A thermometer graduated according to a linear scale reads a value x_0, when in contact with boiling

Q: Why do we use the ratio of temperature differences instead of just direct conversion?

For any linear scale, the physical expansion (like the length of a mercury column) is directly proportional to the temperature change. Taking the ratio of differences cancels out the proportionality constant, allowing us to compare different scales directly without needing to know the specific physical properties of the thermometer.

Q: What are the fixed points for the Celsius scale? The question didn't provide them.

The fixed points for the Celsius scale are standard knowledge you are expected to know. The lower fixed point (ice point) is $0^\circ\text{C}$, and the upper fixed point (steam point) is $100^\circ\text{C}$.

Q: Can I use this method to convert Celsius to Fahrenheit?

Absolutely! This is the exact principle used to derive the conversion formula. For Fahrenheit, the lower fixed point is $32^\circ\text{F}$ and the upper is $212^\circ\text{F}$. Equating the ratios $\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32}$ gives you the standard conversion formula.

Q: What if the scale wasn't linear?

If the scale were non-linear (e.g., quadratic or exponential), this simple ratio method wouldn't work. You would need the specific mathematical function relating the thermometric property to the temperature to perform the conversion.

Q: Why did $x_0$ cancel out in the end? Does its value not matter?

$x_0$ acts as a scaling factor for the new thermometer. Because we are taking a ratio of differences measured on the same scale, the scaling factor appears in both the numerator and the denominator and thus cancels out. This makes the final result independent of the specific value of $x_0$.

Question

A thermometer graduated according to a linear scale reads a value $x_0$, when in contact with boiling water and $x_0/3$, when in contact with ice. What is the temperature of an object in $^\circ	ext{C}$, if this thermometer in the contact with the object reads $x_0/2$?

Options:

(A) 35

(B) 60

(C) 40

(D) 25

Sigma Percentile · Accepted Answer

Imagine you are tasked with creating your very own temperature scale. You grab a glass tube with mercury, mark where it stands in freezing water, and mark where it stands in boiling water. You've just created a linear temperature scale! But how do you translate your new, mysterious readings into something everyone understands, like Celsius?

This is where the beautiful and elegant **Principle of Thermometry** comes into play. Let's dive into how we can decode any linear temperature scale.

### The Principle of Thermometry

For any linear temperature scale, the physical property that changes with temperature (like the length of a mercury column) does so uniformly. Because of this uniformity, the ratio of temperature differences remains constant across any two linear scales.

Mathematically, if we take a reading $T$, subtract the Lower Fixed Point ($T_{LFP}$), and divide it by the difference between the Upper Fixed Point ($T_{UFP}$) and the Lower Fixed Point, this ratio is a universal constant for that specific temperature state:

$$\frac{T - T_{LFP}}{T_{UFP} - T_{LFP}} = 	ext{constant}$$

### Setting Up the Equation

Let's apply this to our problem. We are comparing the standard Celsius scale with a new, arbitrary scale.

For the **Celsius Scale**:
- Lower Fixed Point (Ice Point) = $0^\circ	ext{C}$
- Upper Fixed Point (Steam Point) = $100^\circ	ext{C}$
- Let the unknown temperature be $T_C$.

For the **New Scale**:
- Lower Fixed Point (Ice Point) = $\frac{x_0}{3}$
- Upper Fixed Point (Steam Point) = $x_0$
- The given reading for the object is $x = \frac{x_0}{2}$.

Equating the ratios for both scales, we get our master equation:

$$\frac{T_C - 0}{100 - 0} = \frac{x - \frac{x_0}{3}}{x_0 - \frac{x_0}{3}}$$

### The Final Calculation

Now, we substitute the given reading $x = \frac{x_0}{2}$ into our equation:

$$\frac{T_C}{100} = \frac{\frac{x_0}{2} - \frac{x_0}{3}}{x_0 - \frac{x_0}{3}}$$

Let's carefully simplify the fractions on the right side. In the numerator, finding a common denominator of 6 gives us:

$$\frac{x_0}{2} - \frac{x_0}{3} = \frac{3x_0 - 2x_0}{6} = \frac{x_0}{6}$$

In the denominator, finding a common denominator of 3 gives us:

$$x_0 - \frac{x_0}{3} = \frac{3x_0 - x_0}{3} = \frac{2x_0}{3}$$

Substituting these back into our master equation:

$$\frac{T_C}{100} = \frac{\frac{x_0}{6}}{\frac{2x_0}{3}}$$

Notice how the $x_0$ terms elegantly cancel out, leaving us with pure numbers:

$$\frac{T_C}{100} = \frac{1}{6} 	imes \frac{3}{2} = \frac{1}{4}$$

Finally, multiplying both sides by 100 reveals the temperature in Celsius:

$$T_C = \frac{100}{4} = 25^\circ	ext{C}$$

And there we have it! By trusting the geometry of linear scales, we've successfully translated a completely arbitrary reading into a standard temperature.

Select Answer:

Visualized Solution

\text{Thermometry Setup}

\text{Principle of Thermometry}

\text{Equating the Scales}

\text{Substituting the Given Value}

\text{Simplifying the Expression}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Thermal Expansion

The Principle of Thermometry

Setting Up the Equation

The Final Calculation

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