Question

Suppose that a new temperature scale has been devised on which the melting point of ethanol \(\left(-117.3^{\circ} \mathrm{C}\right)\) and the boiling point of ethanol \(\left(78.3^{\circ} \mathrm{C}\right)\) are taken as \(0^{\circ} \mathrm{S}\) and \(100^{\circ} \mathrm{S},\) respectively, where \(\mathrm{S}\) 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 \(25^{\circ} \mathrm{C} ?\)

Short answer

The thermometer reads approximately \(72.78^{\circ} \text{S}\) at \(25^{\circ} \text{C}\).

Step-by-step solution

Identify the Key Points on Each Scale

First, identify that on the Celsius scale, the melting and boiling points of ethanol are \(-117.3^{\circ} \text{C}\) and \(78.3^{\circ} \text{C}\) respectively. On the new scale (\(S\)): - Melting point corresponds to \(0^{\circ} \text{S}\)- Boiling point corresponds to \(100^{\circ} \text{S}\).

Establish Linear Conversion Relationship

The relationship between Celsius and the new scale \(S\) can be described by using the linear formula:\[ S = mC + c \]where \(m\) is the slope and \(c\) is the intercept.

Calculate the Slope (m)

Calculate the slope \(m\) using the formula for the slope between two points:\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{100 - 0}{78.3 - (-117.3)} \]Hence,\[ m = \frac{100}{78.3 + 117.3} \approx \frac{100}{195.6} = 0.5112 \]

Calculate the Intercept (c)

Using the linear equation \(S = mC + c\) for one known point, substitute the melting point values:\[ 0 = 0.5112(-117.3) + c \]Solve for \(c\):\[ c = 0 + 0.5112 \times 117.3 \approx 60 \]

Formulate the Final Conversion Equation

Now that we have the slope \(m = 0.5112\) and intercept \(c = 60\), substitute these values into the linear equation:\[ S = 0.5112C + 60 \]

Convert the Celsius Reading to the New Scale

Substitute \(25^{\circ} \text{C}\) into the conversion equation:\[ S = 0.5112 \times 25 + 60 \approx 12.78 + 60 = 72.78 \]The thermometer reads \(72.78^{\circ} \text{S}\) at \(25^{\circ} \text{C}\).

Key concepts

Ethanol Melting Point

Ethanol is a common alcohol with a known melting point of \(-117.3^{\circ} \mathrm{C}\). This is the temperature at which ethanol changes state from solid to liquid. When designing temperature scales, known points such as the melting and boiling points of substances can be used as reference. In this exercise, the ethanol melting point is set to \(0^{\circ} \mathrm{S}\), as it serves as a key reference point for converting between the Celsius and the new temperature scale. Understanding these fixed points helps create a reliable and accurate conversion between scales. By defining the melting point on a new scale, it establishes the starting point for that scale's temperature range.

Ethanol Boiling Point

The boiling point of ethanol is at \(78.3^{\circ} \mathrm{C}\). This is the temperature where ethanol evaporates to form a gas, transitioning from liquid to vapor. Ethanol's boiling point is crucial in defining the upper limit on our new temperature scale, where it is set at \(100^{\circ} \mathrm{S}\). Therefore, by assigning specific values to both the melting and boiling points on the new scale, we create an interval that mirrors the range of Celsius for a specific substance like ethanol. Assigning the boiling point helps establish the scale's span and allows anyone using it to predict and understand temperature relations effectively.

Celsius to New Scale Conversion

Converting temperatures between scales requires a systematic approach. In this exercise, we developed a new temperature scale by selecting ethanol's physical state constant points. This involves identifying two known temperatures on both the Celsius and the new scale \(S\). Knowing that the melting and boiling points correspond directly between the scales allows us to formulate a linear relationship. This conversion is particularly helpful in fields like chemistry, where precise thermal measurements are frequently necessary. It allows you to transform thermal readings from Celsius to the new scale efficiently, providing flexibility in various scientific and practical applications.

Linear Conversion Equation

The process of converting temperatures between the Celsius scale and the new scale \(S\) utilizes a linear conversion equation. This is expressed as: \[ S = mC + c \]where \(m\) is the slope and \(c\) is the intercept. The slope is calculated by taking the difference in the new scale values divided by the differences in Celsius values of the known points (melting and boiling). In this case, it solved to be approximately \(0.5112\). The intercept \(c\) is found by calculating the offset of where this line of conversion crosses the new scale at zero Celsius, solved to be around \(60\). By plugging these values into the equation, we establish a formula that efficiently translates temperatures from Celsius to the new scale, ensuring consistency and accuracy in thermal readings.