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(a) A hot-water heater is rated at 4.28 x103 W and operates at 2.40 x 102 V. Find the resistance in the heating element and the current.

(b) How long does it take to heat 125 L of water from 15.9Ã¯Â¿Â½C to 46.7Ã¯Â¿Â½C, neglecting conduction and other losses? (The specific heat of water is 4.186 J/g Ã¯Â¿Â½ K.)

(c) How much does it cost at $0.11/kWh? Edit Added Mon, 03 Aug '15 Community 1 Comment Add a Comment Solutions 0 a) P = v^2/R ==> R = v^2/P = 240^2/4280 = 13.46 ohms b)P*t = m*c*dT t = m*c*dT/P = 125*4186*(46.7-15.9)/4280 = 3765.44 s = 62.75 min c) Energy required, E = 125*4186*(46.7-15.9) = 1.6116*10^7 J = 1.6116*10^7/3.6*10^6 kWh = 4.477 kWh total cose = 4.477*0.11 =$0.492
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