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 step by step, we will follow the given formula for resistance and the provided data points to find the temperature when the resistance is 123.4 ohms.

### Step 1: Identify the known values
We have the following known values:
- Resistance at the triple point of water (R₀): 101.6 Ω at T₀ = 273.16 K
- Resistance at the normal melting point of lead (R₁): 165.5 Ω at T₁ = 600.5 K
- Resistance we want to find the temperature for (R): 123.4 Ω

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