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In an unknown temperature scale freezing point of water is 0°X and boiling point of water is 1000°X. The absolute zero in degree X is
1. -2731.5
2. 2731.5
3. 0
4. -273.15

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Correct Answer - Option 1 : -2731.5


Absolute zero temperature: The temperature at which the enthalpy and entropy of a gas reach their minimum value taken as zero. All the particles stop moving and all the disorders disappear.

The value of absolute zero temperature is 0K or -273.15°C

Celsius scale: Celsius sale also called centigrade scale is based on 0°C for the freezing point of water and 100°C  for boiling pint for water.

Methods of temperature measurement:

1)Two reference point system:

In this method, two reference points are used

  • Ice point (0°C) = Ti 
  • Steam point (100°C) = Ts

In this we consider the basic equation T = a + b × X

\({\bf{T}}\; = \;\frac{{ - 100 × {{\bf{X}}_{\bf{i}}}}}{{{{\bf{X}}_{\bf{s}}} - {{\bf{X}}_{\bf{i}}}}} + \;\frac{{100}}{{{{\bf{X}}_{\bf{s}}} - {{\bf{X}}_{\bf{i}}}}} × {\bf{X}}\)

where, Xi = ice point in new temperature scale, Xs = steam point in new temperature scale, X = required temperature on the new temperature scale corresponding to the given standard temperature scale

2) Single reference point system:

In this method single reference points used i.e triple point of water (273.15K)

Here we consider the basic equation T = a × X

\({\bf{T}} = 273.15 \times \;\frac{{\bf{X}}}{{{{\bf{X}}_{{\bf{tp}}}}}}\)

where, X = required temperature on the new temperature scale, Xtp = triple point temperature on the new temperature scale, T = given standard temperature



The freezing point of water (Xi) = 0°X and boiling point of water (Xs) = 1000°X in an unknown temperature scale 

The freezing point of water (Ti) = 0°C and boiling point of water (Xs) = 100°C on the Celsius scale 

we know the basic equation for two reference point system

T = a + b × X        .......(A)

Ti  = a + b × Xi        .......(1)

Ts  = a + b × Xs        .......(2)

substituting the above values in equation 1 and 2

0 = a + b × 0

100 = a + b × 1000

We get a = 0  and  b = 0.1,  then substituting  absolute zero temperature (T) = -273.15 and values of and in equation A

T = a + b × X

-273.15 = 0 + 0.1 × X

∴ X = -2731.5°X   

If two thermometers are agreed at ice point and steam point it does not mean that the intermediate temperature is also the same and all scales are arbitrary

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