If the temperature of a liquid can be measured in kelvin units as `x^(@)K` or in fahrenheit units as `y^(@)F`, the relation between the two systems of measurement of temperature is given by the linear equation.
`y=(9)/(5)(x- 273)+32`
(i) find the temperature of the liquid in fahrenheit, if the temperature of the liquid is 313 K.
(ii) If the temperature in kelvin.

Explanation

The given relation is , `F= 9/5 ( K - 273) + 32.`
(i) Putting K = 313 in (A) , we get
`F= 9/5 xx (313 - 273) + 32 `
`rArr= ( 9/5 xx40)+ 32 = ( 72 + 32) = 104.`
`:. K= 313 rArr F = 104`
Hence, `313 K = 104^(@)F.`
(ii) Putting F = 158 in (A), we get
`9/5 ( K-273) + 32 = 158`
`rArr 9/5 ( K - 273) = ( 158-32) = 126 `
`rArr(K-273) = ( 126 xx 5/9) = 70`
`rArr K = ( 70 + 273) = 343.`
`:. F= 158 rArr K = 343.`
Hence, `158 ^(@)F = 343 K.`
(iii) We have, `F=9/5 ( K- 273) + 32 " "` ...(A)
Putting K= 273 in (A) , we get
`F= 9/5 ( 273 - 273) + 32 = ( 0+ 32) = 32.`
Putting K= 283 in (A), we get
` F= 9/5 ( 283 - 273) + 32 + ( 18 + 32) = 50.`
Putting K= 293 in (A) , we get
`F=9/5 ( 293 - 273) + 32 = ( 36 + 32) = 68.`
Putting K= 298 in (A), we get
`F= 9/5 ( 298 - 273) + 32 = ( 45 + 32) = 77.`
Thus, we have the following table:
`{:(K,273,283,293,298),(F,32,50,68,77):}`
On a graph paper, we draw X' OX and YOY' as the x- axis and the y- axis respectively. We take the values of K along the x- axis and those of F along the y- axis.
As we start the values of K from, 270, we make a kink at the origin and start with 270.
On the above graph, we plot the points `A(273,32),B(283,50),C(293,68) and D ( 298, 77)`.
Join AB, BC and CD to get the graph line ABCD, which is extended in both the directions, .

(iv) Along the y- axis, take a point P, showing `95^(@)F.` Draw PQ|| X' OX, meeting the graph line ABCD at Q. Draw `QR bot X'OX`, meeting the x-axis at R. Clearly , R represents 308 K.
Hence, `95^(@)F= 308 K.`
Further, let G be a point representing 300 k.
Draw `GH bot X'OX`, meeting ABCD at the point H(300, 80).
Hence, `300 K = 80^(@)F.`