21st century liquid

Classical Mechanics Level 3

A scientist's notes read so:

$\large{'}$ At last I invented the liquid in my lab. For a certain mass of the liquid it had such a firm volume-temperature relation for all range of temperature. It obeys

$\huge{V = (10^{-6})e^\frac{T}{100}}$

Where $\large{T}$ (in °C) is the temperature $\large{V}$ (in $m^{3}$ ) is the volume.

Now I need to make a thermometer with this . But how can I make the "temperature" to vary linearly with "height" of the liqiud column with such a weird $\large{T-V}$ relation?

I decided to make glass container with circular cross section such that radius of the cross sectional circle vary with height of the container such that when liquid expands the height of liquid column increases linearly with temperature.The container consists of a bulb at the bottom that contained certain mass of liquid and a glass expansion chamber with required shape. For every $\large{1°C }$ change in temperature the height raises by $\large{0.1 \hspace {0.1 cm}mm}$ . The chamber cross sectional radius varied as

$\huge{R = x e^{ yH}}$

Where $\large{R}$ is the radius of cross section at height $\large{H '}$ .

$\large{Find\hspace {0.1cm}xy.}$ up to 2 decimal places.

$\large \underline {Details \ and\ Assumptions}$

Bulb contains just the volume of the liquid at $\large{0°C}$ .

Neglect this voulme with respect to the container.

The answer is 0.28.

1 solution

Hari Govind Sekhar
Feb 26, 2018

Given V(T) =a $e^{bT}$

and kH=T ( $height \propto temperature$ )

V(H)=a $e^{bkH}$

$\frac{dV(H)}{d(H)}= abke^{bkH}$

V(H)= $\int_{0}^{H}$ abk $e^{bkH}$ d(H) $\longrightarrow$ $\boxed{1}$

Let F(H) be the radius at height H.

V(H)= $\int_{0}^{H} \pi F(H)^{2}dH \longrightarrow$ $\boxed{2}$
From $\boxed{1}$ and $\boxed{2}$ F(H)= $\sqrt{\frac{abk}{\pi}}e^{\frac{bkH}{2}}$

$a=10^{-6},b= \frac{1}{100},k=10^{4}$ (1°C for $10^{-4}m$ raise in height )

$\sqrt{\frac{abk}{\pi}} \frac{bkH}{2}= 0.28$

21st century liquid

The answer is 0.28.

1 solution

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