Question

Liquid nitrogen has a boiling point of $ - {195.81^o}C$ at atmospheric pressure. Express this temperature in
(A) Degrees Fahrenheit
(B) Kelvins

Answer 1

Hint We are given the boiling point temperature of liquid nitrogen at atmospheric pressure and are asked to convert the value into degrees Fahrenheit and kelvin. Thus, we will use the generic conversion formulation for temperature conversion and will use the formula for conversion of Celsius to kelvin.
Formulae Used:
$\dfrac{{{T_{new}} - {L_{New}}}}{{{U_{New}} - {L_{New}}}} = \dfrac{{{T_{Old}} - {L_{Old}}}}{{{U_{Old}} - {L_{Old}}}}$
Where, ${T_{New}}$ is the temperature reading in the new scale, ${L_{New}}$ is the lower scale limit of the new scale, ${U_{New}}$ is the upper scale limit of the new scale, ${T_{Old}}$ is the temperature reading in the old scale, ${L_{Old}}$ is the lower scale limit of the old scale and ${U_{Old}}$ is the upper scale limit of the old scale.
${T_{Kelvin}} = {T_{Celsius}} + 273.15$
Where, ${T_{Kelvin}}$ is the temperature reading in Kelvin scale and ${T_{Celsius}}$is the temperature in Celsius scale.

Complete Step By Step Solution
Here,
We are given the temperature as$ - {195.81^o}C$.
Now,
First, we will have to convert the given temperature to degrees Fahrenheit.
For that, we will use the generic temperature conversion formula.
Now,
We have, for the Fahrenheit scale,
$\Rightarrow$ ${L_{Fahrenheit}} = {32^o}F$
$\Rightarrow$ ${U_{Fahrenheit}} = {212^o}F$
$\Rightarrow$ ${L_{Celsius}} = {0^o}C$
${U_{Celsius}} = {100^o}C$
${T_{Old}} = - {195.81^o}C$
Thus,
$\dfrac{{{T_{new}} - {L_{Fahrenheit}}}}{{{U_{Fahrenheit}} - {L_{Fahrenheit}}}} = \dfrac{{{T_{Old}} - {L_{Celsius}}}}{{{U_{Celsius}} - {L_{Celsius}}}}$
Substituting the value, we get
\[\dfrac{{{T_{new}} - 32}}{{212 - 32}} = \dfrac{{( - 195.81) - 0}}{{100 - 0}}\]
Further, we get
\[\dfrac{{{T_{new}} - 32}}{{180}} = \dfrac{{ - 195.81}}{{100}}\]
Then,
Cancelling the denominators, we get
\[\dfrac{{{T_{new}} - 32}}{9} = \dfrac{{ - 195.81}}{5}\]
Taking ${T_{new}}$ on one side, we get
\[{T_{new}} = ( - 39.162)(9) + 32\]
Finally, we get
$\Rightarrow$ ${T_{new}} = - 352.458 + 32$
Thus, we get
$\Rightarrow$ ${T_{new}} = - {384.458^o}F$
Now,
For the conversion of temperature into kelvin, we will use the kelvin conversion formula.
$\Rightarrow$ ${T_{Kelvin}} = {T_{Celsius}} + 273.15$
Substituting the values, we get
$\Rightarrow$ ${T_{Kelvin}} = ( - 195.81) + 273.15$
After further calculation, we get
$\Rightarrow$ ${T_{Kelvin}} = 77.34K$

Note We have used the difference of the upper limit of the scales in the denominator. This difference is called the scale range of the temperature scale. We have used the ratio of the scale range indirectly as for the conversion of units of any parameter, we always take the ratio of the defining parameter for the units.