Problem 45
Question
You decide to establish a new temperature scale on which the melting point of mercury \(\left(-38.9^{\circ} \mathrm{C}\right)\) is \(0^{\circ} \mathrm{M},\) and the boiling point of mercury \(\left(356.9^{\circ} \mathrm{C}\right)\) is \(100^{\circ} \mathrm{M} .\) What would be (a) the boiling point of water in \(^{\circ} \mathrm{M} ;\) and \((\mathrm{b})\) the temperature of absolute zero in \(^{\circ}\text{M}\)?
Step-by-Step Solution
Verified Answer
To answer the results are as follows: (a) The boiling point of water in M degrees is just the value you obtained from substituting T_C = 100 into the equation. (b) The absolute zero temperature in M degrees is the result you got from substituting T_C = -273.15 into the equation applied.
1Step 1: Define the linear equation for the conversion
First, form a linear equation using the given mercury points. Since we know that at -38.9°C, the temperature is 0M, and at 356.9°C, it's 100M, the equation will be in the form: \(T_M = a T_C + b\). But for T_C = -38.9°C, T_M = 0M and for T_C = 356.9°C, T_M = 100M.
2Step 2: Calculation of the equation coefficients
To find the coefficients a and b, we'll substitute the given values into the equation. Starting with the known point (-38.9,0), we substitute these as T_C and T_M in the equation, obtaining: \(0 = a*(-38.9) + b\). And then, using the point (356.9,100): \(100=a*356.9 +b\). Solving these simultaneous equations gives us the values of a and b for the equation.
3Step 3: Calculating temperatures in Mercury scale
With the linear equation established, we can find the Mercury equivalent for any Celsius temperature. For example, to solve for part (a), where the boiling point of water is 100°C, substitute T_C = 100 into the equation to find T_M. For part (b), consider that absolute zero is -273.15°C in the Celsius scale, so substitute T_C = -273.15 to find T_M.
Key Concepts
Linear Equation DerivationMelting and Boiling PointsAbsolute Zero Temperature
Linear Equation Derivation
When creating a new temperature scale, it's crucial to define how the temperatures will convert from one scale to the other. This conversion is typically represented by a linear equation of the form:
To derive this equation, we use two known points: the melting point of mercury \((-38.9^{\circ} \text{C}, 0^{\circ} \text{M})\) and its boiling point \( (356.9^{\circ} \text{C}, 100^{\circ} \text{M})\). These points give us a system of equations:
- \( T_M = a \cdot T_C + b \)
To derive this equation, we use two known points: the melting point of mercury \((-38.9^{\circ} \text{C}, 0^{\circ} \text{M})\) and its boiling point \( (356.9^{\circ} \text{C}, 100^{\circ} \text{M})\). These points give us a system of equations:
- \( 0 = a \cdot (-38.9) + b \)
- \( 100 = a \cdot 356.9 + b \)
Melting and Boiling Points
Temperature scales, like the newly created Mercury Scale, need defined reference points for consistency. For most scales, the melting and boiling points of a substance are chosen because they are fixed and repeatable. These points determine the scale's intervals.
For the Mercury Scale, the melting and boiling points of mercury were chosen, with \(-38.9^{\circ}\text{C}\) set to \(0^{\circ}\text{M}\) and \(356.9^{\circ}\text{C}\) set to \(100^{\circ}\text{M}\). By establishing these points:
For the Mercury Scale, the melting and boiling points of mercury were chosen, with \(-38.9^{\circ}\text{C}\) set to \(0^{\circ}\text{M}\) and \(356.9^{\circ}\text{C}\) set to \(100^{\circ}\text{M}\). By establishing these points:
- \(-38.9^{\circ}\) as the baseline or starting point (\(0^{\circ}\text{M}\)).
- \(356.9^{\circ}\) as the endpoint (\(100^{\circ}\text{M}\)).
Absolute Zero Temperature
Absolute zero is a crucial reference in understanding temperature because it represents the theoretical lowest temperature possible. At this point, all molecular motion ceases, and it is equivalent to \(-273.15^{\circ}\text{C}\) in the Celsius scale.
Understanding this concept helps in expanding the Mercury Scale to measure a broader range of temperatures. Since absolute zero is a fundamental benchmark across all temperature scales, converting it into our Mercury Scale involves substituting \( T_C = -273.15 \) in our derived linear equation. By doing this, we find the Mercury Scale equivalent of the absolute zero point.
Absolute zero helps in studying various physical properties and behaviors of substances at extremely low temperatures. Although it is not achievable in practice, establishing it in a newly defined scale provides a complete understanding of temperature measurements from the coldest possible to hotter temperatures like the boiling point of mercury.
Understanding this concept helps in expanding the Mercury Scale to measure a broader range of temperatures. Since absolute zero is a fundamental benchmark across all temperature scales, converting it into our Mercury Scale involves substituting \( T_C = -273.15 \) in our derived linear equation. By doing this, we find the Mercury Scale equivalent of the absolute zero point.
Absolute zero helps in studying various physical properties and behaviors of substances at extremely low temperatures. Although it is not achievable in practice, establishing it in a newly defined scale provides a complete understanding of temperature measurements from the coldest possible to hotter temperatures like the boiling point of mercury.
Other exercises in this chapter
Problem 43
The absolute zero of temperature is \(-273.15^{\circ} \mathrm{C}\) Should it be possible to achieve a temperature of \(-465^{\circ} \mathrm{F} ?\) Explain.
View solution Problem 44
A family/consumer science class is given an assignment in candy-making that requires a sugar mixture to be brought to a "soft-ball" stage \(\left(234-240^{\circ
View solution Problem 46
You decide to establish a new temperature scale on which the melting point of ammonia \(\left(-77.75^{\circ} \mathrm{C}\right)\) is \(0^{\circ}\) A and the boil
View solution Problem 47
A 2.18 L sample of butyric acid, a substance present in rancid butter, has a mass of 2088 g. What is the density of butyric acid in grams per milliliter?
View solution