Question

The coil of a resistance thermometer is made of platinum wire and has a resistance of 45 at 20° C. When immersed in a vessel of a boiling liquid the resistance increases to 90. Calculate the boiling point of the liquid. [Temperature coefficient of platinum wire is 3.92 × 10–3 °C– 1]

Answer 1

QUESTION:

The coil of a resistance thermometer is made of platinum wire and has a resistance of 45 Ohm at $20{}^{\circ}$ C. When immersed in a vessel of a boiling liquid the resistance increases to 90 Ohm.

Calculate the boiling point of the liquid. [Temperature coefficient of platinum wire is $3.92 \times 10^{-3}$ °C$^{-1}$]

SOLUTION:

The resistance R of metal wire depends on temperature according to the next law:

$$R(t) = R_0 (1 + \alpha \cdot t)$$

Where

$R_0$ is the resistance of metal wire at $0{}^{\circ}$ C

t is the temperature

$\alpha$ is the temperature coefficient of resistance.

So, at $20{}^{\circ}$ C the resistance of a coil is 45 Ohm, hence

$$45 = R_0 (1 + 3.92 \cdot 10^{-3} \cdot 20)$$ $$R_0 = \frac{45}{1.0784} = 41.728 \text{ Ohm}$$

When immersed in a boiling liquid the resistance of a coil is 90 Ohm, so

$$90 = 41.728 (1 + 3.92 \cdot 10^{-3} \cdot t_{\text{boiling}})$$ $$\frac{90}{41.728} = 1 + 3.92 \cdot 10^{-3} \cdot t_{\text{boiling}}$$ $$\frac{90}{41.728} - 1 = 3.92 \cdot 10^{-3} \cdot t_{\text{boiling}}$$ $$1.1568 = 3.92 \cdot 10^{-3} \cdot t_{\text{boiling}}$$ $$t_{\text{boiling}} = \frac{1.1568}{3.92 \cdot 10^{-3}} = 295.1{}^{\circ} \text{C}$$

Maybe the liquid is trinitrotoluene.

ANSWER:

295.1 °C