Ice weighing 2 kg, at the melting temperature, is melted on an electric stove. In the hotplate

Ice weighing 2 kg, at the melting temperature, is melted on an electric stove. In the hotplate, two spirals with a resistance of 20 ohms are connected in series and connected to a 220 V network. All the ice melted in 12 minutes. Determine the efficiency of the process.

The tile resistance R is equal to the sum of two 20 Ohm resistances connected in series:
R = 40 ohms.
Effective current through the tile:
I = U / R = 220 V / 40 Ohm = 5.5 A.
Tile power: P = U * I = 220 V * 5.5 A = 1210 W.
Energy consumed by the tiles:
W = P * t = 1210 W * (12 min * 60 s / min) = 1210 W * 720 s = 871200 J.
The amount of heat used to melt the ice:
Q = L * m = 3.4 * 10 ^ 5 J / kg * 2 kg = 680,000 J,
where L is the specific heat of ice melting.
The efficiency is equal to the ratio of the amount of heat consumed to melt ice to the total energy expended:
K = (Q / W) * 100% = (680,000 J / 871200 J) * 100% = 78%.
Answer: 78%