Question
Question: In a random temperature scale Z, water freezes at \({20^ \circ }Z\)and boils at \(220^\circ Z\). Fin...
In a random temperature scale Z, water freezes at 20∘Zand boils at 220∘Z. Find the boiling point of the liquid in this scale if it boils at 60∘C.
Solution
In this given numerical problem we have to change the temperature given in one particular scale to another. This can be done by employing the formula for change of temperature from one scale to another by calculating the upper and lower limits and their difference.
Formula Used:
In order to change the temperature from one scale to another, we can use the formula given below:
UP−LPX−LP=Constant for all temperature scales
In the above mathematical expression,
X=The temperature that is unknown
LP=Lower Point
UP=Upper Point
Complete step by step answer:
Now we have to calculate the respective upper and lower points of the two temperature scales that is the Celsius scale and the new Z scale.
In the Celsius scale:
The Upper point of the scale is =100∘C
The Lower point of the scale is =0∘C
Now their difference is =100∘C−0∘C=100∘C
In the Z scale:
The Upper point of this particular scale is =220∘Z
The Lower point of this particular scale is =20∘Z
Now their difference is =220∘Z−20∘Z=200∘Z
Now, we can use the above mathematical expression to find the unknown temperature. Let us denote this unknown temperature as Z.
Substituting the above values in the mathematical expression, we get:
200Z−20=10060−0 ⇒Z=120+20
Thus, the boiling point of the given liquid on the Z is equal to 140∘C.
Note: While solving the given numerical problem it is important to keep in mind that the upper point and the lower point of both the temperature scales should be of the same substance. In the solution, the upper point if both the temperature scales denote the boiling points of water, while the lower point of both the temperature scales denotes the freezing point of water.