What amount of heat will be released in a 20 ohm wire spiral at a current strength of 5A per 100 s?

To find out the required amount of heat released in the indicated wire spiral, we use the formula: Q = I ^ 2 * R * t, where I is the current in the wire spiral (I = 5 A); R – resistance (R = 20 Ohm); t – the duration of the spiral (t = 100 s).

Let’s perform the calculation: Q = I ^ 2 * R * t = 5 ^ 2 * 20 * 100 = 5 * 10 ^ 4 J.

Answer: The amount of heat released in the indicated wire spiral is 50 kJ.