# Liquid nitrogen is used by physicians to burn moles off of skin. Liquid nitrogen boils at 77 K. What is its boiling point in °C?

Jun 25, 2016

$- {196}^{\circ} \text{C}$

#### Explanation:

The problem wants you to use the conversion factor that takes you from Kelvin to degree Celsius to calculate the boiling point of liquid nitrogen in degrees Celsius.

As you know, the Celsius and Kelvin temperature scales differ from each other in the placement of the zero mark. In other words, an increase of ${1}^{\circ} \text{C}$ is equivalent to an increase of $\text{1 K}$. More specifically, the zero mark on a Celsius scale is set on ${0}^{\circ} \text{C}$ and the zero mark on the Kelvin scale is set on $\text{273.15 K}$.

The conversion factor that takes you from degrees Celsius to Kelvin and vice versa looks like this

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{t \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, in order to go from Kelvin to degrees Celsius, you must subtract $273.15$, since going from degrees Celsius to Kelvin requires you to add $273.15$.

You will thus have

t[""^@"C"] = "77 K" - 273.15 = -196.15^@"C"

Rounded to the number of decimal places of $\text{77 K}$, the answer will be

t[""^@"C"] = color(green)(|bar(ul(color(white)(a/a)color(black)(-196^@"C")color(white)(a/a)|)))