/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 On a Strange temperature scale, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On a Strange temperature scale, the freezing point of water is \(-15.0^{\circ} \mathrm{S}\) and the boiling point is \(+60.0^{\circ} \mathrm{S} .\) Develop a linear conversion equation between this temperature scale and the Celsius scale.

Short Answer

Expert verified
The conversion equation between the Celsius scale and the Strange temperature scale is \(S = (4/3)C + 20\).

Step by step solution

01

Identify the known points

From the problem we can determine two points in each scale that represent the same temperature: (-15°C, 0°C) corresponds to the freezing point of water and (60°C, 100°C) corresponds to the boiling point.
02

Formulate equations

Let's assume the conversion involves a simple linear relation between Celsius (C) and Strange temperature (S) scales: S = mC + b, where m is the slope, and b the y-intercept (temperature on the Strange scale when Celsius temperature is zero). For the two known points we can write two equations: for freezing point, 0 = m(-15) + b and for boiling point, 100 = m60 + b.
03

Solve the system of equations

This system of linear equations can be solved either by substitution or by elimination method. However, it is simpler here to use the elimination method: subtract the second equation from the first, \( 0 - 100 = m(-15 - 60) + b - b \). This simplifies to \( -100 = -75m \) giving \( m = 100 / 75 = 4 /3 \). Substitute \( m = 4/3 \) into the first equation to find b: \( 0 = (4/3)(-15) + b -> b = 20 \).
04

Write the conversion equation

The final step is to enter the values for m and b obtained in the previous step into the original conversion equation. The final equation that transforms Celsius into Strange scale is: \( S = (4/3)C + 20 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Linear Conversion Equation
Understanding linear conversion equations is essential when we want to translate values from one measurement system to another. In our case, we aim to convert temperatures between a known scale, Celsius, and an unknown 'Strange' scale.

A linear conversion equation is, at its core, a simple linear equation of the form: \( y = mx + b \), where \( y \) represents the value in the new scale, \( x\) is the value in the original scale, \( m\) is the slope of the line (also known as the rate of conversion), and \( b\) is the y-intercept (the value of \( y\) when \( x = 0\)).

In the context of temperature conversion, the slope \( m\) tells us how much the temperature on one scale changes for a one-degree change on the other scale. The y-intercept \( b\) indicates the value on one scale when the temperature on the other scale is zero. By figuring out these constants, \( m\) and \( b\), we can move back and forth between the Strange and Celsius scales.
Celsius Scale
The Celsius scale, also known as centigrade, is a temperature scale based on the boiling and freezing points of water, which are precisely defined as 100°C and 0°C respectively, at sea level and standard atmospheric pressure.

It's a widely adopted scale that is used for everyday temperature measurements in most of the world, except for a few countries that use the Fahrenheit scale. When performing conversions between temperature scales, understanding the reference points of the Celsius scale is crucial, as they often serve as the foundation for creating conversion equations such as the one needed to translate temperatures to and from the Strange scale.

Additionally, the Celsius scale has a direct relationship with the Kelvin scale used in scientific measurements, with the formula \( K = C + 273.15 \) to switch between them.
System of Linear Equations
A system of linear equations consists of two or more linear equations with the same set of variables, which we aim to solve simultaneously. For temperature scale conversion, we establish a system based on known reference points that correspond to the same physical temperatures across scales. In this problem, we identify the freezing and boiling points of water on both scales and express these as linear equations.

To solve the system, various methods can be used, such as substitution, graphing, or elimination. The elimination method often proves efficient when the system is simple and involves two equations and two variables, as it allows for the quick elimination of a variable by adding or subtracting the equations. By manipulating the equations, we solve for one variable at a time, ultimately obtaining the values needed for the linear conversion equation ('m' and 'b').

Understanding the concept of systems of linear equations and being fluent in solving them is a valuable skill, not only for converting temperatures but also for various scientific and mathematical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A liquid has a density \(\rho\). (a) Show that the fractional change in density for a change in temperature \(\Delta T\) is \(\Delta \rho / \rho=-\beta \Delta T\). What does the negative sign signify? (b) Fresh water has a maximum density of \(1.0000 \mathrm{g} / \mathrm{cm}^{3}\) at \(4.0^{\circ} \mathrm{C} .\) At \(10.0^{\circ} \mathrm{C},\) its density is \(0.9997 \mathrm{g} / \mathrm{cm}^{3} .\) What is \(\beta\) for water over this temperature interval?

In a chemical processing plant, a reaction chamber of fixed volume \(V_{0}\) is connected to a reservoir chamber of fixed volume \(4 V_{0}\) by a passage containing a thermally insulating porous plug. The plug permits the chambers to be at different temperatures. The plug allows gas to pass from either chamber to the other, ensuring that the pressure is the same in both. At one point in the processing, both chambers contain gas at a pressure of 1.00 atm and a temperature of \(27.0^{\circ} \mathrm{C}\). Intake and exhaust valves to the pair of chambers are closed. The reservoir is maintained at \(27.0^{\circ} \mathrm{C}\) while the reaction chamber is heated to \(400^{\circ} \mathrm{C}\). What is the pressure in both chambers after this is done?

The pressure gauge on a tank registers the gauge pressure, which is the difference between the interior and exterior pressure. When the tank is full of oxygen \(\left(\mathrm{O}_{2}\right),\) it contains \(12.0 \mathrm{kg}\) of the gas at a gauge pressure of \(40.0 \mathrm{atm} .\) Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 25.0 atm. Assume that the temperature of the tank remains constant.

A popular brand of cola contains \(6.50 \mathrm{g}\) of carbon dioxide dissolved in \(1.00 \mathrm{L}\) of soft drink. If the evaporating carbon dioxide is trapped in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C}\) what volume does the gas occupy?

A beaker made of ordinary glass contains a lead sphere of diameter \(4.00 \mathrm{cm}\) firmly attached to its bottom. At a uniform temperature of \(-10.0^{\circ} \mathrm{C},\) the beaker is filled to the brim with \(118 \mathrm{cm}^{3}\) of mercury, which completely covers the sphere. How much mercury overflows from the beaker if the temperature is raised to \(30.0^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.