Question

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 \, \Omega$ and $10 \, \Omega$ respectively. After inserting in a hot bath of temperature $400^\circ \text{C}$, the resistance of the platinum wire is:

• A. $2 \, \Omega$
• B. $16 \, \Omega$
• C. $8 \, \Omega$
• D. $10 \, \Omega$

Answer 1

Answer: (B)

$16 \, \Omega$

Answer 2

Use the Temperature-Resistance Relationship for a Platinum Resistance Thermometer:
The resistance $R$ of a platinum resistance thermometer at a temperature $T$ (in °C) is given by:
$R_T = R_0(1 + \alpha \Delta T)$ where:
$R_0$ is the resistance at 0°C (ice point),
$\alpha$ is the temperature coefficient of resistance of platinum,
$\Delta T$ is the temperature difference from the ice point.

Given Data:
$R_0 = 8 \, \Omega$ (resistance at 0°C)
$R_{100} = 10 \, \Omega$ (resistance at 100°C)
Temperature of the hot bath: $T = 400°C$

Calculate the Temperature Coefficient $\alpha$:
Using the resistance values at 0°C and 100°C:
$R_{100} = R_0(1 + 100\alpha)$
Substitute $R_{100} = 10 \, \Omega$ and $R_0 = 8 \, \Omega$:
$10 = 8(1 + 100\alpha)$ $1 + 100\alpha = \frac{10}{8} = 1.25$
$100\alpha = 0.25$ $\alpha = \frac{0.25}{100} = 0.0025 \, °C^{-1}$

Calculate the Resistance at 400°C:
Now, using $T = 400°C$:
$R_{400} = R_0(1 + \alpha \times 400)$
Substitute $R_0 = 8 \, \Omega$ and $\alpha = 0.0025$:
$R_{400} = 8 \times (1 + 0.0025 \times 400)$
$R_{400} = 8 \times (1 + 1) = 8 \times 2 = 16 \, \Omega$

Conclusion:
The resistance of the platinum wire at 400°C is $16 \, \Omega$.