The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
`R = R_(0) [1 + alpha (T - T_(0))]`
The resistance is `101.6 Omega` at the triple-point of water `273.16 K , " and " 165.5 Omega` at the normal melting point of lead `(600.5 K)`. 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 information:
- Resistance at the triple point of water: \( R_0 = 101.6 \, \Omega \)
- Temperature at the triple point of water: \( T_0 = 273.16 \, K \)
- Resistance at the melting point of lead: \( R_1 = 165.5 \, \Omega \)
- Temperature at the melting point of lead: \( T_1 = 600.5 \, K \)
...