How long does it take for an electric heater to boil 2.2 kg of water, the initial temperature of which is 10 ° C?

How long does it take for an electric heater to boil 2.2 kg of water, the initial temperature of which is 10 ° C? The current in the heater is 7 A, the voltage in the network is 220 V, the efficiency of the heater is 45%.

m = 2.2 kg.

C = 4200 J / kg * ° C.

t1 = 10 ° C.

t2 = 100 ° C.

I = 7 A.

U = 220 V.

Efficiency = 45%.

T -?

Efficiency = Ap * 100% / Az.

Az = I * U * T, where I is the current in the heater, U is the voltage in the heater, T is the operating time of the heater.

Ap = C * m * (t2 – t1), where C is the specific heat capacity of water, m is the mass of heated water, t2, t1 are the final and initial water temperatures.

Efficiency = C * m * (t2 – t1) * 100% / I * U * T.

The water heating time T will be determined by the formula: T = C * m * (t2 – t1) * 100% / I * U * efficiency.

T = 4200 J / kg * ° C * 2.2 kg * (100 ° C – 10 ° C) * 100% / 7 A * 220 V * 45% = 1200 s = 20 min.

Answer: it takes time T = 20 minutes to heat the water.