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Standard XI
Physics
Scales of Temperature
Question

Liquid nitrogen has a boiling point of $$-195.81^0C$$ at atmospheric pressure. Express this temperature (a) in degrees Fahrenheit and (b) in kelvins.

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Solution
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a) By Equation,
$$T_F=\dfrac{9}{5}T_C+32.0^0F=\dfrac{9}{5}(-195.81^0C)+32.0=-320^0F$$
b) Applying Equation,
$$T=T_c+273.15=-195.81^0C+273.15=77.3K$$

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