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8 June, 23:07

A well-insulated electric water heater warms 131 kg of water from 20.0°C to 51.0°C in 31.0 min. Find the resistance (in Ω) of its heating element, which is connected across a 240 V potential difference. 6.34 Correct: Your answer is correct. Ω (b) What If? How much additional time (in min) would it take the heater to raise the temperature of the water from 51.0°C to 100°C? 28 Incorrect: Your answer is incorrect. min (c) What would be the total amount of time (in min) required to evaporate all of the water in the heater starting from 20.0°C?

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  1. 9 June, 01:16
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    Heat required to warm the water from 20 degree to 51 degree

    = mct

    = 131 x 4150 x (51 - 20)

    = 16853150 J

    Power of heating element

    = v² / R

    Heat generated in 31 min

    = (v² / r) x 31 x 60 = 16853150

    r = (240 x 240 x 31 x 60) / 16853150

    6.35 ohm

    In this case heat required will change so time will also change

    Heat required =

    131 x 4150 x (100-51)

    = 26638850 J

    If time required be t hour

    Energy consumed

    Power x time

    = (v² / r) x t x 60 = 26638850

    t = 26638850 x 6.35 / (240 x 240 x 60)

    = 48.95 h

    Heat required to evaporate water at 100 degree

    = mass x latent heat

    = 131 x 2260000

    = 296060000 J

    Total heat required

    = 296060000 + 26638850 + 16853150

    = 339552000 J

    time required = 339552000 x 6.35 / (240 x 240 x 60)

    = 623.88 h.
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