Find the specific resistance of the substance, of which the spiral length is 0.3 m

Find the specific resistance of the substance, of which the spiral length is 0.3 m, the cross-sectional area is 0.5 mm squared, the voltage is 220 volts and the power is 500 watts.

1. We will not translate the cross-sectional area into the si system, because in reference books the resistivity is given in Ohm · mm2 / m
2. Let’s write the expression for determining the resistance of the wire:
R = ρ * l / Ssection, where ρ is the resistivity of the wire, l is the length of the wire, Ssection is the cross-sectional area.
3. Let us express the required value from it:
ρ = Ssection * R / l
4. Let’s write down expressions for determining the power of the heater:
P = U * I, where I is the current consumed, U is the voltage.
Let’s write the expression for the current I:
I = U / R, substituting into the formula for power, we have:
P = (U ^ 2 / R), we express the resistance:
R = (U ^ 2 / P)
Substitute in the formula from point 3:
ρ = Ssection * R / l = ρ = (Ssection * U ^ 2) / (P * l)
Now we substitute the numbers, we get:
ρ = (0.5 * 220 ^ 2) / (500 * 0.3) = 161.3 Ohm mm2 / m
Answer: 161.3 Ohmmm2 / m