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If a nonhuman civilization were to develop on Saturn's largest moon, Titan, its scientists might well devise a temperature scale based on the properties of methane, which is much more abundant on the surface than water is. 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?

Respuesta :

Answer:

- 329.93 °M.

Explanation:

 freezing point = -182.6 °C

= 273 - 182.6

= 90.4 K

Boiling point = -155.2 °C

= 273 - 155.2

= 117.8 K

New scale

freezing point = 0°M

Boiling point = 100°M

Relation between the two scale

M / 100 =  ( K - 90.4 ) / ( 117.8 - 90.4 )

For K = 0

M / 100 =   - 90.4  / ( 117.8 - 90.4 )

M / 100 = - 90.4  / 27.4

M = - 329.93 °M.

Answer:

-329.9°M corresponds to 0°K.

Explanation:

First we need to know the freezing and boiling points of the Methane in the Kelvin scale. Since the Kelvin scale is related to the Celsius scale by:

[tex]K=C+273[/tex]

Then, the freezing and boiling points of the Methane are:

[tex]T_F=(-182.6+273)\°K=90.4\°K\\\\T_B=(-155.2+273)\°K=117.8\°K[/tex]

So, now we know the following relationship between Methane and Kelvin scales:

[tex]0\°M=90.4\°K\\\\100\°M=117.8\°K[/tex]

Now, if we think of that temperatures as the ordered pairs (0, 90.4) and (100, 117.8), we can obtain the equation of the line that contains that two points and find where the "y" (in this case, the temperature in Kelvin scale) is zero.

First, we find the slope:

[tex]m=\frac{117.8-90.4}{100-0}=0.274[/tex]

Now, we use the general formula of a line (remembering that y should be zero):

[tex]y-y_0=m(x-x_0)\\\\0-90.4=0.274(x-0)\\\\x=-\frac{90.4}{0.274}=-329.9[/tex]

It means that the temperature of -329.9°M corresponds to absolute zero in Kelvin scale.

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