# An electric soldering iron with a resistance of R = 48 Ohm is powered by a current of I = 4A.

**An electric soldering iron with a resistance of R = 48 Ohm is powered by a current of I = 4A. How long does it take for its copper part weighing m = 0.2 kg to heat up to t = 500C if the efficiency of the soldering iron is η = 50%?**

Initial data: R (resistance of an electric soldering iron) = 48 Ohm; I (current in the soldering iron) = 4 A; m (mass of the copper part of the soldering iron) = 0.2 kg; Δt (change in temperature of the copper part) = 500 ºС; η (soldering iron efficiency) = 50% (0.5).

Reference data: Cm (specific heat capacity of copper) = 400 J / (kg * ºС).

The operating time of an electric soldering iron is expressed from the formula: η = Qp / Qz = Cm * m * Δt / (I ^ 2 * R * t), whence t = Cm * m * Δt / (I ^ 2 * R * η).

Calculation: t = 400 * 0.2 * 500 / (4 ^ 2 * 48 * 0.5) ≈ 104.2 s.