PHY1506 EXAM PACK 2022 | with complete solutions | verified

Question 1 The figure below represents a pV [p in Pa × 104 and V in m3 ] diagram for 7.50 moles of an ideal diatomic gas taken through the cycle a → b → c. (a) Calculate the highest temperature reached by the gas during the cycle (4) (b) What net work does the gas do during the cycle? (4) (c) How much heat is exchanged with the gas during part bc of the cycle? Does it enter or leave the gas? (4) (d) What is the change in the internal (thermal) energy of the gas during part bc of the cycle? (4) (e) What is the change in the internal (thermal) energy of the gas during the entire cycle? (4) [20] Question 2 (a) The figure below shows three electric charges labeled Q1,Q2,Q3, and some electric field lines in the region surrounding the charges.What are the signs of the three charges? (6) [TURN OVER] Page 2 of 5 PHY1506 (b) A 400-g piece of metal at 120.0 ◦C is dropped into a cup containing 450 g of water at 15.0 ◦C. The final temperature of the system is measured to be 40.0 ◦C. Calculate the specific heat of the metal, assuming no heat is exchanged with the surroundings or the cup. The specific heat of water is 4186 J/(kg.K). (6) (c) Why doing a change of phase, heat is absorbed but the temperature does not rise? (3) [15] Question 3 (a) Two positive point charges +4.00 µC and +2.00µC are placed at the opposite corners of a rectangle as shown in the figure.What are the potentials at points A and B (relative to infinity) due to these charges? (10) (b) A heat engine with an efficiency of 30.0% performs 2500 J of work.How much heat is discharged to the lower temperature reservoir? (10) [20] Question 4 (a) The magnetic field of a moving charge is given by B~ = µ0 4π qv sin θ r 2 . (i) Which law does this equation represent? (2) (ii) What do r,µ,v and θ represent? (4) (b) What is an electromagnetic induction? (3) (c) State Faraday’s law (6) [15] [TURN OVER] Page 3 of 5 PHY1506 Question 5 Consider the electric circuit as shown in the figure below, find: (a) the equivalent resistor of the circuit, (5) (b) the current flowing through each resistor, (8) (c) the potential difference across each resistor. (7) ( Please summarize your results in a table for an easy reading) [20] Question 6 An RLC circuit consists of a 10 Ω resistor, a 1.0 mH and a 1.0 µF capacitor, in series and connected to a (10V ) cos ωt.Calculate: (a) the resonance frequency (3), (b) the resistor and capacitor voltages (VR and VC ). (7) [10] TOTAL: 100 Marks [TURN OVER] Page 4 of 5 PHY1506 FORMULAE P V = nRT Q = mC∆T Uel = kq r QH = Wout η QC + Wout = Q H 1 R = 1 R1 + 1 R2 R = R 1 + R 2 ω = 1 √ LC f = ω 2π I = √ ε R2+(X L +X C ) 2 VR = IR VC = IX C X C = 1 ωC QV = nC V ∆T QP = nC P ∆T W = nRT ln (V f /Vi) (isothermal) W = p∆V ∆E th = Q ∆E th = Q − W ∆E th = ηC V ∆T (any process) Pi Vi Ti = Pf Vf Tf Eth = 5 2 nRT (diatomic gas) USEFUL INFORMATION kB = 1.38 × 10−23 J/K NA = 6.02 × 1023 mol−1 e = 1.60 × 10−19 C R = 8.31 J/mol.K K = 9.0 × 10 9 Nm2 /C 2 1 µC = 10−6 C CV = 5 2R, CP = 7 2R (diatomic gas) c UNISA 2017 Page 5 of 5 FIRST SEMESTER 2017, SOLUTIONS (PHY1506) Question 1 (a) The temperature is obtained following PaVa = nRT a ⇒ T a = PaVa nR = 3 × 104 × 0.2 7.5 × 8.31 = 96.27 K = −176.73◦ C PaVa Ta = PbVb Tb ⇒ T b = Ta PbVb PaVa = 96.27 5 × 104 × 0.6 3 × 0.2 ! = 481.35 K = 208.35◦ C Pb Tb = Pc Tc ⇒ T c = PcTb Pb = 3 × 481.35 5 = 288.81 K = 120◦ C (b) The work done will be given by the triangle abc. That is W = 1 2 ab × bc = 1 2 × 0.4 × 2 × 104 = 4 × 103 J = 4 KJ (c) Since on part bc of the cycle the volume is constant, we have QV = Q bc = nC V ∆T = 7.5 × 5 2 R(288.81 − 481.35) = −30 KJ This heat leaves the gas due to the negative sign. (d) The change in the thermal energy is given by ∆ th = Q bc − Wbc = −30 − 0 = −30 KJ (e) The change in the thermal energy during the entire cycle will be ∆E th = ∆E ab + ∆E bc = Q ab − Wab + Qbc − Wbc Question 2 (a) The field lines at Q 1 and Q3 are outgoing, therefore these charges are repulsive (positive charges). The field lines at Q2 are incoming, it is an attractive charge (negative charge). (b) We write the calorimetric equation as Mm Cm (t m − t f ) + M wCw(t w − t f ) = 0 ⇒ M m Cm (t m − t f ) = M wCw(tf − t w) Cm = MwCw(tf − t w) Mm Cm (t m − t f ) = 0.45 × 4186(40 − 15) 0.4(120 − 40) = 1471.61 J/Kg◦C (c) This is due to the latent heat. 1 Question 3 (a) The potentials are calculated as UA = Kq1 r 1 + Kq2 r 2 = (9 × 109 )(2 × 10−6 ) 0.4 + (9 × 109 )(4 × 10−6 ) 0.8 = 45 × 10 3 + 45 × 103 = 9 × 104 V UB = (9 × 109 )(4 × 10−6 ) 0.4 + (9 × 109 )(2 × 10−6 ) 0.8 = 9 × 10 4 + 22.5 × 103 = 1.125 × 105 V (b) We calculate the heat is discharged to the lower temperature reservoir as follows QH = Wout η QC = Q H − Wout = 2500 0.3 − 2500 = 5833, 3 Joules