An electric kettle with an efficiency of 60% contains 0.6 kg of water at 20 degrees. The kettle was plugged into a 220V

| February 18, 2021 | education

An electric kettle with an efficiency of 60% contains 0.6 kg of water at 20 degrees. The kettle was plugged into a 220V network and forgot to turn off. After 11 minutes the water has completely boiled away. Determine the resistance of the heating element.

Efficiency = 60%.

m = 0.6 kg.

U = 220 V.

t1 = 20 ° C.

t2 = 100 ° C.

C = 4200 J / kg * ° C.

q = 2.3 * 10 ^ 6 J / kg.

T = 11 min = 660 s.

U = 220 V.

R -?

Efficiency = Qp * 100% / Qz.

Qп = C * m * (t2 – t1) + q * m.

Qz = U ^ 2 * T / R.

Efficiency = (C * m * (t2 – t1) + q * m) * 100% * R / U2 * T.

The formula for determining the resistance of the electric kettle R will take the form: R = efficiency * U2 * T / (C * m * (t2 – t1) + q * m) * 100%.

R = 60% * (220 V) ^ 2 * 660 s / (4200 J / kg * ° C * 0.6 kg * (100 ° C – 20 ° C) + 2.3 * 10 ^ 6 J / kg * 0.6 kg) * 100% = 12 ohms.

Answer: the resistance of the electric kettle is R = 12 ohms.



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