Question

Suppose that a new temperature scale has been devised on which the melting point of ethanol (-117.3°C and the boiling point of ethanol (78.3°C) are taken as OOS and 100°S, respectively, where S is the symbol for the new temperature scale. Derive an equation relating a reading on this scale to a reading on the Celsius scale. What would this thermometer read at 25°C?

Answer 1

Answer:

Equation S= (°C + 117.3)/1.956     °S(25°C) = 72.25°S

Explanation:

To know the difference between the scales, we must initially know the length of our new scale compared to the scale in Celcius degrees, for this we add the absolute value of our melting point and the boiling point | -117.3 ° C | + | 78.3 ° C | = 195.6 ° C

Because our new scale will be divided into 100 equal parts we can know the unit value of each degree ° C concerning degrees ° S. 195.6 / 100 = 1,956 ° C / ° S

For a linear relationship of the scales, the crossing point at ° S = 0.00 is equal to -117.3 ° C. whereby we can clear our equation to know the temperature in degrees Celcius from ° S

°C= -117.3 + °S * 1.956°C/°S

Clearing °S we obtain

°S = (°C + 117.3)/1.956°C/°S

(ii)

In this new scale the temperature of 25°C will be readed as

°S = (25°C+117.3)/1.956°C/°S = 72.25°S

Answer 2

Answer:

See explanation below

Explanation:

In this case, let's form the equation into something similar to the known equations of temperature, for example:

°F = 9/5°C + 32 (1)

°K = °C + 273 (2)

As you can see, in both equations we need to add something to the celsius grade. We cannot use the equation of °K because if you make the difference between the new scale and the °C for both melting and boiling points, the result do not match. Let's suppose we say that

°S = °C + 117.3 °C

When you apply this formula to the boiling point, the result is not matching the given point of 100, so, let's put it as the equation of Farenheit:

°S = x°C + y (3)

Where "x" and "y" will be the numbers that will be added to the °C to convert it to the new scale of S.

We apply this formula to both melting and boiling point, and solve for x and y:

For melting:

0 = -117.3x + y

y = 117.3x (4)

Now, for boiling, we replace (4):

100 = 78.3x + y

100 = 78.3x + 117.3x

100 = 195.6x

x = 100/195.6

x = 0.5112

Now, let's replace this value to get "y":

y = 117.3 * 0.5112

y = 59.97 and we can round it to 60

Now, let's confirm this data match the given data (or really near):

°S = 0.5112 * (-117.3) + 60

°S = -59.97 + 60

°S = 0.03 °C

This result match the given m.p. of ethanol in the new scale (Remember that we round the value of y to a round number)

Therefore the new equation would be:

°S = 0.5112°C + 60

Now, 25°C in the new scale would be:

°S = 0.5112*25 + 60

°S = 72.78 °S