When electric current passed through the heater coil, made of 80 m long nicklein wire and 0.84 mm2

When electric current passed through the heater coil, made of 80 m long nicklein wire and 0.84 mm2 cross-sectional area, 726,000 J of heat was released in 10 minutes. What is the voltage of the network to which the heater was turned on?

Given: l (length of nickel wire) = 80 m; S (wire section) = 0.84 mm2; t (heater operation time) = 10 min (in SI t = 600 s); Q (released heat) = 725000 J.

Constants: ρ (nickelin resistivity) = 0.4 Ohm * mm2 / m.

To determine the voltage in the network, we use the equality: U2 * t / Q = R = ρ * l / S and U = √ (ρ * l * Q / (S * t)).

Let’s calculate: U = √ (0.4 * 80 * 725000 / (0.84 * 600)) = 214.55 V.

Answer: The voltage in the network is 214.55 V.