The electric lamp emits 900 kJ of heat in 2.5 hours of operation. Determine the power of the lamp

The electric lamp emits 900 kJ of heat in 2.5 hours of operation. Determine the power of the lamp and the amperage in it if the voltage is 200 V.

t = 2.5 h = 9000 s.
Q = 900 kJ = 900000 J.
U = 200 V.
N -?
I -?
According to the Joule-Lenz law, the amount of heat Q that is released in a conductor through which an electric current flows is determined by the formula: Q = I ^ 2 * U * t. Where I is the current in the conductor, U is the voltage at the ends of the conductor, t is the current flow time.
I = √ (Q / U * t).
I = √ (900000 J / 200 V * 9000 s) = 0.7 A.
The power of the lamp N, through which the electric current flows, is determined by the formula: N = I * U.
N = 0.7 A * 200 V = 140 W.
Answer: lamp power N = 140 W, current I = 0.7 A.