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

Short Answer

Expert verified
In summary, for the plot of temperature in Fahrenheit (\(T_F\)) vs. temperature in Celsius (\(T_C\)), the slope is 9/5 and the y-intercept is 32. For the plot of temperature in Celsius (\(T_C\)) vs. temperature in Kelvin (\(T_K\)), the slope is 1 and the y-intercept is 273.15.

Step by step solution

01

Convert between temperature units

To solve this problem, we need to know the formulas for converting between Fahrenheit, Celsius, and Kelvin. These formulas are: 1. Celsius to Fahrenheit: \(T_F = (9/5)T_C + 32\) 2. Celsius to Kelvin: \(T_K = T_C + 273.15\) Let's find the slope and y-intercept for each required plot using these conversion formulas.
02

Determine the slope and y-intercept for the \(T_F\) vs. \(T_C\) plot

To find the slope and y-intercept for the plot of \(T_F\) vs. \(T_C\), analyze the formula for converting Celsius to Fahrenheit: \(T_F = (9/5)T_C + 32\) This formula can be written in the form of a linear equation: \(y = mx + b\), where: - y is the temperature in Fahrenheit, \(T_F\) - x is the temperature in Celsius, \(T_C\) - m is the slope - b is the y-intercept Comparing the two equations, we find: - The slope, m, is 9/5. - The y-intercept, b, is 32. Therefore, the slope of the \(T_F\) vs. \(T_C\) plot is 9/5, and the y-intercept is 32.
03

Determine the slope and y-intercept for the \(T_C\) vs. \(T_K\) plot.

To find the slope and y-intercept for the plot of \(T_C\) vs. \(T_K\), analyze the formula for converting Celsius to Kelvin: \(T_K = T_C + 273.15\) This formula can also be written in the form of a linear equation: \(y = mx + b\) , where: - y is the temperature in Kelvin, \(T_K\) - x is the temperature in Celsius, \(T_C\) - m is the slope - b is the y-intercept Comparing the two equations, we find: - The slope, m, is 1. - The y-intercept, b, is 273.15. Therefore, the slope of the \(T_C\) vs. \(T_K\) plot is 1, and the y-intercept is 273.15.

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Key Concepts

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

Linear Equations in Chemistry
In chemistry, linear equations play a vital role in understanding the relationships between various properties, such as temperature conversions. A linear equation typically depicts a straight line on a graph, showing a direct proportionality between two variables. This scenario is clearly illustrated with temperature conversion formulas.
For example, when converting from Celsius to Fahrenheit, the linear equation is represented as:
  • \[T_F = \left(\frac{9}{5}\right)T_C + 32\]
Here, the equation format \(y = mx + b\) is identified, where:
  • \(y\) is the dependent variable, equivalent to \(T_F\), the temperature in Fahrenheit.
  • \(x\) is the independent variable, \(T_C\), the temperature in Celsius.
  • \(m\) is the slope, illustrating how steam rises between the varying degrees, specifically \(9/5\) in this case.
  • \(b\) is the y-intercept, showing where the line crosses the y-axis, at \(32\).
A similar linear equation applies when converting Celsius to Kelvin:
  • \[T_K = T_C + 273.15\]
This maintains the form \(y = mx + b\), where the slope \(m\) is \(1\) and the y-intercept \(b\) is \(273.15\). Such equations make it simpler to visualize and calculate relationships in scientific contexts. Understanding these equations enhances problem-solving skills in chemistry, as linear relationships are frequently encountered in this field.
Fahrenheit to Celsius Conversion
The conversion of temperatures from Fahrenheit to Celsius is a common operation in chemistry to ensure that measurements are standardized, especially when comparing findings globally. This conversion uses the inverse of the formula \(T_F = \left(\frac{9}{5}\right)T_C + 32\). By rearranging the equation, Celsius can be expressed in terms of Fahrenheit:
  • \[T_C = \left(\frac{5}{9}\right)(T_F - 32)\]
Here, understanding the placement of the slope helps clarify changes in temperature scales:
  • By subtracting \(32\), the additional Fahrenheit baseline is removed.
  • The factor \(\left(\frac{5}{9}\right)\) aligns the scale intervals between Fahrenheit and Celsius.
This method ensures precise transitions and accurate molecular behavior predictions, which are pivotal in chemical reactions and experiments. Mastering this conversion equips students with the ability to deftly handle temperature variations across differing scales.
This type of conversion is crucial when reports and studies originate from regions utilizing distinct units, ensuring cross-compatibility and uniformity in scientific communication.
Celsius to Kelvin Conversion
In scientific studies, converting Celsius to Kelvin is essential, particularly in chemistry and physics, where the Kelvin scale is preferred due to its absolute zero reference point. The conversion is straightforward and highlighted by the equation:
  • \[T_K = T_C + 273.15\]
Here, Kelvin is defined by adding \(273.15\) to the Celsius temperature, emphasizing that the Kelvin scale begins from absolute zero鈥攖he point where theoretical molecular motion ceases. This conversion is instrumental in calculations involving thermodynamic properties.
  • It facilitates equations necessitating absolute temperatures, such as the Ideal Gas Law.
  • This conversion emphasizes that each degree increment on a Kelvin scale equates to a degree increment on a Celsius scale, only differing by the integer shift.
Because many chemical processes and properties depend on accurate thermal calculations, utilizing Kelvin ensures coherence and consistency in data measurement and interpretation. Understanding this conversion can significantly improve comprehension in practical and theoretical chemistry applications, laying a foundational block in the study of thermal dynamics.

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