How much heat will be released in 20 seconds in a 40 Ohm coil if the current in it is 3.1 A?
| February 16, 2021 | education
Data: t (duration of current passage through the spiral) = 20 s; R (resistance of the taken spiral) = 40 Ohm; I (value of the current in the spiral) = 3.1 A.
The amount of heat released on the spiral can be calculated using the Joule-Lenz law: Q = I ^ 2 * R * t.
Calculation: Q = 3.1 ^ 2 * 40 * 20 = 7688 J.
Answer: In 20 seconds, 7688 J of heat will be released in the spiral.
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