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Thermal Properties of Matter - Measurement of Temperature

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Temperature is a macroscopic physical quantity that determines the degree of hotness or coldness of an object and dictates the direction of heat flow between two bodies in contact.

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The Zeroth Law of Thermodynamics states that if two systems, AA and BB, are separately in thermal equilibrium with a third system CC, then AA and BB are in thermal equilibrium with each other. This law provides the basis for the measurement of temperature.

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A thermometer utilizes a thermometric property—a physical property that changes linearly with temperature. Examples include the volume of a liquid, pressure of a gas at constant volume, or electrical resistance of a wire.

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The Celsius scale defines the ice point (Lower Fixed Point) as 0∘C0^\circ\text{C} and the steam point (Upper Fixed Point) as 100∘C100^\circ\text{C} at standard atmospheric pressure.

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The Fahrenheit scale defines the ice point as 32∘F32^\circ\text{F} and the steam point as 212∘F212^\circ\text{F}.

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The Kelvin scale (Absolute scale) is the SI unit of temperature. The relationship is based on the triple point of water, which is defined as 273.16 K273.16\text{ K}. Absolute zero (0 K0\text{ K}) is the temperature at which molecular motion ceases.

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The Triple Point of water is the unique condition of temperature (273.16 K273.16\text{ K}) and pressure (611.65 Pa611.65\text{ Pa}) where water exists simultaneously in solid, liquid, and gaseous states.

📐Formulae

TC−0100=TF−32180\frac{T_C - 0}{100} = \frac{T_F - 32}{180}

C5=F−329\frac{C}{5} = \frac{F - 32}{9}

T(K)=t(∘C)+273.15T(\text{K}) = t(^{\circ}\text{C}) + 273.15

Reading−LFPUFP−LFP=Constant\frac{\text{Reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{Constant}

Rt=R0(1+αt)R_t = R_0(1 + \alpha t) (Resistance Thermometer)

T=273.16(PPtr)T = 273.16 \left( \frac{P}{P_{tr}} \right) (Constant Volume Gas Thermometer)

💡Examples

Problem 1:

At what temperature do the Celsius and Fahrenheit scales show the same numerical reading?

Solution:

Let the common temperature be xx. Using the relation: x5=x−329\frac{x}{5} = \frac{x - 32}{9} 9x=5(x−32)9x = 5(x - 32) 9x=5x−1609x = 5x - 160 9x−5x4x\begin{array}{r} 9x \\ - 5x \\ \hline 4x \end{array} 4x=−1604x = -160 x=−40x = -40 Therefore, −40∘C=−40∘F-40^\circ\text{C} = -40^\circ\text{F}.

Explanation:

By setting the variables for both scales equal to xx in the conversion formula, we solve a linear equation to find the intersection point of the two scales.

Problem 2:

A faulty thermometer has its fixed points marked as 5∘5^\circ and 95∘95^\circ. What is the correct temperature in Celsius when this thermometer reads 59∘59^\circ?

Solution:

Using the general formula: Reading−LFPUFP−LFP=C−0100−0\frac{\text{Reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0} Given Reading=59\text{Reading} = 59, LFP=5\text{LFP} = 5, UFP=95\text{UFP} = 95: 59−595−5=C100\frac{59 - 5}{95 - 5} = \frac{C}{100} 5490=C100\frac{54}{90} = \frac{C}{100} 0.6=C1000.6 = \frac{C}{100} C=60∘CC = 60^\circ\text{C}

Explanation:

Any linear temperature scale follows the ratio of (measured value minus lower fixed point) to (fundamental interval). By equating the ratio of the faulty scale to the standard Celsius scale, the true temperature can be calculated.