A wire spiral, the resistance of which in a heated state is 60 ohms, is connected to a network with a voltage of 120 V.

A wire spiral, the resistance of which in a heated state is 60 ohms, is connected to a network with a voltage of 120 V. What amount of heat will this spiral release in 2.5 minutes?

To determine the amount of thermal energy released by a given wire spiral, we apply the formula: Q = U ^ 2 * t / R.

Values of variables: U – mains voltage (U = 120 V); t is the time of turning on the spiral into the network (t = 2.5 min; in SI t = 150 s); R – resistance (R = 60 Ohm).

Let’s perform the calculation: Q = U ^ 2 * t / R = 120 ^ 2 * 150/60 = 36 * 10 ^ 3 J (36 kJ).

Answer: In 2.5 minutes, a given wire spiral should have emitted 36 kJ of thermal energy.