What amount of heat will be released in 30 minutes by a wire spiral with a resistance of 20 ohms at a current of 5 A?
| August 6, 2021 | education
The amount of heat that will be released by a wire spiral in 30 minutes can be calculated using the Joule-Lenz law:
Q = I² * R * t, where I is the value of the current on the wire spiral (I = 0.5 A), R is the resistance value of the wire spiral (R = 20 Ohm), t is the operating time of the electric current (t = 30 min = 1800 s).
Let’s calculate the amount of heat:
Q = I² * R * t = 5² * 20 * 1800 = 900,000 J = 900 kJ.
Answer: Within 30 minutes, the spiral wire will release 900 kJ of heat.
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