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Dessins/Moteur1_backup.svg
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Dessins/Moteur1_backup.svg
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xml:space="preserve"
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y="36.583683">Etape3 : isochore</tspan></text>
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After Width: | Height: | Size: 17 KiB |
@ -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
|
||||||
|
|||||||
@ -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 ?
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\smallskip
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\smallskip
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Réponse~:
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Réponse~:
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@ -294,7 +295,7 @@ Q_m&=m\cdot c_{eau}\cdot (\theta_{\acute{e}q}-\theta_{froide})\\
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&=m\cdot 4,18\cdot 10^3\cdot (40-0)\\
|
&=m\cdot 4,18\cdot 10^3\cdot (40-0)\\
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&=167'200\cdot m
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&=167'200\cdot m
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\end{align*}
|
\end{align*}
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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*}
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||||||
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})\\
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||||||
&=\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)\\
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@ -316,7 +317,8 @@ Q_m+Q_{chaude}+Q_{thermos}&=0\\
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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.
|
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\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
|
||||||
|
|||||||
Loading…
Reference in New Issue
Block a user