The aim of this question is to learn the inter-conversion of temperature between different scales.
There are three scales used for the measurement of temperature. These are Celsius, Fahrenheit, and Kelvin, named after their inventors. The inter-conversion of these scales is very common in solving scientific problems.
The relationship for interconversion between these scales is given by the following mathematical formulae:
Celsius to Kelvin Conversion: $ T_K = T_C + 273.15 $
Kelvin to Celsius Conversion: $ T_C = T_K – 273.15 $
Fahrenheit to Celsius Conversion: $ T_C = \dfrac{ 5 }{ 9 } ( T_F – 32 ) $
Celsius to Fahrenheit Conversion: $ T_F = \dfrac{ 9 }{ 5 } T_C + 32 $
Fahrenheit to Kelvin Conversion: $ T_K = \dfrac{ 5 }{ 9 } ( T_F – 32 ) + 273.15 $
Kelvin to Fahrenheit Conversion: $ T_F = \dfrac{ 9 }{ 5 } ( T_K – 273.15 ) + 32 $
Where $ T_F $, $ T_C $, and $ T_K $ are the temperature measurements in Fahrenheit, Celsius and Kelvin scales, respectively.
Expert Answer
Part (a) – For day time:
\[ T_K \ = \ 700 \ K \]
For Kelvin to Fahrenheit Conversion:
\[ T_F \ = \ \dfrac{ 9 }{ 5 } ( T_K – 273.15 ) + 32 \ = \dfrac{ 9 }{ 5 } ( 700 – 273.15 ) + 32 \]
\[ T_F \ = \ \dfrac{ 9 }{ 5 } ( 426.85 ) + 32 \ = \ 768.33 + 32 \]
\[ T_F \ = \ 800.33 \ F \]
For Kelvin to Celsius Conversion:
\[ T_C \ = \ T_K – 273.15 \ = \ 700 \ – \ 273.15 \]
\[ T_C \ = \ 426.85 \ C \]
Part (b) – For night time:
\[ T_K \ = \ 90 \ K \]
For Kelvin to Fahrenheit Conversion:
\[ T_F \ = \ \dfrac{ 9 }{ 5 } ( T_K – 273.15 ) + 32 \ = \dfrac{ 9 }{ 5 } ( 90 – 273.15 ) + 32 \]
\[ T_F \ = \ \dfrac{ 9 }{ 5 } ( -183.15 ) + 32 \ = \ -183.15 + 32 \]
\[ T_F \ = \ -214.15\ F \]
For Kelvin to Celsius Conversion:
\[ T_C \ = \ T_K – 273.15 \ = \ 90 \ – \ 273.15 \]
\[ T_C \ = \ -183.15 \ C \]
Numerical Result
Part (a) – For day time: $ T_K \ = \ 700 \ K, \ T_F \ = \ 269.138 \ F, \ T_C \ = \ 426.85 \ C $
Part (b) – For night time: $ T_K \ = \ 90 \ K, \ T_F \ = \ 3.55 \ F, \ T_C \ = \ -183.15 \ C $
Example
Given that the boiling point of water is 100 C, what is the value of temperature in Fahrenheit and Kelvin Scales?
For Celsius to Kelvin Conversion:
\[ T_K \ = \ T_C \ + \ 273.15 \ = \ 100 \ + \ 273.15 \ = 373.15 \ K \]
For Celsius to Fahrenheit Conversion:
\[ T_F \ = \ \dfrac{ 9 }{ 5 } T_C + 32 \ = \ \dfrac{ 9 }{ 5 } 100 + 32 \ = \ 212 \
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