Answer:
1176.01 °C
Explanation:
Using Ohm's law,
V = IR................. Equation 1
Where V = Voltage, I = current, R = Resistance when the bulb is on
make R the subject of the equation
R = V/I.................. Equation 2
R = 4.3/0.32
R = 13.4375 Ω
Using
R = R'(1+αΔθ)............................. Equation 3
Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature
make Δθ the subject of the equation
Δθ = (R-R')/αR'.................. Equation 4
Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹
Substitute into equation 4
Δθ = (13.4375-1.6)/(1.6×0.0064)
Δθ = 11.8375/0.01024
Δθ = 1156.01 °C
But,
Δθ = T₂-T₁
T₂ = T₁+Δθ
Where T₂ and T₁ = Final and initial temperature respectively.
T₂ = 20+1156.01
T₂ = 1176.01 °C
Answer: 1.16*10^3°C
Explanation:
It is known that resistance depends on temperature
Recalling ohms law of v = ir
R = V/I
R = 4.3/0.32
R = 13.44
R = R•(1 + α(T - T•))
13.44 = 1.6(1 + 6.5*10^-3(T - 20))
13.44/1.6 = 1 + 6.5*10^-3T - 0.13
8.4 = 0.87 + 0.0065T
7.53 = 0.0065T
T = 1158.46°C
T = 1.16*10^3°C
