If a temperature scale were based off benzene.

Question

If a temperature scale were bassed on the freezing point (5.5 degrees Celsius) and boiling point (80.1 degrees Celsius) of benzene, and the temperature difference between these two points was divided into 50 units called degrees X, what would be the freezing and boiling points of water in degrees X?

Answer 1

If I understand your question correctly, you are trying to create a new temperature scale based on the freezing and boiling points of Benzene.

Let's look at the corresponding points on each scale (X and Celsius)

Freezing Point: 0 X = 5.5 C

Boiling Point: 50 X = 80.1 C

The range in X degrees is 50, but in Celsius it is 74.6 degrees (=80.1-5.5)

That means that the X degrees are larger than the Celsius degrees, since there are fewer of them to cover the same temperature range. 1 X degree = 74.6/50 = 1.492 C degrees.

So now you can create a linear relationship between the two scales.

Recall from algebera, the slope of a line, m, can be expressed m = (Y1-Y2)/(X1-X2)

What that means is that the ratio of the two degree ranges from the freezing point to the boiling point of benzene is the same as the slope of the line relating the two scales.

Let's replace Y with C, to represent degrees C, and let the X's represent degrees X. We know the freezing and boiling points lie on the line, so lets use them for Y1, Y2, X1, X2:

m = (80.1-5.5)/(50-0) = 1.492, which agrees with what we found above.

For other points, we are going to replace one Y and one X with a known value and  an unknown.

For example, say we want to know how many degrees X is 20 C.

We could use either of these equations.

1.492 = (80.1- 20)/ (50 - X)

1.492 = (20- 5.5)/ (X - 0)

They both return the same answer when solved for X, but both are a little awkward (20 C = 9.718 X) So, we take one of them and rearrange it to be nice and easy. The bottom one is the easiest to work with. It becomes:

Temp (X) = (Temp C - 5.5) / 1.492

I will leave it to you to rearrange for the conversion of X to C.

I hope this helps.

 

 

Answer 2

Hi Angela A.,

Actually, both answers above make an assumption, which was NOT stated in the original question! And that assumption is, that the benzene-based temperature scale starts with 0 as the freezing value of benzene.

It wouldn't have to; obviously the Fahrenheit scale doesn't zero at the freezing point of pure water. In fact, the Fahrenheit scale was initially intended to range from water freezing at 30F to "normal" body temperature at 90F, neither one of which was exactly borne out with further data! Only later was it reset to 0F = salt/brine/ice equilibrium, and 180 degrees range up to the boiling point of water. And nowadays we prefer the triple point of water, for technical experimental reasons. If you're interested in details comment back, and I'll provide.

So, technically, the question is incompletely specified, hence not strictly answerable, although most people would go with the two answers supplied above (though "Calculus" is a bit of overkill?)..

Answer 3

Whenever you want to derive a temperature scale or make-up a new entirely, it is easiest to use some basic Calculus.  First, the difference of the temperature in Celcius is (80.1 - 5.5 = 74.6), while the difference in the new unit is X is simply 50 because the problem tells you that's how many units of X there'll be.  As a result, the derivative of degrees X with respect to degres C is given by:

dX/dC = (50)/(74.6) = 0.67

Moving dC to the right side of the equation and integrating gives you:

Integral(dX) = Integral(0.67dC)

X = 0.67C + constant

Now, you let the the melting point of benzene be 0 degrees X or X = 0 and since the melting point in Celsius is 5.5 or C = 5.5, you can solve for the constant:

0 = 0.67(5.5) + constant

- 3.7 = constant

Thus the equation for converting to degrees Celcius to degrees X is:

X = 0.67C - 3.7

Now, using this equation we can determine the melting and boiling points of water by plugging 0 and 100 for C and solving for X:

X(mp H2O) = 0.67(0) - 3.7 = -3.7

X(bp H2O) = 0.67(100) - 3.7 = 63.3