# What amount of heat is released in 2 hours in the filament of an electric lamp with a resistance of 25 Ohm

What amount of heat is released in 2 hours in the filament of an electric lamp with a resistance of 25 Ohm if the current in it is 0.2 A?

Data: Δt (duration of the taken electric lamp) = 2 h (7200 s); R (filament resistance) = 25 ohms; I (current in the filament of the electric lamp taken) = 0.2 A.

According to the Joule-Lenz law, the amount of heat that will have time to be released during the operation of the electric lamp can be determined by the formula: Q = I ^ 2 * R * Δt.

Calculation: Q = 0.2 ^ 2 * 25 * 7200 = 7200 J (7.2 kJ).

Answer: For 2 hours of operation of an electric lamp, 7.2 kJ of heat will be released.

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