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Unformatted text preview: Assignment 4 (Solution)
Question 1: (a) X L = 2fL = 2 560 90 10 3 = 316.7 XC = 1 1 = = 18.9 2fC 2 560 15 10 6 Z = ( X L  X C ) 2 + R2 = (316.7  18.9) 2 = 307 I = V/Z = 105/307 = 0.342 A (b) VR = IR = 0.342*78 = 26.7V VL = IXL = 0.342*316.7 = 108.3V VC = IXC = 0.342*18.9 = 6.46 V Question 2: (a) Work done: W = pV = (2*105) [(20*103) (10*103)] = 2*103 J (b) Since the temperature at the beginning and the end of the process is same, no change in internal energy: U = Q W 0 = Q W, Q = W = 2.0*103 J Question 3: At the initial condition: Temperature (T1) = 20 0C = 20+273= 293K Volume (V1) = Length (cm) * Area (A) cm2 = 30* A cm3 = 30A cm3 For final condition: Temperature (T2) = Unknown Volume (V2) (cm3) = Length (cm)*Area (A) (cm2) = 20A cm3 If no heat loss the process must be adiabatic and will conform to the relation: PV = Constant Another method we can write this relationship is: TV1 = Constant Relating the condition 1 to condition 2: T1 (V1) 1 = T2 (V2) 1 T2 = T1 (V1/V2) 1 = 293 (30/20)1.41 = 293 (1.5)0.4 = 293*1.176 = 344.6 K Temperature rise: T2 = 344.6 293 = 51.6K Question 4: l= kT 2 (d p )
2 = (1.38 10 23 )(273 + 20) 2 (2.0 10
10 )(1.013 10 )
5 = 22.5 10 8 m = 225nm Question 5: Efficiency of engine = = (QH QL)/QH = (90004000)/9000 = 0.56 Maximum possible efficiency is given by Carnot engine:= ideal = (TH TL)/TH = (375225)/375 = 0.4 Manufacturer's claims violate the second law of thermodynamics. ...
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 Spring '08
 Sahoo

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