 # The planet mercury’s surface temperature varies from 700K during the day to 90K at night. What are these values in Celsius and Fahrenheit? 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.

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:

$T_F \ = \ \dfrac{ 9 }{ 5 } T_C + 32 \ = \ \dfrac{ 9 }{ 5 } 100 + 32 \ = \ 212 \ F$