1. Find the change in entropy if 125 g of water at 50.0oC is raised to the boiling point and then turned into steam.
  2. Find the change of entropy of 2.00 mol of a monoatomic a gas if it undergoes a process from A to B where TB=TA/2, and VB=3VA.
  3. A freezer is a Carnot Refrigerator that makes ice at -4.00oC. Its coefficient of performance is 5.50. Also 220.0 J of heat is extracted from the freezer every cycle.
    (a) What is the room temperature outside the freezer?
    (b) How much work does the freezer do in a cycle?
    (c) How much energy is delivered to the room every cycle.

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Constants 1 N A = 6.02 × 10 23 mol 1st Law of Thermodynamics J R = 8.31 mol⋅K k = R / NA −19 e = 1.60 × 10 C me = 9.11× 10−31 kg mp = 1.67 × 10−27 kg = 8.99 × 10 ε 0 = 8.85 × 10 µ0 = 4π × 10 c = 3.00 × 10 pV = nRT vrms = Work QH = W + QL εC = QH T ε C = 1− L TH Q K= L W TL K= TH − TL ∫ p dV Vi J/(kg K) T K L kJ/kg T K L kJ/kg Cu 386.0 1356 205 2840 5069 Fe 461.0 1808 96.0 3023 870 Al 900.0 933 398 2750 11400 Ag 233.0 1235 88.0 2485 2330 Au 126.0 1337 63.0 3080 1580 Pb 128.0 621 23.0 2022 871 H 20 4186 (2108 ice) 273 334 373 2257 Constant Quantity Q Q =W pV γ W = nRT ln Ideal gas Q=0 W = −ΔEint p Q = nC P ΔT W = pΔV V Q = nCV ΔT W = 0 ( ) ΔE Vi V= 2 C = KC0 ΔE Vf C = Q / ΔV Ceq = C1 + C2 + ….  p = qd    τ = p×E   U = −p⋅E T Electric Potential 1 1 1 = + + … Ceq C1 C2 Electric Dipole int =0 U = qEs Capacitance U = CV ! ! Φ E = ∫ E ⋅ dA W 1 q1q2 4πε 0 r 1 2 σ E= 2ε 0 W U= C = ε0 A / d 1 q E= 4πε 0 r 2 1 λ E= 2πε 0 r Heat Engines Vf ! F E= q 3RT M f CV = R 2 γ = C p / CV Q = mcΔT Q = mL Electric ! Field S = k lnW CP = CV + R ΔL = LαΔT T N! W= n1 !n2 ! K avg = 23 kT Heat W= 1 q1q2 F = 4πε 0 r 2 ΔS = nR ln Vif + nCV ln Tif Kinetic Theory of Gases Electric Potential Energy Electric Force f dQ ΔS = dEint = dQ − dW ∫i T ΔEint = Q − W V 9 Nm 2 C2 −12 C 2 Nm 2 −7 N A2 8 m s 1 4 πε 0 Entropy ΔEint = nCV ΔT c PHYS 240 Cheat Sheet Resistance ρ =E/J R = ρL / A V = IR Req = R1 + R2 + … 1 1 1 = + + … Req R1 R2 P = IV 1 q 4πε 0 r V =U /q ΔV = −Ed   ΔV = − ∫ E ⋅ dl b a Ex = − ∂V / ∂x DC Circuits Q f = Q0 = CV I0 = V / R ( q = Q f 1 − e −t/RC q = Q0 e −t/RC τ = RC ) PHYS 240 Cheat Sheet Magnetic Force ! ! ! F = qv × B ! ! ! F = IL × B Lµ I I F= 0 1 2 2π r Faraday’s Law   Φ B = ∫ B ⋅ dA B ε = − dΦ dt Magnetic Field ! ! µ0 Idl × r̂ dB = 4π r 2 µ0 I B= 2π r di dt U = 12 LI 2 VL = L VS N S I P = = Vp N P I S RL Circuits τ =L/R I0 = I f = ε / R i = I f (1 − e −t/τ ) i = I 0 e −t/τ Maxwell’s Equations µ = IA    ! ! Qenc E ⋅ d A = “∫ ε0 ! ! B ⋅ d A = 0 “∫ ! ! dΦ E B ⋅ d l = µ I + µ ε 0 enc 0 0 “∫ dt ! ! dΦ B E ⋅ d l = − “∫ dt τ =µ×B   U = −µ ⋅ B Magnetic Applications B = µ0 nI B= ( µ0 R 2 I 2 x +R Inductors NΦ B L= i Magnetic Dipole 2 ε = vBL ) 2 3/2 v=E/B mv R= qB AC Circuits Z = R 2 + (X L − XC )2 X L − XC tan(φ ) = R V = IZ XL = ω L 1 XC = ωC ω0 = 1 LC Table of Integrals VL = IX L VC = IXC I rms I = 2 ∫ ∫ ( ( ∫ dy y +a 2 2 y dy y +a 2 2 ) 3/2 ) 3/2 dy y +a 2 2 = y a 2 =− = ln ( y +a 2 ∫ 2 1 ∫ y 2 + a2 y +a + y 2 2 ) y dy y 2 + a2 = y 2 + a2 dy 1 −1 ⎛ y ⎞ = tan ⎜ ⎟ 2 2 a ⎝ a⎠ y +a ∫(y y dy 2 + a2 ) 1 = ln( y 2 + a 2 ) 2

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