A wire spiral, the resistance of which in a heated state is 55 ohms, is connected to a network

A wire spiral, the resistance of which in a heated state is 55 ohms, is connected to a network with a voltage of 127 V. What amount of heat does this spiral generate in 1 minute?

Given:

R = 55 Ohm – resistance of the wire spiral;

U = 127 Volts – voltage of the electrical network;

t = 1 minute = 60 seconds – time span.

It is required to determine Q (Joule) – the amount of heat that the spiral emits over a period of time t.

Determine the strength of the current in the spiral:

I = U / R.

Then the power of the spiral will be equal to:

W = U * I = U * U / R = U ^ 2 / R.

The amount of heat will be equal to:

Q = W * t = U2 * t / R = 127 ^ 2 * 60/55 = 16129 * 60/55 = 967740/55 = 17595 Joules (approximately 17.6 kJ).

Answer: 17.6 kJ will be released through the spiral in one minute