Question

On a new scale of temperature ( which is linear ) and called the $W$ scale, the freezing and boiling points of water are \[39W\] and \[239W\] respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^\circ C$ on the Celsius scale?
\[(A)\;200\;W\]
\[(B)139\;W\]
\[(C)78\;W\]
\[(D)117\;W\]

Answer 1

Hint: The boiling point and the freezing point of an unknown scale are given. We have to find the given Celcius in terms of that unknown scale. So use the formula of the relation between the readings of two different scales and put the given data and find the required term.

Formula used:
If $p$ be the temperature of the $W$ scale and $q$ be the temperature of the Celsius scale,
Then, $ = \dfrac{{p - {\text{LPF of W scale}}}}{{{\text{UPFof W scale}} - {\text{LPF of W scale}}}} = \dfrac{{q - {\text{LPF of celsius scale}}}}{{{\text{UPF of celsius scale}} - {\text{LPF of celsius scale}}}}$
UPF means the upper fixed point and LPF means the lower fixed point.

Complete step by step answer:
A scale is required to measure the temperature of an object correctly. The accepted protocol about the consideration of the scale is – the hot object has a high temperature and the cold object has a low temperature. That means if a hot and a cold object are kept in touch, the temperature of the hot object decreases, and the temperature of the cold object increases. In the end, they both reach the same temperature i.e in equilibrium.
The temperature scale is generally of four types – Celsius, Fahrenheit, Reaumur, and Kelvin.
The relation between any two scale’s reading of temperature is,
$ \Rightarrow \dfrac{{p - {\text{LPF of W scale}}}}{{{\text{UPF of W scale}} - {\text{LPF of W scale}}}} = \dfrac{{q - {\text{LPF of celsius scale}}}}{{{\text{UPF of celsius scale}} - {\text{LPF of celsius scale}}}}$
Where, $p$ be the temperature of the $W$ scale and $q$ be the temperature of the Celsius scale,
And, UPF means the upper fixed point and LPF means the lower fixed point.
Given, ${\text{UPFof W scale = }}239^\circ W$
$\Rightarrow {\text{LPF of W scale = }}39^\circ W$
$\Rightarrow q = 39^\circ C$
We know, ${\text{UPF of celsius scale = 100}}^\circ C$
$\Rightarrow {\text{UPFof celsius scale = }}0^\circ C$
So from the above relation, we get,
$ \Rightarrow \dfrac{{p - {\text{39}}^\circ {\text{C}}}}{{{\text{239}}^\circ {\text{C}} - {\text{39}}^\circ {\text{C}}}} = \dfrac{{{\text{39}}^\circ {\text{C}} - 0^\circ {\text{C}}}}{{{\text{100}}^\circ {\text{C}} - 0^\circ {\text{C}}}}$
$ \Rightarrow p - {\text{39}}^\circ {\text{C}} = \dfrac{{{\text{39}}^\circ {\text{C}}}}{{100^\circ {\text{C}}}} \times 200^\circ {\text{C}}$
$ \Rightarrow p - {\text{39}}^\circ {\text{C}} = {\text{78}}^\circ {\text{C}}$
$ \Rightarrow p = 117^\circ {\text{C}}$
So, the temperature on the scale $W$, corresponding to a temperature of ${\text{39}}^\circ {\text{C}}$ on the Celsius scale, will be $117^\circ {\text{C}}$.

Hence the right answer is in option $(D)$.

Note: We may show the upper fixed point and the lower fixed point for different temperature scales through a table:

Temperature Scale	LFP	UFP	Sign
Celcius	$0^\circ C$	$100^\circ C$	$^\circ C$
Fahrenheit	$32^\circ F$	$212^\circ F$	$^\circ F$
Reaumur	\[0^\circ R\]	\[80^\circ R\]	\[^\circ R\]
Kelvin	$273K$	$373K$	$K$