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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