In the spiral of an electric stove plugged into a 220 V socket, 690 kJ of heat was released
| May 17, 2021 | education
In the spiral of an electric stove plugged into a 220 V socket, 690 kJ of heat was released at a current of 3.5 A. How long has the tile been plugged in?
Task data: U (mains voltage) = 220 V; I (current in the coil) = 3.5 A; Q (released heat) = 690 kJ (690 * 10 ^ 3 J).
The duration of switching on the used electric stove into the network is determined using the Joule-Lenz law: Q = U * I * t, whence t = Q / (U * I).
Let’s make a calculation: t = 690 * 10 ^ 3 / (220 * 3.5) = 896 s = 14 min 56 s.
Answer: The tile used was switched on for 14 minutes 56 seconds.
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