How long will 1.5 kg of water be heated from 20 to 100 ° C in an electric kettle with a current of 0.2A

How long will 1.5 kg of water be heated from 20 to 100 ° C in an electric kettle with a current of 0.2A and a voltage of 220 V, if its efficiency is 80%? from water = 4200 J / Kg * ° C

Data: m (mass of heated water) = 1.5 kg; t0 (initial temperature) = 20 ºС; t (final temperature) = 100 ºС; I (current) = 0.2 A; U (operating voltage) = 220 V; η (kettle efficiency) = 80% (0.8).

Constants: Sv (heat capacity of water) = 4200 J / (kg * ºС).

To calculate the duration of water heating, we use the formula: η = Cw * m * (t – t0) / (U * I * t), whence we express: t = Cw * m * (t – t0) / (U * I * η ).

Calculation: t = 4200 * 1.5 * (100 – 20) / (220 * 0.2 * 0.8) = 14318 s ≈ 239 min ≈ 4 hours.