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

The Correct Option is B

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

Provided Data:
\( R_0 = 8 \, \Omega \) (resistance at 0°C)
\( R_{100} = 10 \, \Omega \) (resistance at 100°C)
Hot bath temperature: \( T = 400°C \)

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

Resistance Calculation at 400°C:
Applying \( T = 400°C \):
\[ R_{400} = R_0(1 + \alpha \times 400) \]
Substituting \( R_0 = 8 \, \Omega \) and \( \alpha = 0.0025 \):
\[ R_{400} = 8 \times (1 + 0.0025 \times 400) \]
\[ R_{400} = 8 \times (1 + 1) = 8 \times 2 = 16 \, \Omega \]

Result:
The resistance of the platinum wire at 400°C is \( 16 \, \Omega \).