Wire spiral, resistance = 55 ohms, 110 volts. How much heat does this coil release in 1 minute? in half an hour?

Given: R (coil resistance) = 55 Ohm; U (voltage) = 110 V; t1 (first time of operation) = 1 min (in SI t1 = 60 s); t2 (second operating time) = 30 min (in SI t2 = 1800 s).

1) Heat released on the considered wire spiral in 1 min: Q1 = U ^ 2 * t1 / R = 110 ^ 2 * 60/55 = 13.2 * 10 ^ 3 J (13.2 kJ).

2) Heat released in 30 minutes: Q2 = U ^ 2 * t ^ 2 / R = 110 ^ 2 * 1800/55 = 396 * 10 ^ 3 J (396 kJ).

Answer: For 1 minute and 30 minutes, 13.2 kJ and 396 kJ of thermal energy will be released, respectively.