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_{ice}}{X_{steam} - X_{ice}} = \frac{Y - Y_{ice}}{

Vedclass · Accepted Answer

$117$

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(D) The relationship between any two linear temperature scales is given by the formula: $\frac{X - X_{ice}}{X_{steam} - X_{ice}} = \frac{Y - Y_{ice}}{Y_{steam} - Y_{ice}}$.Here,for the Celsius scale $(C)$: $C_{ice} = 0\,^{\circ}C$ and $C_{steam} = 100\,^{\circ}C$.For the $W$ scale: $W_{ice} = 39\,^{\circ}W$ and $W_{steam} = 239\,^{\circ}W$.We want to find $W$ when $C = 39\,^{\circ}C$.Substituting the values: $\frac{39 - 0}{100 - 0} = \frac{W - 39}{239 - 39}$.$\frac{39}{100} = \frac{W - 39}{200}$.$0.39 = \frac{W - 39}{200}$.$W - 39 = 0.39 	imes 200 = 78$.$W = 78 + 39 = 117\,^{\circ}W$.

List-$I$	List-$II$
$(P). PbS + 2PbO \xrightarrow{\Delta} 3Pb + SO_2 \uparrow$	$(1).$ Smelting
$(Q). PbS + \frac{3}{2}O_2 \xrightarrow{\Delta} PbO + SO_2 \uparrow$	$(2).$ Self reduction
$(R). PbCO_3 \xrightarrow{\Delta} PbO + CO_2 \uparrow$	$(3).$ Calcination
$(S). PbO + C \xrightarrow{\Delta} Pb + CO \uparrow$	$(4).$ Roasting

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