# Question 6626e

Oct 27, 2015

$- {28}^{\circ} \text{C}$

#### Explanation:

To go from temperature expressed in degrees Celsius, $^ \circ \text{C}$, to temperature expressed in Kelvin, $\text{K}$, you need the following conversion factor

$\text{t K" = "t"^@"C} + 273.15$

In simple terms, the origin of the Kelvin scale, $\text{0 K}$, also called absolute zero, is set at $- {273.15}^{\circ} \text{C}$. Temperatures expressed in Kelvin cannot be lower than $\text{0 K}$, so always make sure that you don't have negative temperatures in Kelvin.

Now, since you can go from degrees Celsiu to Kelvin by adding $273.15$, it follows that you can go from Kelvin to degrees Celsius by subtracting $273.15$

$\text{t"^@"C" = "t K} - 273.15$

In your case, a temperature of $\text{245 K}$ would correspond to

$\text{t"[""^@"C"] = "245 K" - 273.15 = -28.15^@"C}$

You need to round this off to

"t"[""^@"C"] = color(green)(-28^@"C")#