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On a linear \(X\) temperature scale, water freezes at \(-125.0^{\circ} X\) and boils at \(360.0^{\circ} \mathrm{X}\). On a linear \(\mathrm{Y}\) temperature scale, water freezes at \(-70.00^{\circ} \mathrm{Y}\) and boils at \(-30.00^{\circ} \mathrm{Y}\). A temperature of \(50.00^{\circ} \mathrm{Y}\) corresponds to what temperature on the X scale?

Short Answer

Expert verified
50.00°Y corresponds to 1330.00°X.

Step by step solution

01

Identify Known Data Points

We have the following information:- On the X scale: water freezes at \(-125.0^{\circ} X\) and boils at \(360.0^{\circ} X\).- On the Y scale: water freezes at \(-70.00^{\circ} Y\) and boils at \(-30.00^{\circ} Y\).These will serve as our reference points to create equations that relate the two scales.
02

Establish Linear Equations for Both Scales

The relationship between the X and Y scales can be expressed as a linear equation. Using the formula for a linear conversion:\[ T_X = mT_Y + c \]where \(T_X\) and \(T_Y\) are temperatures in X and Y scales, respectively, \(m\) is the slope, and \(c\) is the intercept that we need to determine.
03

Calculate Slope (m) of the Linear Equation

The slope \(m\) between two scales can be calculated using the formula:\[ m = \frac{\text{change in X}}{\text{change in Y}} = \frac{360.0 - (-125.0)}{-30.00 - (-70.00)} \]\[ m = \frac{360.0 + 125.0}{-30.00 + 70.00} = \frac{485.0}{40.00} = 12.125 \]
04

Calculate Intercept (c) of the Linear Equation

Using one of the known points, substitute the freezing point of water into the linear equation to find \(c\):\[ T_X = mT_Y + c \]Substitute \(T_X = -125.0\) and \(T_Y = -70.0\):\[ -125.0 = 12.125(-70.0) + c \]Calculate:\[ -125.0 = -848.75 + c \]\[ c = 723.75 \]
05

Formulate the Conversion Equation

Now that we have both the slope \(m = 12.125\) and the intercept \(c = 723.75\), assemble the conversion equation:\[ T_X = 12.125T_Y + 723.75 \]
06

Convert 50.00°Y to X Scale

Substitute \(T_Y = 50.00\) into the conversion equation:\[ T_X = 12.125(50.00) + 723.75 \]Calculate:\[ T_X = 606.25 + 723.75 = 1330.00 \]Thus, 50.00°Y corresponds to 1330.00°X.

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Key Concepts

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

Linear Equations
Linear equations are an essential concept in mathematics that describe a straight-line relationship between two variables. These equations are expressed in the form: \[ y = mx + c \]where:
  • \( y \) and \( x \) are variables,
  • \( m \) is the slope of the line, indicating the rate of change, and
  • \( c \) is the y-intercept, showing the point where the line intersects the y-axis.
The simplicity of linear equations makes them invaluable for describing relationships that depend on two varying factors. In the context of temperature conversions, this linear model allows for a systematic method to translate values from one temperature scale to another. By identifying the reference points and changes between freezing and boiling, we derive an equation that explains the exact conversion from one system to another.
Temperature Scales
Temperature scales are the frameworks used to measure and communicate warmth or coldness. Popular scales include Celsius, Fahrenheit, and Kelvin. Each scale has standard reference points that help define it in comparison to others. For arbitrary scales such as X and Y used in the problem, water's freezing and boiling points act as custom reference points to interrelate temperatures between the two scales. Understanding temperature scales and their relative points is vital. They allow us to create equations that can transfer values from one scale to another, as seen with the conversion formula between scales X and Y. Grasping how these scales interconnect helps in forming accurate conversions suitable to the context provided.
Slope and Intercept
In a linear equation, the slope and intercept are crucial for understanding the equation's characteristics. The **slope** \( m \) signifies the steepness or incline of the line. It’s found using the formula:\[ m = \frac{\text{change in Y}}{\text{change in X}} \]This indicates how much one unit on the X scale changes relative to a unit on the Y scale. In our exercise, the slope \( m = 12.125 \) shows how steep the conversion is between both temperature scales.The **intercept** \( c \) reveals where the line crosses the vertical axis. It tells us the starting point of the linear equation in a temperature context. Our intercept \( c = 723.75 \) was calculated by applying a known reference, allowing us to form the complete conversion equation. These components enable linear equations to be tailored to diverse scenarios, offering a meaningful transition between different temperature measures.

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Most popular questions from this chapter

Suppose that on a linear temperature scale \(X\), water boils at \(-72.0^{\circ} \mathrm{X}\) and freezes at \(-123.0^{\circ} \mathrm{X}\). What is a temperature of \(59.0 \mathrm{~K}\) on the X scale? (Approximate water's boiling point as \(373 \mathrm{~K}\).)

A small electric immersion heater is used to heat \(170 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "180 watts" (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from \(23.0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\), ignoring any heat losses.

A steel rod is \(3.000 \mathrm{~cm}\) in diameter at \(-10.00^{\circ} \mathrm{C}\). A brass ring has an interior diameter of \(2.992 \mathrm{~cm}\) at \(-10.00^{\circ} \mathrm{C}\) At what common temperature will the ring just slide onto the rod?

Evaporative cooling of beverages A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces The container and beverage have temperature \(T=10^{\circ} \mathrm{C}\), the environment has temperature \(T_{\mathrm{cuv}}=32^{\circ} \mathrm{C}\), and the container is a cylinder with radius \(r=2.2 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). Approximate the emissivity as \(\varepsilon=1\), and neglect other energy exchanges. At what rate \(d m / d t\) is the container losing water mass?

At \(20^{\circ} \mathrm{C}\), a rod is exactly \(20.05 \mathrm{~cm}\) long on a steel ruler. Both are placed in an oven at \(250^{\circ} \mathrm{C}\), where the rod now measures \(20.11\) \(\mathrm{cm}\) on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made?

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