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A student asserts that a suitable unit for specific heat is $1m^{2}/s^{2}\cdot ºC$. Is the correct? Why or why not?

My solution:

Specific heat units:

$$\frac{J}{kg\cdot ºC}$$

Joule units:

$$J=\frac{kg\cdot m^{2}}{s^{2}}$$

Therefore,

$$=\frac{\frac{kg\cdot m^{2}}{s^{2}}}{kg\cdot ºC}$$

$$=\frac{kg\cdot m^{2}}{s^{2}\cdot kg\cdot ºC}$$

$$=\frac{m^{2}}{s^{2}\cdot ºC}$$

So, it's possible to say that the student is correct... But, how to interpret the derived relation for describing specific heat?

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closed as off-topic by knzhou, Kyle Kanos, Jon Custer, AccidentalFourierTransform, heather Feb 5 '17 at 14:44

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  • $\begingroup$ The derived relation is the same as the relation up top. How do you interpret J/kg/C? $\endgroup$ – Jon Custer Feb 3 '17 at 14:39
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The new dimensions are correct. Mathematically, both the forms are equivalent. However, you must use the one which is easier to understand (the original relation in your case).

The derived relation, though technically correct, is hard to interpret (not sure if someone can even come up with a nice way to interpret it).

Here is one more example,

A unit of pressure is $\frac{N}{m^2}$.

Another unit for pressure is $\frac{J}{m^3}$.

When you are solving problems in mechanics, you are usually referring to pressure as perpendicular force per unit area. In this situation, using the first unit is always better. When you are solving problems from thermodynamics of an ideal gas, you might want to use the second form as we consider the pressure to be $\frac{2}{3}$ of the energy per unit volume.

$E_{per \space unit \space volume} = \frac{3}{2}P$

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It makes perfect sense for ideal gas:

\begin{align*} pV &= NkT \\ &= \frac{1}{3}N m \overline{v^2} \\ c_V &= \frac{\frac{1}{2} Nm \overline{v^2}-\frac{1}{2} Nm \overline{u^2}} {Nm\Delta T} \\ &= \frac{\overline{v^2}-\overline{u^2}}{\Delta T} \\ &= \frac{3k}{2} \end{align*}

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