A wire spiral whose resistance in a heated state is 55 ohms is connected to a network with a voltage of 127 V
A wire spiral whose resistance in a heated state is 55 ohms is connected to a network with a voltage of 127 V. How much heat does the spiral release in 1 minute?
R = 55 ohms.
U = 127 V.
t = 1 min = 60 s.
Q -?
In 1840, the English scientist D. Joule and in 1842 the Russian scientist Lenz experimentally established a law according to which, in a conductor through which a current flows, an amount of heat Q is released.
The value of the amount of heat Q is determined by the formula: Q = U ^ 2 * t / R, where U is the voltage at the ends of the conductor, t is the time the current passes through the conductor, R is the resistance of the conductor.
Q = (127 V) ^ 2 * 60 s / 55 Ohm = 17595 J.
Answer: Q = 17595 J of thermal energy was released in the wire spiral.
