What amount of heat will be released in 30 minutes by a 20 Ohm wire spiral with a current of 5 A?
| March 20, 2021 | education
According to the Joule-Lenz law, the amount of heat that is released on the wire spiral can be calculated using the formula:
Q = I ^ 2 * R * t, where I is the current strength (I = 5 A), R is the resistance of the wire spiral (R = 20 Ohm), t is the time of passage of the electric current (t = 30 min = 30 * 60 s = 1800 s).
Let’s calculate the released amount of heat:
Q = I ^ 2 * R * t = 5 ^ 2 * 20 * 1800 = 25 * 20 * 1800 = 900000 J = 900 kJ.
Answer: 900 kJ of heat will be released.
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