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Chapter 21: Problem 3

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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Most popular questions from this chapter

An automobile tire has a volume of \(988 \mathrm{in} .^{3}\) and contains air at a gauge pressure of \(24.2 \mathrm{lb} / \mathrm{in}^{2}\) where the temperature is \(-2.60^{\circ} \mathrm{C}\). Find the gauge pressure of the air in the tire when its temperature rises to \(25.6^{\circ} \mathrm{C}\) and its volume increases to \(1020 \mathrm{in} .{ }^{3}\). (Hint: It is not necessary to convert from British to SI units. Why? Use \(p_{\text {atm }}=14.7 \mathrm{lb} / \mathrm{in}^{2} .\) )

(a) Prove that the change in rotational inertia \(I\) with temperature of a solid object is given by \(\Delta I=2 \alpha I \Delta T .(b)\) A thin uniform brass rod, spinning freely at 230 rev/s about an axis perpendicular to it at its center, is heated without mechanical contact until its temperature increases by \(170 \mathrm{C}^{\circ}\). Calculate the change in angular velocity.

A \(1.28-\mathrm{m}\) -long vertical glass tube is half-filled with a liquid at \(20.0^{\circ} \mathrm{C}\). How much will the height of the liquid column change when the tube is heated to \(33.0^{\circ} \mathrm{C}\) ? Assume that \(\alpha_{\text {glass }}=1.1 \times 10^{-5} / \mathrm{C}^{\circ}\) and \(\beta_{\text {liquid }}=4.2 \times 10^{-5} / \mathrm{C}^{\circ}\)

Oxygen gas having a volume of \(1130 \mathrm{~cm}^{3}\) at \(42.0^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa}\) expands until its volume is \(1530 \mathrm{~cm}^{3}\) and its pressure is \(106 \mathrm{kPa}\). Find \((a)\) the number of moles of oxygen in the system and ( \(b\) ) its final temperature.

( \(a\) ) Prove that the change in period \(P\) of a physical pendulum with temperature is given by \(\Delta P=\frac{1}{2} \alpha P \Delta T .(b)\) A clock pendulum made of invar has a period of \(0.500 \mathrm{~s}\) and is accurate at \(20^{\circ} \mathrm{C}\). If the clock is used in a climate where the temperature averages \(30^{\circ} \mathrm{C}\), what approximate correction to the time given by the clock is necessary at the end of 30 days?
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