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Repeat Exercise 1, except choose the new temperature scale \(\mathrm{Q}\) so that absolute zero is \(0^{\circ} \mathrm{Q}\) and \(T_{\mathrm{bp} \text { , water }}-T_{\text {mp, water }}=\) \(100 \mathrm{Q}^{\circ} .(a)\) What is the conversion formula from Celsius to Q? (b) What is \(T_{\text {bp, water }}\) and \(T_{\text {mp,water }}\) in Q? ( \(c\) ) This scale actually exists. What is the official name?

Short Answer

Expert verified
The conversion formula from Celsius to Q is \(T_Q = T_C + 273.15\). The melting point of water is 273.15°Q and the boiling point is 373.15°Q. This Q scale is officially known as the Kelvin scale.

Step by step solution

01

Determine Conversion Factor

First, express the Q scale in terms of the Celsius scale, taking into account that the difference between the boiling point and melting point of water is 100 degrees on both scales. Let's denote the Celsius temperature as \(T_C\) and the Q temperature as \(T_Q\). We know that \(T_{\mathrm{bp, water}} ^Q - T_{\mathrm{mp, water}} ^Q = 100 ^{\circ}\) and also \(T_{\mathrm{bp, water}} ^C - T_{\mathrm{mp, water}} ^C = 100 ^{\circ}\). This implies that the conversion factor is 1.
02

Define the Conversion Formula

Since absolute zero on the Q scale is 0, and absolute zero on the Celsius scale is -273.15°C, the conversion formula from Celsius to Q can be expressed as: \(T_Q = T_C + 273.15\). This formula states that to convert a temperature from the Celsius scale to the Q scale, add 273.15.
03

Compute Q temperatures for Water

Now, we use the conversion formula to find \(T_{\mathrm{bp, water}} ^Q\) and \(T_{\mathrm{mp, water}} ^Q\). For the melting point of water (0°C in Celsius scale), \(T_{\mathrm{mp, water}} ^Q = 0 + 273.15 = 273.15 ^{\circ} \mathrm{Q}\). For the boiling point of water (100°C in Celsius scale), \(T_{\mathrm{bp, water}} ^Q = 100 + 273.15 = 373.15 ^{\circ} \mathrm{Q}\).
04

Identify the Official Name

Lastly, the Q temperature scale in this exercise corresponds to the Kelvin scale in real life. The Kelvin scale is an absolute thermodynamic temperature scale used widely in the physical sciences and engineering.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Conversion
Temperature conversion is all about changing temperature measurements from one scale to another. Since different temperature scales exist, it’s important to understand how to switch between them to accurately convey thermal conditions.
For instance, common temperature scales include Celsius, Fahrenheit, and Kelvin, each with its own reference points.
  • The Celsius scale is based on the freezing point (0°C) and boiling point (100°C) of water.
  • The Fahrenheit scale uses 32°F for water's freezing point and 212°F for boiling.
  • The Kelvin scale starts at absolute zero, the theoretically lowest temperature possible, equal to -273.15°C.

To convert temperatures effectively, formulas like the one used to change Celsius to Kelvin or Fahrenheit can be applied. Understanding these formulas help in switching between scales and is essential in scientific measurements and everyday thermometer use.
Celsius to Kelvin
Converting Celsius to Kelvin is straightforward and essential in scientific calculations, especially when dealing with thermodynamic temperatures.
The Kelvin scale, unlike Celsius, starts at absolute zero, where all molecular motion ceases, making it very useful in scientific settings.
Formula for Celsius to Kelvin Conversion:
  • The formula to convert Celsius ( T_C) to Kelvin ( T_K) is: \( T_K = T_C + 273.15 \)
  • This means, if you have a temperature in Celsius, simply add 273.15 to find its equivalent in Kelvin.

For instance, water's freezing point is 0°C. Converted to Kelvin, it becomes 273.15 K. Similarly, water boils at 100°C, which equals 373.15 K.
This conversion is crucial as Kelvin is often used in areas such as physics and chemistry. Unlike Celsius, Kelvin provides an absolute starting point with no negative temperatures.
Thermodynamic Temperature Scale
The thermodynamic temperature scale is essential in measuring absolute temperatures based on fundamental physical principles. Amongst the available scales, the Kelvin scale stands out as the standard thermodynamic temperature scale.
Why Kelvin Scale?
  • The Kelvin scale starts at absolute zero, which is the lowest possible temperature where molecular motion stops entirely.
  • This makes it a true reflection of energy levels, allowing scientists to easily compute thermal energies.
  • Using the Kelvin scale bypasses the issue of negative temperatures, simplifying many thermodynamic calculations.

The Kelvin scale is widely used in scientific research and industry because it relates directly to absolute energy measurements. It ensures that calculations in fields like thermodynamics, chemistry, and physics are consistent and universally understood.
Mastery of this scale and the understanding of thermodynamic principles makes scientific inquiry on temperature much more precise.

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Most popular questions from this chapter

(a) Calculate the volume occupied by \(1.00 \mathrm{~mol}\) of an ideal gas at standard conditions - that is, pressure of \(1.00 \mathrm{~atm}\) \(\left(=1.01 \times 10^{5} \mathrm{~Pa}\right)\) and temperature of \(0^{\circ} \mathrm{C}(=273 \mathrm{~K}) \cdot(b)\) Show that the number of molecules per cubic centimeter (the Loschmidt number) at standard conditions is \(2.68 \times 10^{19}\).

A quantity of ideal gas at \(12.0^{\circ} \mathrm{C}\) and a pressure of \(108 \mathrm{kPa}\) occupies a volume of \(2.47 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? \((b)\) If the pressure is now raised to \(316 \mathrm{kPa}\) and the temperature is raised to \(31.0^{\circ} \mathrm{C}\), how much volume will the gas now occupy? Assume there are no leaks.

The amplification or gain of a transistor amplifier may depend on the temperature. The gain for a certain amplifier at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) is \(30.0\), whereas at \(55.0^{\circ} \mathrm{C}\) it is 35.2. What would the gain be at \(28.0^{\circ} \mathrm{C}\) if the gain depends linearly on temperature over this limited range?

A resistance thermometer is a thermometer in which the electrical resistance changes with temperature. We are free to define temperatures measured by such a thermometer in kelvins (K) to be directly proportional to the resistance \(R\), measured in ohms \((\Omega) .\) A certain resistance thermometer is found to have a resistance \(R\) of \(90.35 \Omega\) when its bulb is placed in water at the triple-point temperature \((273.16 \mathrm{~K}) .\) What temperature is indicated by the thermometer if the bulb is placed in an environment such that its resistance is \(96.28 \Omega ?\)

The best vacuum that can be attained in the laboratory corresponds to a pressure of about \(10^{-18} \mathrm{~atm}\), or \(1.01 \times 10^{-13} \mathrm{~Pa}\). How many molecules are there per cubic centimeter in such a vacuum at \(22^{\circ} \mathrm{C} ?\)

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