In the copper bath, 1.98 grams of copper were released in 20 minutes. Determine the power of the electrolyte going

In the copper bath, 1.98 grams of copper were released in 20 minutes. Determine the power of the electrolyte going to heating the resistance of the bath solution of 0.8 Ohm. Electrical equivalent of copper 0.329 * 10 ^ -6

Initial data: t (duration of copper evolution) = 20 min (1200 s); m (mass of released copper) = 1.98 g (1.98 * 10 ^ -3 kg); R (bath solution resistance) = 0.8 ohm.

Reference values: k (electrochemical equivalent of copper) = 32.9 * 10 ^ -8 kg / C.

1) Determine the current in the bath using the first Faraday’s law: m = k * I * t, whence I = m / (k * t) = 1.98 * 10 ^ -3 / (32.9 * 10 ^ -8 * 1200) ≈ 5 A.

2) Let’s calculate the current power: P = I ^ 2 * R = 5 ^ 2 * 0.8 = 20 W.

Answer: The current power is 20 W.