**Temperature Scales**

**Temperature **is ordinarily defined as the measure of hotness or coldness of an object. In physics, it refers to the average kinetic energy of the atoms and molecules in a substance. It is also a property that determines the amount of heat transferred to and from a substance. It is observed that, between two substances, the one with the higher temperature transfers heat to the one with the lower temperature.

Temperature is measured using different scales. There are three known scales namely – Kelvin Scale, Celsius Scale, and Fahrenheit Scale.

The Kelvin scale also called as the absolute zero is named after Sir William Thompson (Lord Kelvin) who devised it. This scale is often used in measurements involving gases. The 0 K or called as the absolute zero is the temperature at which the molecular energy of matter is at minimum. 273 K is the freezing point and 373 K is the boiling point of water.

The Celsius scale which is commonly known as the Centigrade scale was devised by Anders Celsius, based on the properties of water. The boiling point of water at 1 atm is 100 ºC and the freezing point is 0 ºC.

The Fahrenheit scale, devised by Daniel Gabriel Fahrenheit, is often used in the United States of America and England in measuring temperature and other purposes. In such scale, the boiling point of water is 212 ºF and its freezing point is 32 ºF.

A given temperature reading can be converted from one scale to another. To do this, we consider the following equations also called as the conversion formulae.

Celsius to Kelvin | T_{K} = T_{C} + 273 |

Kelvin to Celsius | T_{C} = T_{K} – 273 |

Celsius to Fahrenheit | T_{F} = \(9 \over 5\)T_{C} + 32 |

Fahrenheit to Celsius | T_{C} = \(5 \over 9\)(T_{F} – 32) |

**Example 1.**

What is 450 K in ºC?

Solution:

To convert Kelvin to Celsius, we use the formula **T _{C} = T_{K} - 273**. Substituting the given value in K, we have

\(T_C = 450 - 273 = 177\; ^\circ C\)

Thus, 450 K is equivalent to 177 ºC.

**Example 2.**

Convert 100 ºF to ºC.

Solution:

\(T_C = {5 \over 9}(T_F – 32)={5 \over 9}(200 – 32)={5 \over 9}(168)=93.33\;^\circ C\)

**Example 3.**

The normal body temperature is said to be 37 C. What is this temperature in K? in F?

Solution #1:

\(T_K = 37 + 273=310\;K\)

Solution #2:

\(T_F= {9 \over 5}(37) + 32=98.6\;^\circ F\)

**Example 4.**

The pan is heated from 45 ºC to130 ºC. What is the change in temperature in a) ºC and b) K?

Solutions:

a) \(\Delta T=T_f - T_i =130^\circ C-45^\circ C=\mathbf {85^\circ C}\).

b) First, convert the given temperature from ºC to K.

\(T_i (^\circ C \rightarrow K) = 45^\circ C+273=\mathbf{318\;K}\)

\(T_f (^\circ C \rightarrow K) = 130^\circ C+273=\mathbf{403\;K}\)

\(\Delta T=T_f - T_i =403\; K-318\;K=\mathbf {85\;K}\)