Two spirals of an electric stove with a resistance of 10 ohms each are connected in series and connected to a 220V

Two spirals of an electric stove with a resistance of 10 ohms each are connected in series and connected to a 220V network. How long will it take for water weighing kg to boil on this hotplate, poured into an aluminum pan weighing 0.3 kg, if their initial temperature was 20C? Neglect energy losses for heating the ambient air

R1 = R2 = 10 ohms.

U = 220 V.

mk = 0.3 kg.

Ca = 920 J / kg * ° C.

mw = 1 kg.

Cw = 4200 J / kg * ° C.

t1 = 20 ° C.

t2 = 100 ° C.

T -?

The amount of thermal energy Q, which is released in the tile, is expressed by the Joule-Lenz law: Q = U2 * T / R, where U is the voltage, T is the current flow time, R is the total resistance of the spirals.

Since the spirals are connected in series, their total resistance R will be the sum: R = R1 + R2.

The amount of heat energy Q required for heating water and a saucepan is expressed by the formula: Q = Ca * mk * (t2 – t1) + Cw * mw * (t2 – t1).

U2 * T / (R1 + R2) = Ca * mk * (t2 – t1) + Cw * mw * (t2 – t1).

T = (Ca * mk * (t2 – t1) + Cw * mw * (t2 – t1)) * (R1 + R2) / U2.

T = (920 J / kg * ° C * 0.3 kg * (100 ° C – 20 ° C) + 4200 J / kg * ° C * 1 kg * (100 ° C – 20 ° C)) * (10 Ohm + 10 Ohm) / (220 V) 2 = 148 s.

Answer: the water will boil in a time T = 148 s.