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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 $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ 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 $273.16 \mathrm{~K}$. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (kelvin scale) $T$ is related to temperature $t_c$ on the Celsius scale by $t_c=T-273.15$. Why do we have $273.15$ in this relation, 0 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?
(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 $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ 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 $273.16 \mathrm{~K}$. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (kelvin scale) $T$ is related to temperature $t_c$ on the Celsius scale by $t_c=T-273.15$. Why do we have $273.15$ in this relation, 0 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?
Solution:
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Verified Answer
(a) Triple point of water has a unique value $(273.16 \mathrm{~K}$ ) at fixed values of pressure and volume. But melting point of ice and boiling point of water do not have unique value as they depend both on pressure and temperature.
(b) The other fixed point is the absolute zero.
(c) On Celsius scale, $0^{\circ} \mathrm{C}$ corresponds to melting point of ice at normal pressure and the same temperature in Kelvin scale is $273.15 \mathrm{~K}$. The triple point of water in Kelvin scale is $273.16 \mathrm{~K}$.
$\therefore \quad$ By the relation, $t_c=T-273.15$,
the triple point of water on Celsius scale $=273.16$ $-273.15=0.01^{\circ} \mathrm{C}$
(d) Relation between temperature in Fahrenheit scale and absolute scale is
$\frac{\mathrm{F}-32}{180}=\frac{\mathrm{K}-273.15}{100}$
For another set of temperature, $\frac{\mathrm{F}^{\prime}-32}{180}=\frac{\mathrm{K}^{\prime}-273.15}{100}$
Subtracting (i) from (ii), we get
$$
\begin{aligned}
& \frac{F^{\prime}-F}{180}=\frac{K^{\prime}-K}{100} \\
\Rightarrow F^{\prime}-F &=\frac{180}{100} \times\left(K^{\prime}-K\right)
\end{aligned}
$$
If $K^{\prime}-K=1 K$, then
$$
\mathrm{F}^{\prime}-\mathrm{F}=\frac{180}{100} \times 1=\frac{9}{5}
$$
$\therefore \quad$ The triple point of water $(273.16 \mathrm{~K})$ in the new scale is $273.16 \times \frac{9}{5}=491.69$.
(b) The other fixed point is the absolute zero.
(c) On Celsius scale, $0^{\circ} \mathrm{C}$ corresponds to melting point of ice at normal pressure and the same temperature in Kelvin scale is $273.15 \mathrm{~K}$. The triple point of water in Kelvin scale is $273.16 \mathrm{~K}$.
$\therefore \quad$ By the relation, $t_c=T-273.15$,
the triple point of water on Celsius scale $=273.16$ $-273.15=0.01^{\circ} \mathrm{C}$
(d) Relation between temperature in Fahrenheit scale and absolute scale is
$\frac{\mathrm{F}-32}{180}=\frac{\mathrm{K}-273.15}{100}$
For another set of temperature, $\frac{\mathrm{F}^{\prime}-32}{180}=\frac{\mathrm{K}^{\prime}-273.15}{100}$
Subtracting (i) from (ii), we get
$$
\begin{aligned}
& \frac{F^{\prime}-F}{180}=\frac{K^{\prime}-K}{100} \\
\Rightarrow F^{\prime}-F &=\frac{180}{100} \times\left(K^{\prime}-K\right)
\end{aligned}
$$
If $K^{\prime}-K=1 K$, then
$$
\mathrm{F}^{\prime}-\mathrm{F}=\frac{180}{100} \times 1=\frac{9}{5}
$$
$\therefore \quad$ The triple point of water $(273.16 \mathrm{~K})$ in the new scale is $273.16 \times \frac{9}{5}=491.69$.
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