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Citrus2011 [14]

3 years ago

14

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 it

s 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?

1 answer:

Nookie1986 [14]3 years ago

3 0

Answer:

Explanation:

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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