Question

Convert $-30^{\circ}C$ into (a) kelvin$\;K$, (b) $^{\circ}F$.

Answer 1

Hint: We know that the freezing point of water in Kelvin scale is $273.15\;K$ degrees and the freezing point of water in Fahrenheit scale is $32^{\circ}F$. In such a case, determine how ahead the freezing point of water is in these scales in comparison to the Celsius scale and arrive at a suitable expression for the appropriate temperature conversion. Remember to account for the difference in the distribution of the divisions between successive readings of the Celsius and Fahrenheit scales.
Formula Used:
$T(K) = T(^{\circ} C) + 273.15$
$T({}^{\circ}F) = \left(T({}^{\circ}C) \times \dfrac{9}{5}\right) +32$

Complete answer:
We know that $kelvin\;(K),\; Celsius\;(^{\circ} C)$ and $Fahrenheit\;(^{\circ} F)$ are units of temperature, which is a quantity used to define the degree of hotness or coldness of a body.
In the Kelvin Scale, the freezing and boiling points of water are $273.15\;K$ and $373.15\;K$ respectively.
In the Fahrenheit Scale, the freezing and boiling points of water are $32^{\circ}F$ and $212^{\circ}F$ respectively.
In the Celsius Scale, the freezing and boiling points of water are $0^{\circ}C$ and $100^{\circ}C$ respectively.
We see that with such varied definitions in units of temperature, we require conversion formulas while obtaining equivalent temperatures in different units.

(a).To convert $-30^{\circ} C$ to kelvin (K).
From our definitions, we see that the freezing point of water in Kelvin is$\;273.15$ units ahead of the freezing point of water in Celsius, i.e.,
$T(K) = T(^{\circ} C) + 273.15$
Taking $T(^{\circ}C) = -30^{\circ}C$, we get the equivalent temperature in Kelvin as:
$T(K) = -30 + 273.15 = 243.15\;K$

(b).To convert $-30^{\circ}C$ to Fahrenheit ($^{\circ}$F).
From our definitions, we see that, the difference between the freezing and boiling points of water in the Celsius scale is:
$T({}^{\circ}C) = 100^{\circ} C – 0^{\circ}C = 100^{\circ}C$
And the difference between the freezing and boiling points of water in the Fahrenheit scale is:
$T({}^{\circ}F) = 212^{\circ} F-32^{\circ}F = 180^{\circ}F$
Therefore, the ratio of $T({}^{\circ}F)$ to $T({}^{\circ}C)$ will be:
$\dfrac{T({}^{\circ}F)}{T({}^{\circ}C)} = \dfrac{180}{100} = \dfrac{9}{5}$
$\Rightarrow T({}^{\circ}F) = T({}^{\circ}C) \times \dfrac{9}{5}$
Additionally, we see that the freezing point of water in Fahrenheit is$\;32$ units ahead of the freezing point of water in Celsius, i.e.,
$T({}^{\circ}F) = \left(T({}^{\circ}C) \times \dfrac{9}{5}\right) +32$
Plugging in $T({}^{\circ}C) = -30^{\circ}C$, we get the equivalent temperature in Fahrenheit as:
$T({}^{\circ}F) = \left(-30\times \dfrac{9}{5}\right) +32 = -54 +32= -22^{\circ}F$

Note:
The difference between temperature scales arises from the difference in the reference points chosen as zero degrees and the magnitude of incremental units on the scale or the difference between successive readings/markings on a scale.
Note that in addition to the commonly used scales discussed above, there are other temperature scales such as:
Rankine: $T({}^{\circ}R) = \left(T({}^{\circ}C) + 273.15\right) \times \dfrac{9}{5}$
Delile: $T({}^{\circ}De) = \left(100 – T({}^{\circ}C)\right) \times \dfrac{3}{2}$
Newton: $T({}^{\circ}N) = T({}^{\circ}C) \times \dfrac{33}{100}$
Reamur: $T({}^{\circ}Re) = T({}^{\circ}C) \times \dfrac{4}{5}$
Romer: $T({}^{\circ}Ro) = T({}^{\circ}C) \times \dfrac{21}{40} +7.5$