A wire spiral, the resistance of which in a heated state is 55 Ohm, is included in a network with a voltage

A wire spiral, the resistance of which in a heated state is 55 Ohm, is included in a network with a voltage of 127 V. What amount of heat does this spiral release in 1 minute; in 0.5 h?

According to the Joule-Lenz law, the amount of heat that will be released on the wire spiral:
Q = U ^ 2 * t / R, where U is the voltage in the network (U = 127 V), t is the time of passage of the current (s), R is the resistance of the wire spiral (R = 55 Ohm).
The amount of heat in 1 min (t1 = 1 min = 60 s):
Q1 = U ^ 2 * t1 / R = (127 ^ 2) * 60/55 = 17595 J ≈ 17.6 kJ.
The amount of heat in 0.5 h (t2 = 0.5 min = 0.5 * 60 * 60 s = 1800 s):
Q2 = U ^ 2 * t2 / R = (127 ^ 2) * 1800/55 = 527858 J ≈ 527.9 kJ.
Answer: 17.6 kJ and 527.9 kJ of heat will be released.