Question

Answer the following :

1. The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
2. There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
3. The absolute temperature (Kelvin scale) T is related to the temperature \(t_c\) on the Celsius scale by
   \(t_c\) = T – 273.15 
   Why do we have 273.15 in this relation, and not 273.16 ?
4. What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

Answer 1

(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

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(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

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(c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.
Hence, absolute temperature (Kelvin scale) T, is related to temperature \(t_c\), on Celsius scale as:
\(t_c\) = T – 273.15

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(d) Let \(T_F\) be the temperature on Fahrenheit scale and \(T_K\) be the temperature on absolute scale. Both the temperatures can be related as:
\(\frac{T_F - 32 }{180}\) = \(\frac{T_K -273.15}{100}\).........(i)

Let \(T_{F1}\) be the temperature on Fahrenheit scale and \(T_{K1}\) be the temperature on absolute scale. Both the temperatures can be related as:
\(\frac{T_{F1}-{32}}{180}\) = \(\frac{T_{K1}-{273.15}}{100}\) .........(ii)

It is given that:
\({T_{K1}- T_K}\) = 1 K
Subtracting equation (i) from equation (ii), we get: 
\(\frac{{T}_{F1}-T_F}{180}\)= \(\frac{{T}_{K1}-T_K}{100}\) = \(\frac{1}{100}\)

\(T_{F1} = T_F\) = \(\frac{1\times{180}}{100}\)= \(\frac{9}{5}\)

Triple point of water = 273.16 K

∴ Triple point of water on absolute scale = \(273.16\times\frac{9}{5}\) = 491.69