# The electric furnace, which has a spiral made of nickel wire with a cross section of 1.7 mm2 and a length of 51 m

**The electric furnace, which has a spiral made of nickel wire with a cross section of 1.7 mm2 and a length of 51 m, is connected to a network with a voltage of 220 V. Determine the power of the furnace and the amount of heat generated by the heating element in 1 hour. Nickelin resistivity is 0.4 Ohm * mm2 / m**

S = 1.7 mm2.

L = 51 m.

U = 220 V.

t = 1 h = 3600 s.

ρ = 0.4 Ohm * mm2 / m.

N -?

Q -?

The power of the electric current N is determined by the formula: N = U ^ 2 / R, where U is the voltage of the current, R is the resistance of the spiral of the electric furnace.

R = ρ * L / S, where ρ is the resistivity of nickelin, L is the length of the spiral, S is the cross-sectional area of the spiral.

N = S * U ^ 2 / ρ * L.

N = 1.7 mm2 * (220 V) ^ 2 / 0.4 Ohm * mm2 / m * 51 m = 4033 W.

The amount of heat in the Q spiral is expressed by the Joule-Lenz law: Q = U ^ 2 * t / R = S * t * U2 / ρ * L.

Q = 1.7 mm2 * 3600 J * (220 V) ^ 2 / 0.4 Ohm * mm2 / m * 51 m = 14518800 J.

Answer: the power of the electric furnace is N = 4033 W, the amount of heat is released in it Q = 14518800 J.