The electric iron warmed up in 2 minutes from a network with a voltage of 220 V at a current of 1 A.

The electric iron warmed up in 2 minutes from a network with a voltage of 220 V at a current of 1 A. How many conduction electrons passed through the cross section of the wire of the heating element during this time and what energy was released?

t = 2 min = 120 s.
U = 220 V.
I = 1 A.
q = 1.6 * 10 ^ -19 Cl.
N -?
Q -?
According to the Joule-Lenz law, the amount of heat Q in a conductor is determined by the formula: Q = I * U * t, where I is the current in the conductor, U is the voltage at the ends of the conductor, t is the time of passage of the electric current.
Q = 1 A * 220 V * 120 s = 26400 J.
According to the definition, the current strength is determined by the formula: I = Q / t, where Q is the electric charge that has passed through the cross section of the conductor, t is the charge transit time.
Q = N * q.
I = N * q / t.
N = I * t / q.
N = 1 A * 120 s / 1.6 * 10 ^ -19 Cl = 75 * 10 ^ 19.
Answer: Q = 26400 J of thermal energy will be released, N = 75 * 10 ^ 19 electrons will pass.