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Sol. Volume of water boiled = 4 * 200 cc = 800 cc T base 1 = 25°C T base 2 = 100°C ⇒ T base 2 - T base 1 = 75°C Mass of water boiled = 800 * 1 = 800 gm = 0.8 kg Q(heat req.) = MS∆θ = 0.8 * 4200 * 75 = 252000 J. 1000 watt – hour = 1000 * 3600 watt-sec = 1000* 3600 J No. of units = 252000/1000 * 3600 = 0.07 = 7 paise (b) Q = mS∆T = 0.8 * 4200 * 95 J No. of units = 0.8 * 4200 *95/1000 *3600 = 0.0886 = 0.09 Money consumed = 0.09 Rs = 9 paise.
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