91Ó°ÊÓ

There is room temperature water in the bottle. The goal is to predict how much the temperature of the water will drop in a few minutes of shaking the container.

Start with some assumptions. If you try to shake the bottle$3$times in a second, you will succeed. So it $3$times up and$3$times down, for a total of $6$moves. If you pay closer attention to the shaking, you'll see that the distance between each up and down movement is around$20\mathrm{cm}$.

localid="1648469916086" $\begin{array}{rcl}\mathrm{v}& =& \frac{\mathrm{s}}{\mathrm{t}}\\ & =& \frac{0.2}{1/6}\\ & =& 1.2\frac{\mathrm{m}}{\mathrm{s}}\end{array}$.

Energy equation,

$\frac{1}{2}·{\mathrm{v}}^2=\mathrm{C}·\mathrm{Δ°Õ}$

Need to find Specific heat$\mathrm{C}$.

$\mathrm{C}=4.18\frac{\mathrm{kJ}}{\mathrm{molK}}$

So,

localid="1648469974559" role="math" $\begin{array}{rcl}\mathrm{Δ°Õ}& =& \frac{{\mathrm{v}}^2}{2·\mathrm{C}}\\ & =& \frac{1.2^2}{2·4180}\\ & =& 1.72·{10}^4\mathrm{s}\end{array}$

To figure out how much it will change in a matter of minutes. Let's say$5$minutes have passed. After$5$minutes, calculate the temperature change.

localid="1648469963214" $\begin{array}{rcl}\mathrm{Δ°Õ}& =& 1.72·{10}^{-4}·5·60\\ & =& 0.05\mathrm{K}\end{array}$.