A wire 3.00 m long and .450 mm2 in cross-sectional area has a resistance of 41.0Ω at 20 degrees C. If its resistance increases to 41.4Ω at 29.0 C, what is the temperature coefficient of resistivity?
why are you posting under different names?
changeresistane=tempcoeff*deltTemp
.4=tempcoeff*9C
solve for tempcoefficent
oh its a group of six friends working together and i didn't know they all posted it twice...sorry about that
To find the temperature coefficient of resistivity (α), we can use the formula:
α = (1/R) * ((ΔR) / (ΔT))
Where:
- R is the resistance at the initial temperature (20 °C)
- ΔR is the change in resistance (41.4 Ω - 41.0 Ω)
- ΔT is the change in temperature (29.0 °C - 20 °C)
Let's plug in the values and calculate:
R = 41.0 Ω
ΔR = 41.4 Ω - 41.0 Ω = 0.4 Ω
ΔT = 29.0 °C - 20 °C = 9.0 °C
α = (1/41.0 Ω) * (0.4 Ω / 9.0 °C)
First, we calculate (0.4 Ω / 9.0 °C):
ΔR/ΔT = 0.044 Ω/°C
Then, we calculate (1/41.0 Ω) * (0.044 Ω/°C):
α = 0.001073 Ω/°C
Therefore, the temperature coefficient of resistivity (α) is approximately 0.001073 Ω/°C.