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Chapter 1: Problem 78

Suppose you decide to define your own temperature scale with units of \(\mathrm{O},\) using the freezing point \(\left(13^{\circ} \mathrm{C}\right)\) and boiling point \(\left(360^{\circ} \mathrm{C}\right)\) of oleic acid, the main component of olive oil. If you set the freezing point of oleic acid as \(0^{\circ} \mathrm{O}\) and the boiling point as \(100^{\circ} \mathrm{O},\) what is the freezing point of water on this new scale?

Short Answer

Expert verified
The freezing point of water on the new temperature scale (O) is approximately -3.75 °O.

Step by step solution

01

Identify given values and temperature scales

We are given a problem that is defined in two different temperature scales: Celsius (C) and the new scale called O. Given values: - Freezing point of oleic acid in Celsius: \(13^{\circ}\mathrm{C}\) - Boiling point of oleic acid in Celsius: \(360^{\circ}\mathrm{C}\) - Freezing point of oleic acid in new scale O: \(0^{\circ}\mathrm{O}\) - Boiling point of oleic acid in new scale O: \(100^{\circ}\mathrm{O}\) - Freezing point of water in Celsius: \(0^{\circ}\mathrm{C}\) (from general knowledge) We need to find the freezing point of water in the new scale O.
02

Find the ratio between two scales

The first step in finding the relationship between two temperature scales is to find the ratio between their units. To do this, we will use the difference between the values of the two points in both scales. In Celsius: - Difference between freezing and boiling points of oleic acid: \(360 - 13 = 347^{\circ}\mathrm{C}\) In new scale O: - Difference between freezing and boiling points of oleic acid: \(100 - 0 = 100^{\circ}\mathrm{O}\) Therefore, the ratio between the two scales is: Ratio = \(\frac{\Delta O}{\Delta C} = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\)
03

Find the relationship between the two scales

Now, we need to create a linear relationship between the Celsius and O temperature scales. Let's denote the temperature in Celsius by T(C) and in O by T(O). Since we know the ratio between the two scales, we can set up an equation: \(T(O) - T_O(0) = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(T(C) - T_C(0))\) where \(T_O(0)\) is the freezing point of oleic acid in O (0 °O), and \(T_C(0)\) is the freezing point of oleic acid in Celsius (13 °C).
04

Find the freezing point of water in the new scale

Now, we can use the relationship we found in Step 3 to find the freezing point of water in the new O scale. We'll input the known freezing point of water in Celsius (0 °C) and solve for T(O). \(T(O) - 0^{\circ}\mathrm{O} = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(0^{\circ}\mathrm{C} - 13^{\circ}\mathrm{C})\) \(T(O) = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(-13^{\circ}\mathrm{C})\) \(T(O) = -\frac{1300^{\circ}\mathrm{O}}{347}\) \(T(O) \approx -3.75^{\circ}\mathrm{O}\) Therefore, the freezing point of water on this new temperature scale is approximately -3.75 °O.

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Most popular questions from this chapter

Carry out the following operations and express the answers with the appropriate number of significant figures. $$ \begin{array}{ll}{\text { (a) } 14.3505+2.65} & {\text { (b) } 952.7-140.7389} \\\ {\text { (c) }\left(3.29 \times 10^{4}\right)(0.2501)} & {\text { (d) } 0.0588 / 0.677}\end{array} $$

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Is the use of significant figures in each of the following statements appropriate? (a) The 2005 circulation of National Geo graphic was \(7,812,564\) . (b) On July 1, 2005, the population of Cook County, Illinois, was \(5,303,683 .(\mathbf{c})\) In the United States, 0.621\(\%\) of the population has the surname Brown. (\boldsymbol{d} ) ~ Y o u ~ c a l c u l a t e ~ y o u r ~ g r a d e ~ p o i n t ~ a v e r a g e ~ t o ~ b e ~ 3.87562

Round each of the following numbers to four significant figures and express the result in standard exponential notation: (a) \(102.53070,\) (b) \(656.980,\) (c) 0.008543210 ,(d) \(0.000257870 (\mathbf{e})-0.0357202\)

Water has a density of 0.997 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C} ;\) ice has a density of 0.917 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(-10^{\circ} \mathrm{C}\) (a) If a soft-drink bottle whose volume is 1.50 \(\mathrm{L}\) is completely filled with water and then frozen to \(-10^{\circ} \mathrm{C},\) what volume does the ice occupy? (b) Can the ice be contained within the bottle?
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