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# A 1 500-kg car is moving at 20.0 m/s. The driver brakes to a stop. The brakes cool off to the temperature of the surrounding air, which is nearly constant at 20.0$^ { \circ } \mathrm { C }$. What is the total entropy change?

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### 1 Concepts and Principles

1- The $\textbf{kinetic energy}$ of an object of mass $\color{#c34632}{m}$ moving with a speed $\color{#c34632}{v}$ is defined as:

$\begin{equation*} K=\dfrac{1}{2}mv^2\tag{1} \end{equation*}$

2- The $\textbf{entropy change}$ $\textcolor{#c34632}{\Delta S}$ for an irreversible process that takes a system from an initial state $\textcolor{#c34632}{i}$ to a final state $\textcolor{#c34632}{f}$ is equal to the entropy change $\textcolor{#c34632}{\Delta S}$ for any reversible process that takes the system between those same two states:

$\begin{equation*} \Delta S=\int_{i}^{f}\dfrac{dQ}{T}\tag{2} \end{equation*}$

For a reversible isothermal process, the previous equation reduces to:

$\begin{equation*} \Delta S=\dfrac{Q}{T}\tag{3} \end{equation*}$

where $\textcolor{#c34632}{Q}$ is the energy transferred as heat to or from the system during the process and $\textcolor{#c34632}{T}$ is the temperature of the system in kelvins during the process.

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