An electric stove with a current of 3A for 5s released 3600J of heat. What is the spiral resistance of the tile.

Initial data: I (current in the electric stove) = 3 A; t (duration of the hotplate) = 5 s; Q (released amount of heat on the spiral) = 3600 J.

The resistance of the electric stove spiral is determined using the Joule-Lenz law: Q = I ^ 2 * R * t, whence R = Q / (I ^ 2 * t).

Let’s calculate: R = 3600 / (3 ^ 2 * 5) = 80 Ohms.

Answer: The coil of the hotplate has a resistance of 80 ohms.