A current flows through the boiler, the strength of which is equal to 3A. Determine the amount
| April 3, 2021 | education
A current flows through the boiler, the strength of which is equal to 3A. Determine the amount of heat released by the coil of the boiler for 20 minutes of operation if the electrical resistance of the coil is 30 Ohm
Initial data: I (current that passes through the boiler) = 3 A; t (duration of the boiler operation) = 20 minutes (1200 s); R (resistance of the coil of the boiler) = 30 ohms.
The amount of heat that will be released on the coil of the boiler is determined by the Joule-Lenz law: Q = I ^ 2 * R * t.
Let’s make the calculation: Q = 3 ^ 2 * 30 * 1200 = 324,000 J (324 kJ).
Answer: 324 kJ of heat will be released on the coil of the boiler.
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