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Chapter 13: Problem 5

Aliens from the planet Jeenkah have based their temperature scale on the boiling and freezing temperatures of ethyl alcohol. These temperatures are \(78^{\circ} \mathrm{C}\) and \(-114^{\circ} \mathrm{C}\) respectively. The people of Jeenkah have six digits on each hand, so they use a base- 12 number system and have decided to have \(144^{\circ} \mathrm{J}\) between the freezing and boiling temperatures of ethyl alcohol. They set the freezing point to \(0^{\circ} \mathrm{J} .\) How would you convert from " \(\mathrm{J}\) to \(^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Question: Convert a given temperature in Jeenkahn (°J) to Celsius (°C) using the provided conversion equation. Answer: To convert a given temperature in Jeenkahn (°J) to Celsius (°C), use the conversion equation: \(T_C = -114^{\circ}\mathrm{C} + (\frac{4}{3})T_J\)

Step by step solution

01

Identify the known values

We know the following values in Celsius and Jeenkahn: 1. Freezing temperature of ethyl alcohol: \(-114^{\circ}\mathrm{C}=0^{\circ}\mathrm{J}\) 2. Boiling temperature of ethyl alcohol: \(78^{\circ}\mathrm{C}=144^{\circ}\mathrm{J}\)
02

Determine the temperature difference in both scales

In Celsius, the difference between boiling and freezing temperatures is: \(78^{\circ}\mathrm{C}-(-114^{\circ}\mathrm{C})=192^{\circ}\mathrm{C}\) In Jeenkahn, the difference between boiling and freezing temperatures is: \(144^{\circ}\mathrm{J}-0^{\circ}\mathrm{J}=144^{\circ}\mathrm{J}\)
03

Find the conversion factor

To find the conversion factor, we need to determine how many degrees Celsius correspond to one degree Jeenkahn. So, we divide the Celsius temperature difference by the Jeenkahn temperature difference: Conversion factor = \(\frac{192^{\circ}\mathrm{C}}{144^{\circ}\mathrm{J}}=\frac{4}{3}\) Hence, one degree \(\mathrm{J}\) is equal to \(\frac{4}{3}^{\circ}\mathrm{C}\).
04

Establish the conversion equation

To establish the conversion equation, let \(T_J\) be the given temperature in Jeenkahn, and \(T_C\) be the unknown temperature in Celsius. We need to find \(T_C\). Our reference point is the freezing point of ethyl alcohol which corresponds to: \(-114^{\circ}\mathrm{C}=0^{\circ}\mathrm{J}\) The additional degrees in Celsius from the reference point (\(-114^{\circ}\mathrm{C}\)) are given by the multiplication of the conversion factor and the additional degrees in Jeenkahn (since \(T_J - 0 = T_J\)): Additional Degrees Celsius = \((\frac{4}{3})T_J\) The conversion equation is then: \(T_C = -114^{\circ}\mathrm{C} + (\frac{4}{3})T_J\)
05

Use the conversion equation

To convert any given temperature in Jeenkahn (in \(^{\circ}\mathrm{J}\)) to Celsius (in \(^{\circ}\mathrm{C}\)), just plug the Jeenkahn temperature value into \(T_J\) in the equation above: \(T_C = -114^{\circ}\mathrm{C} + (\frac{4}{3})T_J\) By following these steps, a student will be able to convert temperatures between the Jeenkahn and Celsius scales.

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Most popular questions from this chapter

A scuba diver has an air tank with a volume of \(0.010 \mathrm{m}^{3}\) The air in the tank is initially at a pressure of \(1.0 \times 10^{7} \mathrm{Pa}\) Assuming that the diver breathes \(0.500 \mathrm{L} / \mathrm{s}\) of air, find how long the tank will last at depths of (a) \(2.0 \mathrm{m}\) and (b) $20.0 \mathrm{m} .\( (Make the same assumptions as in Example \)13.6 .)$
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