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Question: On a temperature scale Y, water freezes at \(-160^o\) Y and boils at \(-50 ^o\). On this Y scale, a ...

On a temperature scale Y, water freezes at −160o-160^o Y and boils at −50o-50 ^o. On this Y scale, a temperature of 340 K is :
A.−106.3oY B.−96.3oY C.−86.3oY D.−76.3oY \begin{aligned} & A. -{{106.3}^{o}}Y \\\ & B. -{{96.3}^{o}}Y \\\ & C. -{{86.3}^{o}}Y \\\ & D. -{{76.3}^{o}}Y \\\ \end{aligned}

Explanation

Solution

Hint: Use the general formula for the conversion of temperature scales and substitute the temperature in kelvin given to get the value of temperature in Y scale OR use the fact that the number of divisions on any scale, for a given temperature difference, is the same.

Step by step solution :
The given temperature in kelvin scale is 340 K
The freezing point of the water on the Y scale is −160o-160^o
The given boiling point of the water on the Y scale is −50o-50^o
We know the scale conversion formula for all the scales is given as

X−MpBp−Mp=K−273373−273\dfrac{X -M_p}{B_p -M_p} = \dfrac{K-273}{373-273}
Y−(−160)−50−(−160)=K−273100=Y+160110\dfrac{Y - (-160)}{-50 -(-160)} = \dfrac{K-273}{100} = \dfrac{Y +160}{110}
Y=1110(K−273)−160=1110(340−273)−160Y = \dfrac{11}{10}(K - 273)-160 = \dfrac{11}{10}(340 - 273)-160

We get

Y=−86.3oY = -86.3^o

Thus using the general formula for the scale conversion we got the corresponding temperature on the Y scale as Y=−86.3oY = -86.3^o

Additional Information :
We have used the general scale conversion formula. This formula is based on the simple rule that the number of divisions on any scale, for a given temperature difference, is the same.

Note: The possible mistake one can make in this kind of problem is that we go on calibrating the Y scale with the given limits of freezing and boiling points then finally one may take the corresponding value of the calibration on the Kelvin scale which may take much time. So in this kind of problem, we can make use of the general scale conversion formula and get the unknown quantities required.