3.2: Temperature
- Page ID
- 101157
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you should be able to:
- Understand the physical properties described by the temperature
Calculate conversions between scales of temperature
Temperature
Temperature is operationally defined as the quantity of what we measure with a thermometer. The average kinetic energy of a molecule is directly proportional to its absolute temperature. Differences in temperature maintain heat transfer through different systems. Heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature.\(^{[1][2]}\)
Assigned scales are based on physical measurements:
Celsius (°C) – based on water freezing (0 °C) and boiling (100 °C) at atmospheric pressure. Absolute zero at -273.15°C.
Fahrenheit (°F) – based on water freezing (32 °F) and boiling (212 °F) at atmospheric pressure. Absolute zero at -459.67°F.
There are also other absolute scales with 0 being absolute zero:
Kelvin (K) – Same size of a degree as Celsius
Rankine (°R) – Same size of a degree as Fahrenheit
Defining Equations
Celsius to Kelvin: \(T (K) = T (°C) + 273.15\)
Fahrenheit to Rankine: \(T (°R) = T (°F) + 459.67\)
Kelvin to Rankine: \(T (°R) = 1.8·T (K)\)
Celsius to Fahrenheit: \(T (°F) = 1.8·T (°C) + 32\)
Some solar plants use molten salt as a heat transfer agent that can be stored efficiently. This allows the plant to operate after the sun has set by extracting energy from the molten salt they have stored during the day. The temperature difference is important to know as it will dictate the energy obtainable from the salt. If the molten salt is heated to 1000°F on a given day and if the environment is at 25°C what is the temperature difference in Kelvin?
Solution
Add example text here.
First, convert the two temperatures to the same units. Because the question asks for the temperature difference in Kelvin, it is more convenient to compare on a Celsius scale because the sizes of 1 degree are the same for Celcius/Kelvin.
Step 1: Convert the temperature in Fahrenheit to Celsius:
\[T(°C)=\frac{T(°F)-32°F}{1.8\frac{°C}{°F}}=\frac{1000°F-32°F}{1.8\frac{°C}{°F}}=537.8°C\]
Step 2: Calculate the difference in temperature in Celsius, which is equal to the difference in Kelvin (you can prove this to yourself if desired if you convert both temperatures above from Celsius to Kelvin).
\[\Delta T(°C)=537.8°C-25°C=512.8°C\]
We can use Newton’s second law \(F=ma\) to find mass, with \(a\) being the acceleration of gravity. Since the mass we are looking for is “per square metre”, we can also divide the force by area(in \(m^2\)), which gives us pressure on the left side of the equation.
\[mass\;(per\;m^2)=\frac{P}{a}=\frac{1.013×10^5Pa}{9.81\frac{m}{s^2}}=\frac{1.013×10^5\frac{kg}{ms^2}}{9.81\frac{m}{s^2}}=10326\frac{kg}{m^2}\]
References
[1] OpenStax University Physics Volume 2. 2016. 1.1 Temperature and Thermal Equilibrium. [online]<https://openstax.org/books/university-physics-volume-2/pages/1-1-temperature-and-thermal-equilibrium> [Accessed 13 May 2020].
[2] OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature, and RMS Speed. [online]<https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed> [Accessed 13 May 2020].