An electric boiler with a resistance of 160 ohms was connected to a network under a voltage of 220V and a vessel

An electric boiler with a resistance of 160 ohms was connected to a network under a voltage of 220V and a vessel containing water weighing 0.5 kg at 20 degrees Celsius was placed. Efficiency = 80%. Determine the time it takes for the water to boil.

R = 160 ohms.

U = 220 V.

m = 0.5 kg.

t1 = 20 ° C.

t2 = 100 ° C.

С = 4200 J / kg * ° C.

Efficiency = 80%.

T -?

Efficiency = Qw * 100% / Qn, where Qw is the amount of heat required to heat water, Qn is the amount of heat energy that is generated in the boiler.

Qw = C * m * (t2 – t1).

Qн = U ^ 2 * T / R.

Efficiency = C * m * (t2 – t1) * 100% * R / U ^ 2 * T.

T = C * m * (t2 – t1) * 100% * R / U ^ 2 * efficiency.

T = 4200 J / kg * ° C * 0.5 kg * (100 ° C – 20 ° C) * 100% * 160 Ohm / (220 V) ^ 2 * 80% = 694 s.

Answer: water will boil in time T = 694 s.