The electrical resistance in ohms of a certain thermometer varies with temperature ac cording to the approximate law: `R =R_(0)[1+alpha(T-T_(0))]`
The resistances is `101.6 Omega` at the triple-point of water `273.16K`, and `165.5 Omega` at the normal melting point of lead `(600.5K)`. What is the temperature when the resistance is `123.4 Omega`?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
We have the following resistances and their corresponding temperatures:
- At the triple point of water (T0 = 273.16 K), R0 = 101.6 Ω
- At the normal melting point of lead (T1 = 600.5 K), R1 = 165.5 Ω
- We need to find the temperature (T2) when the resistance is R2 = 123.4 Ω.

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