The electric cooker is connected to a 220V network. the current flowing in the spiral of the tile is 5A

The electric cooker is connected to a 220V network. the current flowing in the spiral of the tile is 5A. it is required to determine how much heat will be released by that in 30 mt

Initial data: U (mains voltage) = 220 V; I (the current that flows along the spiral of the tile in question) = 5 A; t (tile duration) = 30 min (1800 s).

The heat that the tile spiral will release is determined by the formula: Q = U * I * t.

Let’s make the calculation: Q = 220 * 5 * 1800 = 1,980,000 J (1.98 MJ).

Answer: The spiral of the tile will release 1.98 MJ of heat in 30 minutes.