Question

If a non-human civilization were to develop on Saturn's largest moon, Titan, its scientists might well develop a temperature scale based on
the properties of methane, which is much more abundant on the surface than water. Methane freezes at -182.6°C on Titan, and boils at
-155.2°C Taking the boiling point of methane as 100.0°M (degrees Methane) and its freezing point as 0°M, what temperature on the
Methane scale corresponds to the absolute zero point of the Kelvin scale?

Answer 1

Answer: $-330.47 \°M$

Explanation:

Let's begin by explaining that the relation between the Celsius scale and Kelvin scale is:

$\°C +273.15=K$

This means the absolute zero point of the Kelvin scale is $0 K=-273.15 \°C$

Keeping this in mind, we have the freezing point and boling point of Methane as:

Freezing point: $0 \°M=-182.6 \°C=90.55 K$

Boiling point: $100 \°M=-155.2 \°C=117.95 K$

According to this, there is a linear relation between the methane scale ($\°M$) and the Kelvin scale in the form:

$T=a+bt_{M}$ (1)

Where:

$T$ is the temperature in Kelvin

$t_{M}$ is the temperature in degrees Methane

Firstly, we need to find the value of $a$ and $b$ with the two given points ($90.55 K$ and $117.95 K$):

When $T=90.55$ and $t_{M}=0$:

$90.55=a+b(0)$

$a=90.55$ (2)

When $T=117.95$ and $t_{M}=100$:

$117.95=90.55+b(100)$

$b=0.274$ (3)

Now we have the linear equation:

$T=90.55+0.274t_{M}$ (4)

Isolating $t_{M}$:

$t_{M}=\frac{T-90.55}{0.274}$ (5)

Evaluating for $T=0 K$:

$t_{M}=\frac{0-90.55}{0.274}$ (6)

Finally:

$t_{M}=-330.47 \°M$

This means $0 K=-330.47 \°M$