural log must equal one hence Tf = √ √ √ Wmax = mc(T1 − 2 T1 T2 + T2 ) = mc( T1 − T2 )2 (b) The eﬃciency is just W/QH and QH = mc(T − Tf ) = mc(T − e= √ √ W T1 − 2 T1 T2 + T2 T1 T2 − T2 T2 √ √ = =1− <1− QH T1 T1 − T1 T2 T1 − T1 T2 √ √ T1 T2 √ T1 T2 ) Let us try to prove the inequality: 1− T1 T2 − T2 √ T1 − T1 T2 √ T1 T2 − T2 √ T1 − T1 T2 < > > > > > 1− T2 ⇒ T1 √ √ Hence the inequality is equivalent to the statement ( T1 − T2 )2 > 0 which is obviously true as long as T1 = T2 . Therefore, our engine is less eﬃcient than a carnot engine even though it is reversible. The reason for this is that any heat engine does the most work it can do when it operates reversibly, however, in order to reach the maximum possible theoretical eﬃciency, the carnot eﬃciency, the temperature of the hot reservoir has to remain at its initial value and not go down as heat is removed from it, and the temperature of the cold reservoir has to remain at its initial value and not increase as heat is added to it. √ T1 T1 T2 − T1 T2 √ (T1 + T2 ) T1 T2 − 2T1 T2 √ √ T1 T2 (T1 − 2 T1 T2 + T2 ) √ √ ( T1 − T2 )2 0 0 0 √ T1 T2 − T2 T1 T2 ⇒ T2 ⇒ T1 ...
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This note was uploaded on 11/19/2010 for the course LECTURE 1 taught by Professor Yildiz during the Fall '10 term at University of California, Berkeley.
- Fall '10