Q.1.30 - An Introduction to Thermal Physics

Question

Put a few spoonfuls of water into a bottle with a tight lid. Make sure everything is at room temperature, measuring the temperature of the water with a thermometer to make sure. Now close the bottle and shake it as hard as you can for several minutes. When you're exhausted and ready to drop, shake it for several minutes more. Then measure the temperature again. Make a rough calculation of the expected temperature change, and compare. 

Step-by-Step Solution

Answer

The expected Temperature is$\mathrm{\Delta T}=0.05 \mathrm{K}$.

Step 1. The assumption from the given question.

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$20 \mathrm{cm}$.

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

Step 2. Energy equation.

Energy equation,

$\frac{1}{2}·{\mathrm{v}}^2=\mathrm{C}·\mathrm{\Delta T}$

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

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

So,

$\begin{array}{rcl}\mathrm{\Delta T}& =& \frac{{\mathrm{v}}^2}{2·\mathrm{C}}\\ & =& \frac{1.2^2}{2·4180}\\ & =& 1.72·{10}^4 \mathrm{s}\end{array}$

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.

$\begin{array}{rcl}\mathrm{\Delta T}& =& 1.72·{10}^{-4}·5·60\\ & =& 0.05 \mathrm{K}\end{array}$.