A tungsten wire has a radius of 0.073 mm and is heated from 20.0 to 1500 °C. The temperature coefficient of resistivity is α = 4.5 10-3 (C°)−1. When 110 V is applied across the ends of the hot wire, a current of 1.6 A is produced. How long is the wire? Neglect any effects due to thermal expansion of the wire.
Resistance of wire at 1500°C is R = 110V/1.6A or R = 68.75 ohms
Electrical resistivity of tungsten at 20°C is rho20 = 5.5*10-8 ohm.m
Electrical resistivity of tungsten at 1500°C is rho1500 = rho20*(1 + alpha*deltaT)
rho1500 = 5.5*10-8*(1 + 4.5*10-3*(1500-20)) = 4.213*10-7 ohm.m
Length of tungsten wire at 1500°C L = rho1500/R or L = 4.213*10-7 ohm.m/68.75ohms ≈ 6*10-9 m
Hmm! Seems very short! Someone needs to check this.
Resistance of wire at 1500°C is R = 110V/1.6A or R = 68.75 ohms
Electrical resistivity of tungsten at 20°C is rho20 = 5.5*10-8 ohm.m
Electrical resistivity of tungsten at 1500°C is rho1500 = rho20*(1 + alpha*deltaT)
rho1500 = 5.5*10-8*(1 + 4.5*10-3*(1500-20)) = 4.213*10-7 ohm.m
Length of tungsten wire at 1500°C L = rho1500/R or L = 4.213*10-7 ohm.m/68.75ohms ≈ 6*10-9 m
Hmm! Seems very short! Someone needs to check this.
Solution:
\(R = \dfrac{110V}{1.6A} = 68.75 {~ohms} \\ \rho_{20} = 55.0*10^{-9} \Omega \cdot M \\ \alpha = \text {Temperature coefficient for tungsten} = 0.0045K^{-1} (K^{-1} \equiv C^{-1})\\ \rho_{1500}= (\rho_{20})*[1+(\alpha\Delta T)]\\ \rho_{1500}= (55.0*10^{-9})*(1+0.0045 C^{-1})*(1500-20)C =8.287125*10^{-5} \text {M} \cdot \Omega \\ A_c = (\pi * r^2) = \pi * (0.073*10^{-6})^2 = 1.67415473*10^{-8} M^2 \text {(Area cross-section)} \\\\ \)
\(R= \rho *(L*A) \rightarrow \text {Solve for length L. } \\ L = \dfrac {RA} {(\rho_{20}*[1 + \alpha * \Delta T])} \\ \hspace {10 mm} \\ \dfrac {(68.75)*(16.7415473*10^{-9})} {(55.0*10^{-9})*(1 + [0.0045 *1480])} =2.73198 ~\text {Meters} \\ \)