Question

1) 150 g of a liquid at 45°C is filled in an insulated metal container at 35°C whose mass is 110 g. The system eventually reaches an equilibrium. Find: (a) the final equilibrium temperature, and (b) estimate total change in entropy of the system (i.e. metal container plus liquid). Specific heat of liquid is 4186 J/kg.Cº and specific heat of metal is 900 J/kg.C. 2) A Carnot engine working between a hot and cool reservoir, extracts 800 J from a 600 K reservoir during each cycle and rejects 500 J to a cooler reservoir. It operates at 20 cycles per second. Find (a) the work done during each cycle. (b) its efficiency, (c) the temperature of the cool reservoir, and (d) its mechanical power output. 3) In the figure below, find the magnitude and direction of the net force acting on the charge located at A. +0 B +O L 4) Find net elctric field (both magnitude and direction) at point A (see figure below). + 1/2 - U2

Answer 1

Te =? Me= 150g, Te = 45°C. To = 35°C m. = 11og, ce=4186J 1 kg C, Cm 900J/kg °C mise (Te-Tp) = m Cm (Tg - Tc) 15 0X 4186145 -TP)= ||0X900(Te -35) (45-Tex (150x4186) =(Tp-35) Ilox 900 6.342(45 -Tp) = TP-35 285.39 -6.34 Tp = T-35

734 T = 285.39 +35 =) Tp =320.39 Te=43.64°C 7.34

B) Effciency. = W 2) TH = 600k , Qu=800 J , Qc = 500J 20 cycles per second a) Woock done during each cycle. W= QH-Q = 800-500 = 300J W=300J 300 Qu 800 (n=0.375 c) Tc =P n=1-Te 0.375=1-Tc TH Tc = 1-0.375 = 0.625 Te = 375k d) mechanical power output. in each cycle work done W=3007 in 2ocycles work done W20 = 20 X300 = 6000J 2ocycles per second will release 600oJ Power output then P=6000J/s. 600 600 FlaFn F2 3) Filosoo P=6000w QF cos60° force due to charge at location F=kQQ =kQ² L 1 tQ L2 L² +Q force due to charge at location 2 F2=kQ²

Fia afrafi 600 360° Ficasso F. Cos 60° 60 net force in nedirection is o As Ficasto f F 6560 are some in magnitude & opposite indirection net force is only in ydirection

Fn = Fisin 60°+F₂ sin 60" F = kQB+kQ 3 2 Fn =3 kQ? direction is in yaxis Ei 4) A E L L r 2 42 +Q 3 -Q 42 +Q EA=P field due to the charge at I E,= kQ direction is away 2² from charge field due to the charge at location 2 E =kQ direction is away from charge field due to the charge at location 3 a= L2 - Ez =kQ 12 المالك =kQ2 32² are some the horizontal component of E,& Ez in magnitude & opposite in direction, net horizontal field is o yo VEz net field is in only y direction Enzy AE2 E, cos60° > ECO5 60°

En = Elzy-E Erzg= { sin 60° + Ersin 60° = {6 + kg 3 — 3kg 22 2 L? 13 - 58

En = 3kQ - 2kQ 2 - 312 33kQ-2Qk 322 En = 3.196kQ >> En = 1.065k Q 3L2 L² En=1.065kQ L² direction is intyanis 2