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Mathematics

If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 95\dfrac{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 is 158 °F, then find the temperature in Kelvin.

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Answer

The given relation is y = 95\dfrac{9}{5}(x - 273) + 32.

(i) When x = 313 K:

Substituting x = 313 in the equation:

y=(95(313−273)+32)=(95(40)+32)=(9×405+32)=(3605+32)=(72+32)=104°Fy = \left(\dfrac{9}{5}(313 - 273) + 32\right) \\[1em] = \left(\dfrac{9}{5}(40) + 32\right) \\[1em] = \left(\dfrac{9 \times 40}{5} + 32\right) \\[1em] = \left(\dfrac{360}{5} + 32\right) \\[1em] = (72 + 32) \\[1em] = 104 \text{°F}

∴ The temperature in Fahrenheit is 104 °F.

(ii) When y = 158 °F:

Substituting y = 158 in the equation:

=158=(95(x−273)+32)⇒158−32=(95(x−273))⇒126=(95(x−273))⇒x−273=(126×59)⇒x−273=(6309)⇒x−273=70⇒x=(70+273)⇒x=343 K\phantom{=}158 = \left(\dfrac{9}{5}(x - 273) + 32\right) \\[1em] \Rightarrow 158 - 32 = \left(\dfrac{9}{5}(x - 273)\right) \\[1em] \Rightarrow 126 = \left(\dfrac{9}{5}(x - 273)\right) \\[1em] \Rightarrow x - 273 = \left(\dfrac{126 \times 5}{9}\right) \\[1em] \Rightarrow x - 273 = \left(\dfrac{630}{9}\right) \\[1em] \Rightarrow x - 273 = 70 \\[1em] \Rightarrow x = (70 + 273) \\[1em] \Rightarrow x = 343 \text{ K}

∴ The temperature in Kelvin is 343 K.

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