On a temperature scale Y,water freezes at -16 | vedclass

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(C) The relationship between any temperature scale $X$ and the Kelvin scale $K$ is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{fre

Vedclass · Accepted Answer

$-86.3$

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(C) The relationship between any temperature scale $X$ and the Kelvin scale $K$ is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{K - 273}{373 - 273}$.Given for scale $Y$: $Y_{freezing} = -160^{\circ} Y$ and $Y_{boiling} = -50^{\circ} Y$.Substituting these values into the formula for $K = 340 \ K$:$\frac{Y - (-160)}{-50 - (-160)} = \frac{340 - 273}{373 - 273}$$\frac{Y + 160}{110} = \frac{67}{100}$$Y + 160 = \frac{67 	imes 110}{100}$$Y + 160 = 73.7$$Y = 73.7 - 160 = -86.3^{\circ} Y$.

On a temperature scale $Y$,water freezes at $-160^{\circ} Y$ and boils at $-50^{\circ} Y$. On this $Y$ scale,a temperature of $340 \ K$ is (in $^{\circ} Y$)

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