Question

an 500W electric kettle heats up 1Ltr of water 25 degree C - 100 degree C in 15 min. How many present of electrical energy is lost​

Answer 1

Answer:

1 litre of water is 1kg

The heat capacity Cp of water is 4.186kJ/kg-K

The heat of vaporization Cvap of water is 2257kJ/kg

So, the energy required to raise the temperature of 1kg of water from 0C to 100C is:

E1 = m•Cp•ΔT = 1 x 4.186 x 100 = 418.6kJ

The energy required to vaporize (boil) the water to produce steam is

E2 = m•Cvap = 1 x 2257 = 2257kJ

So, the total energy required is

Et = 418.6 + 2257 = 2675.6kJ

Power = Energy / time = 2675.6kJ/300s = 8.9186kJ/s = 8918.6J/s

• And this is 8918.6 Watts of power (since 1J/s = 1W)

Explanation:

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