physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

11.4 Answer the following :

(a) 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) ?

(b) 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 ?

(c) 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 ?

(d) 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 ?

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5 months ago

11.4 (a) The triple point of water has a unique value of 273.16 K, irrespective of pressure and volume. Whereas, melting point of ice and boiling point of water, the temperature value depends on pressure and volume.

(b) The other fixed point on Kelvin scale is 0 K.

(c) The temperature 273.16 K is the triple point of water, it is not the melting point of ice. The melting point of ice is specified in Celsius scale as 0 $° C .$ Hence the absolute temperature in Kelvin scale, $T_{k}$ is related to temperature $T_{c}$ in Celsius scale as

$T_{c} = T_{k} - 273.15$

(d) Let $T_{f}$ and $T_{k}$ be the temperature in Fahrenheit and absolute scale. From the co-rela

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(b) The other fixed point on Kelvin scale is 0 K.

$T_{c} = T_{k} - 273.15$

(d) Let $T_{f}$ and $T_{k}$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_{f} - 32}{9}$ = $\frac{T_{k} - 273.15}{5}$ ……(i)

Let $T_{f 1}$ and $T_{k 1}$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_{f 1} - 32}{9}$ = $\frac{T_{k 1} - 273.15}{5}$ ……(ii)

It is given $T_{k 1} - T_{k}$ = 1 K

Now subtracting (i) from (ii), we get

$\frac{T_{f 1} - 32}{9}$ - $\frac{T_{f} - 32}{9}$ = $\frac{T_{k 1} - 273.15}{5} - \frac{T_{k} - 273.15}{5}$

$\frac{T_{f 1} - T_{f}}{9}$ = $\frac{T_{k 1} - T_{k}}{5}$

$T_{f 1} - T_{f}$ = $\frac{9}{5}$

Triple point of water = 273.16 K

Hence triple point of water in the absolute scale = 273.16 $*$ $\frac{9}{5}$ = 491.69

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