The spiral wire is connected to a 120 V network. Its resistance is 80 Ohm. How much heat will this spiral release in 0.5 minutes?

Given:

U = 120 Volts – voltage of the electrical network;

R = 80 Ohm – resistance of the wire spiral;

t = 0.5 minutes – time interval.

It is required to determine Q (Joule) – the amount of heat that this spiral will release during the time interval t.

Let’s convert time units from minutes to seconds:

t = 0.5 minutes = 0.5 * 60 = 30 seconds.

Determine the current strength (Ohm’s law):

I = U / R = 120/80 = 1.5 Amperes.

Then, to determine the amount of energy, you must use the following formula:

Q = I * t = 1.5 * 30 = 45 Joules.

Answer: In 0.5 minutes, the spiral will release an energy equal to 45 Joules.