On a temperature scale Y, water freezes at -1 | vedclass

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(C) The relationship between any temperature scale and the Kelvin scale is given by the formula: $\frac{Y - Y_{ice}}{Y_{steam} - Y_{ice}} = \frac{K -

Vedclass · Accepted Answer

$-86.3$

Vedclass · Answer

(C) The relationship between any temperature scale and the Kelvin scale is given by the formula: $\frac{Y - Y_{ice}}{Y_{steam} - Y_{ice}} = \frac{K - K_{ice}}{K_{steam} - K_{ice}}$.Given for scale $Y$: $Y_{ice} = -160^{\circ} Y$ and $Y_{steam} = -50^{\circ} Y$.For the Kelvin scale: $K_{ice} = 273 \ K$ and $K_{steam} = 373 \ K$.Substituting the values for $K = 340 \ K$:$\frac{Y - (-160)}{-50 - (-160)} = \frac{340 - 273}{373 - 273}$$\frac{Y + 160}{110} = \frac{67}{100}$$Y + 160 = 0.67 	imes 110$$Y + 160 = 73.7$$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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