What amount of heat will the 30 Ohm wire spiral release in 5 minutes if the current is 2 A?

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 = 2 A), R is the resistance of the wire spiral (R = 30 Ohm), t is the time the current passes along the spiral (t = 5 min = 5 * 60 s = 300 s).
Let’s calculate the amount of heat:
Q = I ^ 2 * R * t = (2 ^ 2) * 30 * 300 = 36000 J = 36 kJ.
Answer. Heat will be released on the wire spiral in the amount of 36 kJ.