Question

An electric heater rated at 1kw, 230v has to raise the temperature of 1.5 litres of water from 15°c to boiling point. find the time taken by the heater if its efficiency is 80%

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Answer 1

Answer:

Theoretical Assumptions (not realistic): 1) ALL of the heat energy is transferred to the water, and 2)the water is uniformly heated. 3) The water density is 1.0g/mL.

Because we are only heating it TO the “boiling point”, ONLY the energy to heat the water from 20 to 100’C using the heat capacity or specific heat of water is required. A “Watt” is 1 Joule/second. The specific heat of water is 4.184 J/g-’C. Water mass is 1000g. The energy required to heat it is therefore

1000g⋅4.184(Jg’C)⋅(100−20)’C=334720J

The heater supplies 1200 J/s, so the time to provide this heat to the water is 334720J1200(Js)=278.9s or 4.65 minutes.

Answer 2

Answer:

Theoretical Assumptions (not realistic): 1) ALL of the heat energy is transferred to the water, and 2)the water is uniformly heated. 3) The water density is 1.0g / m * L

Because we are only heating it TO the "boiling point", ONLY the energy to heat the water from 20 to 100^ prime C using the heat capacity or specific heat of water is required. A "Watt" is 1 Joule/second. The specific heat of water is 4.184 J/ g-'C. Water mass is 1000g. The energy required to heat it is therefore

1000g*4.184(Jg^ prime C)*(100-20)^ prime C=334720J

The heater supplies 1200 J / s , so the time to provide this heat to the water is 334720 * J * 1200(Js) = 278.9s or 4.65 minutes.