Physics Solution Manual for 1100 and 2101

20 a 010 a or v r1 r2 020 a 010 a 014

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Unformatted text preview: these expressions for Q and P will allow us to determine the time t. SOLUTION Substituting Equations 12.4 and 20.6c into the expression for the time and recognizing that the normal boiling point of water is 100.0 ºC, we find that Chapter 20 Problems t= = 1071 Q cm∆T Rcm∆T = = P V2 /R V2 (15 Ω ) 4186 J/ ( kg ⋅ C° ) ( 0.50 kg )(100.0 °C − 13 °C ) = 190 s (120 V )2 30. REASONING The total volume of ice melted is the product of the thickness h and the area A of the layer that melts, which permits us to determine the thickness in terms of the volume and the area: hA = Volume h= or Volume A Given the density ρ = 917 kg/m3, we can find the volume of the ice from ρ = (1) m Volume (Equation 11.1), where m is the mass of the ice. Volume = m ρ (2) The mass m of the ice determines the heat Q required to melt it, according to Equation 12.5 Q = mL (12.5) In Equation 12.5, L = 33.5 ×104 J/kg is the latent heat of fusion for water (see Table 12.3). In order to find the maximum thickness of ice the defroster can melt, we will assume that all the heat generated by the defroster goes into melting the ice. The power output P of the defroster is the rate at which it converts electrical energy to heat, so we have that P= Q t or Q = Pt (6.10b) where t is the elapsed time (3.0 minutes). The power output P is, in turn, found from the operating voltage V and current I: P = IV SOLUTION Substituting Equation (2) into Equation (1) yields (20.6a) 1072 ELECTRIC CIRCUITS h= Volume ( m ρ ) m = = A A ρA Solving Q = mL (Equation 12.5) for m, we find that m = Equation (3), we obtain h= (3) Q . Substituting this result into L m (Q L) Q = = ρA ρA Lρ A (4) Substituting Equation (6.10b) into Equation (4) yields h= Q Pt = Lρ A Lρ A (5) We note that the elapsed time t is given in minutes, which must be converted to seconds. Substituting Equation 20.6a into Equation (5), we obtain the maximum thickness of the ice that the defroster can melt: 60 s min Pt IVt 1 min = 3.1×10−4 m h= = = L ρ A L ρ A 33.5 ×104 J/kg 917 kg/m3 0.52 m2 ( 23 A )(12 V ) ( 3.0 ( )( ) )( ) ______________________________________________________________________________ 31. SSM REASONING AND SOLUTION As a function of temperature, the resistance of the wire is given by Equation 20.5: R = R0 1 + α (T − T0 ) , where α is the temperature 2 coefficient of resistivity. From Equation 20.6c, we have P = V / R . Combining these two equations, we have 2 P0 V P= = R0 1 + α (T – T0 ) 1 + α (T – T0 ) 1 2 where P0 = V /R0, since the voltage is constant. But P = P0 , so we find 2 P0 2 = P0 1 + α (T – T0 ) or 2 = 1 + α (T − T0 ) Solving for T, we find 1 + 28° = 250 °C –1 α 0.0045 (C°) ______________________________________________________________________________ T= 1 + T0 = Chapter 20 Problems 1073 32. REASONING Substituting V = 1 V0 into Equation 20.7 gives a result that can be solved 2 directly for the desired time. SOLUTION From Equation 20.7 we have V = 1 V0 = V0 sin 2π f t 2 or 1 2 = sin 2π f t Using the inverse trigonometric sine function, we find 2π f t = sin −1 ( 1 ) = 0.524 2 In this result, the value of 0.524 is in radians and corresponds to an angle of 30.0º. Thus we find that the smallest value of t is t= 0.524 0.524 = = 1.39 × 10−3 s 2π f 2π ( 60.0 Hz ) 33. REASONING a. The average power P delivered to the copy machine is equal to the square of the 2 rms-current Irms times the resistance R, or P = I rms R (Equation 20.15b). Both Irms and R are known. b. According to the discussion in Section 20.5, the peak power Ppeak is twice the average power, or Ppeak = 2 P . SOLUTION a. The average power is 2 2 P = I rms R = ( 6.50 A ) (18.6 Ω ) = 786 W (20.15b) b. The peak power is twice the average power, so Ppeak = 2 P = 2 ( 786 W ) = 1572 W ______________________________________________________________________________ 1074 ELECTRIC CIRCUITS 34. REASONING AND SOLUTION The expression relating the peak current, I0, to the rmscurrent, Irms, is I 2.50 A I rms = 0 = = 1.77 A 2 2 ______________________________________________________________________________ 35. 2 SSM REASONING The average power is given by Equation 20.15c as P = Vrms / R . In this expression the rms voltage Vrms appears. However, we seek the peak voltage V0. The relation between the two types of voltage is given by Equation 20.13 as Vrms = V0 / 2 , so we can obtain the peak voltage by using Equation 20.13 to substitute into Equation 20.15c. SOLUTION Substituting Vrms from Equation 20.13 into Equation 20.15c gives P= 2 Vrms R (V / 2 ) = 0 R 2 = V02 2R Solving for the peak voltage V0 gives V0 = 2 RP = 2 ( 4.0 Ω ) ( 55 W ) = 21 V ______________________________________________________________________________ 36. REASONING Because we are ignoring the effects of temperature on the heater, the resistance R of the heater is the same whether it is plugged into a 120-V outlet or a 230-V outlet. The average power output P of the heater is related to the rms outlet voltage Vrms and the heater’s resistance R by P =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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