Correct Answer - Option 1 : -2731.5

__Concept:__

**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) = T**_{i}
**Steam point (100°C) = T**_{s}

****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, **X**_{i} = ice point in new temperature scale, **X**_{s} = **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, **X**_{tp} = triple point temperature on the new temperature scale**, T = given standard temperature**

__Calculation:__

__Given:__

The freezing point of water **(X**_{i}) = 0°X and boiling point of water **(X**_{s}) = 1000°X in an unknown temperature scale

The freezing point of water **(T**_{i}) = 0°C and boiling point of water **(X**_{s}) = 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 **a **and **b **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