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Guyot 2024-12-17 21:26:39 +01:00
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16
ThermodynamiqueOS/ThermodynamiqueOS.tex
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@ -957,14 +957,14 @@ Par contre, on peut relever que le c\oe fficient \(\gamma\) de l'équation \ref{
\begin{figure*}[th!] \begin{figure*}[th!]
\centering \centering
\psfrag{Etape1~: isochore}{Étape A~: isochore} %\psfrag{Etape1~: isochore}{Étape A~: isochore}
\psfrag{Etape2~: isobare}{Étape B~: isobare} %\psfrag{Etape2~: isobare}{Étape B~: isobare}
\psfrag{Etape3~: isochore}{Étape C~: isochore} %\psfrag{Etape3~: isochore}{Étape C~: isochore}
\psfrag{Etape4~: isobare}{Étape D~: isobare} %\psfrag{Etape4~: isobare}{Étape D~: isobare}
\psfrag{Chauffage}{Chauffage} %\psfrag{Chauffage}{Chauffage}
\psfrag{Chargement}{Chargement} %\psfrag{Chargement}{Chargement}
\psfrag{Déchargement}{Déchargement} %\psfrag{Déchargement}{Déchargement}
\psfrag{Refroidissement}{Refroidissement} %\psfrag{Refroidissement}{Refroidissement}
\subfigure[Chargement\label{moteur1}]{\includegraphics[width=6cm]{Moteur1.eps}}\qquad \subfigure[Chargement\label{moteur1}]{\includegraphics[width=6cm]{Moteur1.eps}}\qquad
\subfigure[Chauffage\label{moteur2}]{\includegraphics[width=6cm]{Moteur2.eps}}\\ \subfigure[Chauffage\label{moteur2}]{\includegraphics[width=6cm]{Moteur2.eps}}\\
\subfigure[Déchargement\label{moteur3}]{\includegraphics[width=6cm]{Moteur3.eps}}\qquad \subfigure[Déchargement\label{moteur3}]{\includegraphics[width=6cm]{Moteur3.eps}}\qquad

36
ThermodynamiqueOS/ThermodynamiqueOS.tex.bak
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@ -153,7 +153,7 @@ Mercure & \num{7e-4}\\
Considérons l'exemple suivant~: Considérons l'exemple suivant~:
\smallskip \smallskip
Une piscine de \SI{10 x 5 x 2}{\metre} est remplie d'eau. Si on suppose que la matière du récipient qui la constitue ne se dilate pas, calculez le volume d'eau qui déborde de celle-ci quand elle est entièrement remplie à \SI{17}{\celsius} et que sa température s'élève à \SI{25}{\celsius}. Une piscine de 10 x 5 x 2 \si{\metre} est remplie d'eau. Si on suppose que la matière du récipient qui la constitue ne se dilate pas, calculez le volume d'eau qui déborde de celle-ci quand elle est entièrement remplie à \SI{17}{\celsius} et que sa température s'élève à \SI{25}{\celsius}.
\smallskip \smallskip
Réponse~: Réponse~:
@ -164,8 +164,8 @@ On commence par calculer le volume initial~:
Puis, on détermine la variation de volume~: Puis, on détermine la variation de volume~:
\begin{align*} \begin{align*}
\Delta V&=100\cdot 2\cdot 10^{-4}\cdot (25-17)\\ \Delta V&=100\cdot 3\cdot 2\cdot 10^{-4}\cdot (25-17)\\
&=\SI{0,16}{\metre\cubed}=\SI{160}{\deci\metre\cubed}=\SI{160}{\liter} &=\SI{0,48}{\metre\cubed}=\SI{480}{\deci\metre\cubed}=\SI{480}{\liter}
\end{align*} \end{align*}
\section{Chaleurs spécifique et latente} \section{Chaleurs spécifique et latente}
@ -282,7 +282,8 @@ Lorsque plusieurs matières, ou états de la matière, à températures différe
Considérons maintenant les deux exemples suivants~: Considérons maintenant les deux exemples suivants~:
\medskip \medskip
Un thermos d'un litre est remplis au deux tiers d'eau chaude à \SI{80}{\celsius}. La température moyenne du thermos est alors de \SI{60}{\celsius}. Si la capacité thermique du thermos vaut \SI{0,4}{\celsius}, quelle doit être en grammes la masse d'eau froide à \SI{0}{\celsius} qu'il faut mettre dans le thermos pour que la température d'équilibre s'établisse à \SI{40}{\celsius} ? Est-ce possible ? Un thermos d'un litre est remplis au deux tiers d'eau chaude à \SI{80}{\celsius}. Quelle doit être en gramme la masse d'eau froide à \SI{0}{\celsius} qu'il faut mettre dans le thermos pour que la température d'équilibre soit de \SI{40}{\celsius} ? Initialement la température du thermos est de \SI{60}{\celsius} et sa capacité thermique \SI{0,4}{\joule\per\celsius}. Est-ce possible ?
\smallskip \smallskip
Réponse~: Réponse~:
@ -294,7 +295,7 @@ Q_m&=m\cdot c_{eau}\cdot (\theta_{\acute{e}q}-\theta_{froide})\\
&=m\cdot 4,18\cdot 10^3\cdot (40-0)\\ &=m\cdot 4,18\cdot 10^3\cdot (40-0)\\
&=167'200\cdot m &=167'200\cdot m
\end{align*} \end{align*}
La chaleur cédée par l'eau chaude à \SI{90}{\celsius} vaut~: La chaleur cédée par l'eau chaude à \SI{80}{\celsius} vaut~:
\begin{align*} \begin{align*}
Q_{chaude}&=V_{chaude}\cdot \rho_{eau}\cdot c_{eau}\cdot (\theta_{\acute{e}q}-\theta_{chaude})\\ Q_{chaude}&=V_{chaude}\cdot \rho_{eau}\cdot c_{eau}\cdot (\theta_{\acute{e}q}-\theta_{chaude})\\
&=\frac{2}{3}\cdot 10^{-3}\cdot 10^3\cdot 4,18\cdot 10^3\cdot (40-80)\\ &=\frac{2}{3}\cdot 10^{-3}\cdot 10^3\cdot 4,18\cdot 10^3\cdot (40-80)\\
@ -316,7 +317,8 @@ Q_m+Q_{chaude}+Q_{thermos}&=0\\
m&=\SI{0,666}{\kilo\gram} m&=\SI{0,666}{\kilo\gram}
\end{align*} \end{align*}
Soit en terme de volume~: \SI{0,666}{\liter}. Il n'est donc pas possible de mettre cette quantité d'eau dans le thermos puisqu'il ne reste qu'un tiers de litre. Soit en terme de volume~: \SI{0,666}{\liter}. Il n'est donc pas possible de mettre cette quantité d'eau dans le thermos puisqu'il ne reste qu'un tiers de litre.
\medskip
\bigskip
L'exemple suivant est plus complexe, car il fait intervenir un changement d'état. Un récipient de capacité thermique négligeable contient un demi-litre d'eau à \SI{10}{\celsius}. On y verse \SI{200}{\gram} de glace à \SI{-20}{\celsius}. Quel est l'état d'équilibre final et quelle est sa température ? L'exemple suivant est plus complexe, car il fait intervenir un changement d'état. Un récipient de capacité thermique négligeable contient un demi-litre d'eau à \SI{10}{\celsius}. On y verse \SI{200}{\gram} de glace à \SI{-20}{\celsius}. Quel est l'état d'équilibre final et quelle est sa température ?
\medskip \medskip
@ -563,7 +565,7 @@ Alors que pour une molécule triatomique rigide, on a trois degrés de liberté
E=\frac{6}{2}\cdot n\cdot R\cdot T E=\frac{6}{2}\cdot n\cdot R\cdot T
\end{equation} \end{equation}
L'épreuve du théorème de l'équipartition de l'énergie va venir de sa confrontation avec les chaleurs spécifiques. En effet, comme on le verra par la suite au paragraphe \ref{par:chalspe, on peut montrer que : L'épreuve du théorème de l'équipartition de l'énergie va venir de sa confrontation avec les chaleurs spécifiques. En effet, comme on le verra par la suite au paragraphe \ref{par:chalspec}, page \pageref{par:chalspec}, on peut montrer que :
\begin{align} \begin{align}
C_V&=\frac{i}{2}\cdot R\\ C_V&=\frac{i}{2}\cdot R\\
C_p&=(\frac{i}{2}+1)\cdot R=\frac{i+2}{2}\cdot R C_p&=(\frac{i}{2}+1)\cdot R=\frac{i+2}{2}\cdot R
@ -594,7 +596,7 @@ Polyatomique & \color{violet}{i=7} & & & & & & & & \\
\end{center} \end{center}
\end{shaded} \end{shaded}
%\end{figure*} %\end{figure*}
\caption{Modèle de l'équipartition de l'énergie\\\color{red}{\(C_V=3R/2\)}, \color{blue}{\(C_V=5R/2\)}, \color{green}{\(C_V=6R/2\)}, \color{violet}{\(C_V=7R/2\)}\\\color{red}{\(C_p=5R/2\)}, \color{blue}{\(C_p=7R/2\)}, \color{green}{\(C_p=8R/2\)}, \color{violet}{\(C_p=9R/2\)}\label{tab:equipartition}} \caption[Modèle de l'équipartition de l'énergie]{Modèle de l'équipartition de l'énergie\\\color{red}{\(C_V=3R/2\)}, \color{blue}{\(C_V=5R/2\)}, \color{green}{\(C_V=6R/2\)}, \color{violet}{\(C_V=7R/2\)}\\\color{red}{\(C_p=5R/2\)}, \color{blue}{\(C_p=7R/2\)}, \color{green}{\(C_p=8R/2\)}, \color{violet}{\(C_p=9R/2\)}\label{tab:equipartition}}
\end{sidewaysfigure*} \end{sidewaysfigure*}
\subsection{Premier principe} \subsection{Premier principe}
@ -916,7 +918,7 @@ Q&=0\\
}} }}
\end{center} \end{center}
\subsection{Chaleurs spécifiques} \subsection{Chaleurs spécifiques}\label{par:chalspec}
Rappelons que la chaleur spécifique molaire est définie comme la quantité de chaleur nécessaire pour élever la température d'une mole d'un gaz de un degré. Soit~: Rappelons que la chaleur spécifique molaire est définie comme la quantité de chaleur nécessaire pour élever la température d'une mole d'un gaz de un degré. Soit~:
\begin{equation} \begin{equation}
C=\frac{Q}{n\cdot \Delta \theta}\;\Rightarrow\;Q=n\cdot C\cdot \Delta \theta C=\frac{Q}{n\cdot \Delta \theta}\;\Rightarrow\;Q=n\cdot C\cdot \Delta \theta
@ -955,14 +957,14 @@ Par contre, on peut relever que le c\oe fficient \(\gamma\) de l'équation \ref{
\begin{figure*}[th!] \begin{figure*}[th!]
\centering \centering
\psfrag{Etape1~: isochore}{Étape A~: isochore} %\psfrag{Etape1~: isochore}{Étape A~: isochore}
\psfrag{Etape2~: isobare}{Étape B~: isobare} %\psfrag{Etape2~: isobare}{Étape B~: isobare}
\psfrag{Etape3~: isochore}{Étape C~: isochore} %\psfrag{Etape3~: isochore}{Étape C~: isochore}
\psfrag{Etape4~: isobare}{Étape D~: isobare} %\psfrag{Etape4~: isobare}{Étape D~: isobare}
\psfrag{Chauffage}{Chauffage} %\psfrag{Chauffage}{Chauffage}
\psfrag{Chargement}{Chargement} %\psfrag{Chargement}{Chargement}
\psfrag{Déchargement}{Déchargement} %\psfrag{Déchargement}{Déchargement}
\psfrag{Refroidissement}{Refroidissement} %\psfrag{Refroidissement}{Refroidissement}
\subfigure[Chargement\label{moteur1}]{\includegraphics[width=6cm]{Moteur1.eps}}\qquad \subfigure[Chargement\label{moteur1}]{\includegraphics[width=6cm]{Moteur1.eps}}\qquad
\subfigure[Chauffage\label{moteur2}]{\includegraphics[width=6cm]{Moteur2.eps}}\\ \subfigure[Chauffage\label{moteur2}]{\includegraphics[width=6cm]{Moteur2.eps}}\\
\subfigure[Déchargement\label{moteur3}]{\includegraphics[width=6cm]{Moteur3.eps}}\qquad \subfigure[Déchargement\label{moteur3}]{\includegraphics[width=6cm]{Moteur3.eps}}\qquad