While repairing the spiral of the electric stove, it was necessary to shorten it by 0.2 lengths.

While repairing the spiral of the electric stove, it was necessary to shorten it by 0.2 lengths. How many times has the amount of heat released in 1 s increased?

These tasks: lр (length of the spiral after the repair) = 0.8ln (initial length of the spiral).

To find out the change in heat release by the indicated spiral (the Joule-Lenz law, but we take into account that only the mains voltage is constant), we use the relation: k = Q2 / Q1 = (U ^ 2 * t / R2) / (U ^ 2 * t / R1 ) = R1 / R2 = (ρ * ln / S) / (ρ * lр / S) = ln / lр.

Let’s perform the calculation: k = ln / lr = ln / 0.8ln = 1 / 0.8 = 1.25 p.

Answer: The release of heat should have increased by 1.25 times.