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When the temperature in degrees Fahrenheit \(\left(T_{\mathrm{F}}\right)\) is plotted vs. the temperature in degrees Celsius \(\left(T_{c}\right)\), a straight-line plot results. A straight-line plot also results when \(T_{\mathrm{C}}\) is plotted vs. \(T_{\mathrm{K}}\) (the temperature in kelvins). Reference Appendix A \(1.3\) and determine the slope and \(y\) -intercept of each of these two plots.

Short Answer

Expert verified
In conclusion, for the Fahrenheit-Celsius plot, the slope is \(9/5\) and the y-intercept is 32. For the Celsius-Kelvin plot, the slope is 1 and the y-intercept is -273.15.

Step by step solution

01

Find the relationship between Fahrenheit and Celsius

We know that the relationship between a temperature in degrees Fahrenheit \((T_F)\) and degrees Celsius \((T_C)\) can be expressed as: \[T_{\mathrm{F}} = (9/5) \times T_{\mathrm{C}} + 32\]
02

Calculate the slope and y-intercept for Fahrenheit-Celsius plot

The relationship between \(T_{\mathrm{F}}\) and \(T_{\mathrm{C}}\) can be rewritten in the form of a linear equation, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. From the formula in Step 1, we can see that the slope (m) and the y-intercept (b) are: - Slope (m) = 9/5 - Y-intercept (b) = 32 So, for the Fahrenheit-Celsius plot, the slope is 9/5, and the y-intercept is 32.
03

Find the relationship between Celsius and Kelvin

The relationship between a temperature in degrees Celsius \((T_{\mathrm{C}})\) and in Kelvins \((T_{\mathrm{K}})\) can be expressed as: \[T_{\mathrm{C}} = T_{\mathrm{K}} - 273.15\]
04

Calculate the slope and y-intercept for Celsius-Kelvin plot

The relationship between \(T_{\mathrm{C}}\) and \(T_{\mathrm{K}}\) can be rewritten in the form of a linear equation, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. From the formula in Step 3, we can see that the slope (m) and the y-intercept (b) are: - Slope (m) = 1 - Y-intercept (b) = -273.15 So, for the Celsius-Kelvin plot, the slope is 1, and the y-intercept is -273.15. In conclusion, the slope and y-intercept for the Fahrenheit-Celsius plot are 9/5 and 32, respectively, and for the Celsius-Kelvin plot, they are 1 and -273.15, respectively.

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Key Concepts

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

Fahrenheit
Fahrenheit is a unit of measurement for temperature frequently used in the United States.
It differs from the Celsius scale in both how intervals are divided and the point they start from.
  • The freezing point of water in Fahrenheit is 32°F.
  • The boiling point of water is 212°F.

To convert a temperature reading from Celsius to Fahrenheit, you use the linear equation:\[T_{F} = \frac{9}{5} \cdot T_C + 32\]This formula is derived from aligning the freezing and boiling points of water between the two scales and calculating the linear relationship.
Celsius
Celsius is a temperature scale based on the freezing and boiling points of water at standard atmospheric pressure. It is widely used in science and around the world for everyday temperature measurement.
  • The freezing point of water in degrees Celsius is 0°C.
  • The boiling point is 100°C.

The conversion of Celsius to Fahrenheit shows a linear relationship between the two, allowing us to easily switch back and forth between systems.
This scale is central when converting to Kelvin as well, due to its direct offset relationship.
Kelvin
The Kelvin scale is an absolute temperature scale used extensively in scientific settings as it begins at absolute zero, the theoretical absence of all thermal energy.
  • The size of the unit in Kelvin is the same as in Celsius.
  • To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.

This results in a linear equation:\[T_{\mathrm{K}} = T_{\mathrm{C}} + 273.15\]Kelvin is not only crucial for scientific calculations but also helps in formulating general gas laws without negative numbers, making complex equations easier to handle.
Linear Equations
Linear equations represent relationships between two variables. They graph as straight lines on a coordinate plane.
The general form is \(y = mx + b\). Here:
  • \(m\) is the slope of the line; it indicates how much \(y\) changes for each unit increase in \(x\).
  • \(b\) is the y-intercept; it is the point where the line crosses the y-axis.

Linear equations are crucial for temperature conversions. For instance, we derive equations for converting between Fahrenheit and Celsius, and Celsius and Kelvin from the alignments of their respective scales.
Slope and Y-Intercept
Understanding the slope and y-intercept is important in interpreting linear relationships.
The slope and y-intercept can provide insights into how variables like temperature relate to each other.
  • The slope is the rate of change between the two variables. A slope like \(\frac{9}{5}\) in the Fahrenheit-Celsius relationship suggests that for every increase of 1°C, the temperature increases by \(\frac{9}{5}\)°F.
  • The y-intercept is the starting point of the line when \(x\) (or in this case, Celsius or Kelvin) is zero. For temperatures, it often reflects the freezing or base point in the equation, like 32°F in Fahrenheit.
Understanding these concepts helps to decipher and predict changes across temperature scales effectively.

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Most popular questions from this chapter

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(\frac{2.526}{3.1}+\frac{0.470}{0.623}+\frac{80.705}{0.4326}\) b. \((6.404 \times 2.91) /(18.7-17.1)\) c. \(6.071 \times 10^{-5}-8.2 \times 10^{-6}-0.521 \times 10^{-4}\) d. \(\left(3.8 \times 10^{-12}+4.0 \times 10^{-13}\right) /\left(4 \times 10^{12}+6.3 \times 10^{13}\right)\) e. \(\frac{9.5+4.1+2.8+3.175}{4}\) (Assume that this operation is taking the average of four numbers. Thus 4 in the denominator is exact.) f. \(\frac{8.925-8.905}{8.925} \times 100\) (This type of calculation is done many times in calculating a percentage error. Assume that this example is such a calculation; thus 100 can be considered to be an exact number.)

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(6.022 \times 10^{23} \times 1.05 \times 10^{2}\) b. \(\frac{6.6262 \times 10^{-34} \times 2.998 \times 10^{8}}{2.54 \times 10^{-9}}\) c. \(1.285 \times 10^{-2}+1.24 \times 10^{-3}+1.879 \times 10^{-1}\) d. \(\frac{(1.00866-1.00728)}{6.02205 \times 10^{23}}\) e. \(\frac{9.875 \times 10^{2}-9.795 \times 10^{2}}{9.875 \times 10^{2}} \times 100(100\) is exact \()\) f. \(\frac{9.42 \times 10^{2}+8.234 \times 10^{2}+1.625 \times 10^{3}}{3}(3\) is exact)

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