On a new scale of temperature (which is linea | vedclass

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(D) The relationship between any two linear temperature scales is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{C

Vedclass · Accepted Answer

$117$

Vedclass · Answer

(D) The relationship between any two linear temperature scales is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{C - C_{freezing}}{C_{boiling} - C_{freezing}}$.For the $W$ scale,the freezing point is $39\,^{\circ}W$ and the boiling point is $239\,^{\circ}W$.For the Celsius scale,the freezing point is $0\,^{\circ}C$ and the boiling point is $100\,^{\circ}C$.Substituting the values into the formula:$\frac{W - 39}{239 - 39} = \frac{39 - 0}{100 - 0}$$\frac{W - 39}{200} = \frac{39}{100}$$W - 39 = \frac{39 	imes 200}{100}$$W - 39 = 39 	imes 2 = 78$$W = 78 + 39 = 117\,^{\circ}W$.

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