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Physical Chemistry : Other Thermodynamic Principles

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Example Question #1 : Other Thermodynamic Principles

If a $140\textup{ g}$ $\displaystyle 140\textup{ g}$ copper wire gains $280\textup{ J}$ $\displaystyle 280\textup{ J}$of heat, what is the change in temperature of the wire?

$C_{\text{copper}}=0.384 \ \frac{\text J}{\text{g}^\circ \text C}$ $\displaystyle C_{\text{copper}}=0.384 \ \frac{\text J}{\text{g}^\circ \text C}$

Possible Answers:

$15^\circ \text{C}$ $\displaystyle 15^\circ \text{C}$

$0.19^\circ \text{C}$ $\displaystyle 0.19^\circ \text{C}$

$5.2^\circ \text{C}$ $\displaystyle 5.2^\circ \text{C}$

$0.77^\circ \text{C}$ $\displaystyle 0.77^\circ \text{C}$

$1.3^\circ \text{C}$ $\displaystyle 1.3^\circ \text{C}$

Correct answer:

$5.2^\circ \text{C}$ $\displaystyle 5.2^\circ \text{C}$

Explanation:

The thermal energy an object contains is given by:

$q=mC\Delta T$ $\displaystyle q=mC\Delta T$

Where $\displaystyle C$ is the specific heat, $\Delta T$ $\displaystyle \Delta T$ is the change in temperature, $\displaystyle m$ is the mass, and $\displaystyle q$ is the amount of energy. We are given everything but the $\Delta T$ $\displaystyle \Delta T$, so we rearrange the equation to $\Delta T= \frac{q}{mC}$ $\displaystyle \Delta T= \frac{q}{mC}$

Plug in known values and solve:

$\Delta T=\frac{280}{140*0.384}=5.2^\circ \text{C}$ $\displaystyle \Delta T=\frac{280}{140*0.384}=5.2^\circ \text{C}$

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Physical Chemistry : Other Thermodynamic Principles

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