Answer:
a) 3.162 m
b) 339.7 W
Explanation:
Assume ρ = 1.50*10^-6 Ωm, and
α = 4.000 10-4(°C)−1 for Nichrome
To solve this, we would use the formula
P = V² / R
So when we rearrange and make R subject of formula, we have
R = V² / P
Resistance of the heating coil, R
R = (110² / 500)
R = 12100 / 500
R = 24.2 ohms
Recall the formula for resistivity of a wire
R = ρ.L/A
Again, in rearranging and making L subject of formula, we have
L = R.A / ρ
To make it uniform, we convert our radius from mm to m.
Diameter, D = 0.5 mm
Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m
We then use this radius to find our area
A = πr²
A = π * 0.00025²
A = 1.96*10^-7 m²
And finally, we solve for L
L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =
L = 3.162 m
(b)
Temperature coefficient of resistance.
R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]
R₁₂₀₀ = R₂₀[1 + α(1180) ]
R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]
R₁₂₀₀ = 24.2[1 + 0.472]
R₁₂₀₀ = 24.2 * 1.472
R₁₂₀₀ = 35.62 ohms
Putting this value of R in the first formula from part a, we have
P = V² / R
P = (110² / 35.62)
P = 12100/ 35.62
P = 339.70 watts