Q.The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5\ \Omega$ and at steam point is $5.23\ \Omega$. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795\ \Omega$. Calculate the temperature of the bath.

Concept: Temperature Dependence of Resistance — for platinum, resistance varies nearly linearly with temperature over this range.

Reasoning:

1. For a platinum resistance thermometer, the temperature $t$ (in °C) is related to resistance $R_t$ by:

where $R_0$ is resistance at $0^\circ\text{C}$ and $\alpha$ is the temperature coefficient.

2. At ice point ($0^\circ\text{C}$): $R_0 = 5\ \Omega$.

   At steam point ($100^\circ\text{C}$): $R_{100} = 5.23\ \Omega$.

3. Find $\alpha$:

   $5.23 = 5(1 + 100\alpha) \quad\Rightarrow\quad \alpha = \frac{5.23 - 5}{5 \times 100} = \frac{0.23}{500} = 4.6 \times 10^{-4}\ {^\circ\text{C}}^{-1}$ …

The key idea here is that platinum resistance thermometers rely on the predictable, nearly linear increase in resistance with temperature. For a pure metal like platinum, over a moderate temperature range, the resistance changes according to:

where $R_0$ is the resistance at $0^\circ\text{C}$ (ice point), $R_T$ is the resistance at temperature $T$ (in °C), and $\alpha$ is the temperature coefficient of resistance — a constant for the material.

We are given two calibration points: the ice point ($0^\circ\text{C}$, $R_0 = 5\ \Omega$) and the steam point ($100^\circ\text{C}$, $R_{100} = 5.23\ \Omega$). These let us determine $\alpha$ for this specific wire. Once $\alpha$ is known, any measured resistance can be converted directly to temperature.

Let’s work through it step by step.

1. Find the temperature coefficient $\alpha$ At the steam point, $T = 100^\circ\text{C}$ and $R_{100} = 5.23\ \Omega$. Using the formula:

Substitute the known values:

Divide both sides by 5:

Subtract 1:

So:

2. Set up the equation for the unknown temperature The hot bath gives a resistance $R = 5.795\ \Omega$. Using the same linear relation: $R = R_0(1 + \alpha T)$ …

Method: Linear Interpolation Using the Platinum Resistance Thermometer (PRT) Relation

This method uses the linear approximation for resistance vs. temperature in platinum resistance thermometers over a limited range (0°C to 100°C). The relation is:

where:

• $R_t$ = resistance at temperature $t$ (°C)
• $R_0$ = resistance at 0°C (ice point)
• $\alpha$ = temperature coefficient of resistance (assumed constant)

Steps

Step 1: Identify given data

• $R_0 = 5\ \Omega$ (at ice point, $0^\circ$C)
• $R_{100} = 5.23\ \Omega$ (at steam point, $100^\circ$C)
• $R_t = 5.795\ \Omega$ (at unknown temperature $t$)

Step 2: Find $\alpha$ using the steam point data

From $R_{100} = R_0 (1 + \alpha \cdot 100)$:

Step 3: Use the same relation for the unknown temperature

$5.795 = 5 (1 + 0.00046 \cdot t)$ …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula for resistance variation

Many students directly apply the linear approximation:

Why this is wrong:

This formula assumes resistance varies linearly with temperature from $0^\circ C$. But the given data has two known points (ice point = $0^\circ C$, steam point = $100^\circ C$). You must use the two-point calibration formula:

How to avoid:

Always check: are you given $R_0$ and $\alpha$, or two known temperatures with their resistances? For a platinum resistance thermometer, use the ratio method unless $\alpha$ is explicitly given.

Mistake 2: Confusing ice point and steam point values

Students sometimes swap $R_0$ and $R_{100}$.

Correct assignment:

• Ice point ($0^\circ C$) → $R_0 = 5\ \Omega$
• Steam point ($100^\circ C$) → $R_{100} = 5.23\ \Omega$

How to avoid:

Memorise: Ice = 0°C, Steam = 100°C. Write them clearly before substituting.

Mistake 3: Arithmetic errors in the ratio

A common slip:

Then forgetting to multiply by 100:

How to avoid:

Write the full formula step-by-step:

Then substitute after simplifying the fraction.

Mistake 4: Not checking if the answer is reasonable …