The chemical formula of water is **H _{2}O**.

The water is a combination of two Hydrogens (**H**) atoms and one oxygen (**O**) atom.

Water is the lifeline of all living creatures including humans.

Let us explore the temperature value of water at different parameters such as, in **degree Celsius, Fahrenheit, Kelvin**.

**Degree Celsius to Degree Fahrenheit**

The formula for calculating Celsius to Fahrenheit

** (Degree Calsius ^{°}C × 9/5) + 32 = ?^{°}F**

**For example**

**0 ^{°}C = 32^{°}F**

If you want to find **1 ^{°}C **in Fahrenheit, put this value in the formula,

**(1 ^{°}C × 9/5) + 32 = 33.8^{°}F**

For finding 10^{°}C

**(10 ^{°}C × 9/5) + 32 = 50^{°}F**

and so on…..

Now it’s your turn,

Find the value of **50**** ^{°}C **in Fahrenheit?

**Degree Fahrenheit to Degree Celsius**

The formula for calculating Fahrenheit to Celsius

**(Degree Fahrenheit ^{°}F − 32) × 5/9 = ?^{°}C**

**For examples**

**0 ^{°}F = −17.78^{°}C**

**(1 ^{°}F − 32) × 5/9 = −17.22^{°}C**

for finding 10^{°}F

**(10 ^{°}F − 32) × 5/9 = −12.22^{°}C**

and so on…..

Now it’s your turn,

Find the value of **41**** ^{°}F **in Celsius?

**The temperature for the Kelvin scale.**

The formula for **Degree Celsius to Kelvin**

**(Degree Celsius ^{°}C +273.15) = ?k**

**For examples**

**0 ^{°}C = 273.15 k**

**(1 ^{°}C +273.15) = 274.15 k**

**(10 ^{°}C +273.15) = 283.15 k**

and so on…..

The formula for **Kelvin to degree Celsius**

**(Kelvin−273.15)** = **? ^{°}C**

**For examples**

**0 k = −273.15 ^{°}C**

**(1−273.15)** = −**272.15 ^{°}C**

**(10−273.15)** = −**263.15 ^{°}C**

**(300−273.15)** = **26.85 ^{°}C**

and so on…..

The formula for calculating **Fahrenheit to kelvin**

**(Degree Fahrenheit**

^{°}F − 32) × 5/9 + 273.15 = ?k**For examples**

**0**

^{°}F = 255.372 k**(1**

^{°}F − 32) ×5/9 + 273.15 = 255.928 k**(10**

^{°}F − 32) ×5/9 + 273.15 = 260.928 kNow it’s your turn,

Find the value of **50**** ^{°}F **in Kelvin?

**kelvin to Fahrenheit**

**(0K − 273.15) × 9/5 + 32 = ?**

^{°}F**For examples**

**0 k = −459.67**

^{°}F**(1k − 273.15) × 9/5 + 32 = −457.87**

^{°}F**(10k − 273.15) × 9/5 + 32 = −441.67**

^{°}F**The Boiling point of water in kelvin.**

The boiling point of water in degree Celsius is **100 ^{°}C**.

Now apply the formula given above of Celsius to the kelvin.

**(Degree Calsius ^{°}C +273.15) = ?k**

**(100 ^{°}C +273.15) = 373.15k**

This is the boiling point of water in kelvin.

If you want to explore boiling point such as, in Fahrenheit, you can see the above formulas for that.

**The Melting point of water in kelvin.**

The melting point of water in degrees Celsius is **0 ^{°}C**.

Simply apply the formula

**(Degree Calsius ^{°}C +273.15) = ?k**

**(0 ^{°}C +273.15) = 273.15k**

You can explore other values by using the above formulas.

**The Freezing point of water in kelvin.**

The freezing point of water in degree Celsius is **0 ^{°}C.**

The formula

**(Degree Calsius ^{°}C +273.15) = ?k**

**(0 ^{°}C +273.15) = 273.15k**

These are the values of boiling, freezing and melting point of water in kelvin.

At last…

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**References**

**https://www.rapidtables.com/convert/temperature/celsius-to-fahrenheit.html****https://en.wikipedia.org/wiki/Kelvin**

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