Centigrade scale of temperature

Centigrade scale of temperature: Of or relating to a temperature scale on which the freezing point of water is 0 degrees and the boiling point of water is 100 degrees.

The fundamental interval is measured and divided into 100 equal divisions called degrees, thereby establishing a temperature scale on the thermometer. Since there are 100 equal divisions on the fundamental interval, this scale is called the “Centigrade scale of temperature”. It is also known as the “Celsius Scale” so named after the scientist who invented it.

Once this scale has been established, the temperature is read by noting the level of the mercury thread and reading it against the scale. When the temperature of the surroundings increases, the mercury forces the mercury thread up the capillary tube. When the mercury in the bulb cools it contracts and the mercury thread falls down the capillary tube towards the bulb.

The Relationship between Centigrade Scale or Celsius and Kelvin Scales of Temperature

All known substances exhibit their lowest possible amount of molecular energy at a temperature of about -273.16℃. No substance can attain a temperature less than -273.16  and this is known as the “absolute zero” of temperature.

The temperature scale used in the SI units is the Kelvin scale, which is also known as ‘Thermodynamic or Absolute Scale’. The lowest point on the Thermodynamic or Absolute Scale is called absolute zero and is given the value of zero Kelvin (0K) subsequently-273.16℃≈0K

∴0℃≅273K And ∴100℃≈373K

Figure 3.7: Comparing Celsius and Kelvin Scales

The size of one degree Celsius 1℃) is the same as one degree Kelvin (1˚K) That is to say a temperature change of 1℃ is equivalent to a temperature of 1˚K. To convert a temperature θ which is in degrees Celsius into Kelvin (T), we simply add 273 to the Celsius reading. That is:

Temperature in Kelvin (T)=Temperature in degree Celsius(θ)+273

Or  T=θ+273

Examples:

1. Express 47 in degrees Kelvin

Solution:

We know that T=θ+273 so as we are given that θ=47℃ then

T=47+273

T=320K

2. Change 483K into degrees Celsius

Solution

T=θ+273

θ=T-273, we are given that    T=320K 
θ=483-273
θ=210℃