Question

hyy ans this friends❗

Answer 1

Using linear approximation for resistance-temperature dependence, and applying two point form,

$\dfrac{5.795-5}{T-0}=\dfrac{5.23-5}{100-0}$

Solving it, we get

$T=345^{\circ}C$

Alternatively, Use  

$R_{100}=R_0(1+\alpha\Delta T)\\\Rightarrow 5.23=5(1+\alpha (100-0))\\5.795=5(1+\alpha (T-0))$

Solving above two equations, you get

$T=345^{\circ}C$