Adjusting the thermometer

Given is,

coefficient of real expansion of mercury, $\gamma_r$ = $.18^{-3} K^{-1}$

Radius of the glass tube, $r= \frac {diameter}{2} = \frac {0.15 mm}{2} = .075 mm$

Minimal height increase required for manual distinction ( Taking this as height of the cylinder made of mercury ), $h= 0.05 mm$

Primary volume of mercury, $V_{\circ} = .255 cm^3 = 255 mm^3$

Minimal temperature increase distinguished manually, $\Delta \theta = ?$

Our known equation about real expansion of liquid ,
$\Delta V = \gamma_rV_{\circ} \Delta \theta$

$\iff \Delta \theta = \frac {\Delta V}{ \gamma_rV_{\circ}}$

$\iff \Delta \theta = \frac { \pi r^2 h}{\gamma_rV_{\circ}}$

$\iff \Delta \theta = \frac { \pi (.075 mm)^2 (.05mm)}{.18^{-3} K^{-1} \times 255 mm^3}$

$\iff \Delta \theta \approx 0.0193 K =0.0193^\circ \mathrm{C}$

And that's the answer!

First, find the minimum volume variation that can be seen: [; \mathbf{ \Delta V = cylinder \ area \times minimum \ height} ;], where cylinder area equals [; \mathbf{ 7.5 \times \pi \ mm^2 } ;] and minimum height equals [; \mathbf{ 0.05 \ mm } ;] [; \mathbf{ \Delta V = \pi \times \left (7.5 \times 10^{-2} \right )^2 \times 0.05 = 8.84 \times 10^{-4} mm^3 } ;]

Then, use the volume variation formula: [; \mathbf{ \Delta V = V_{0} \times \gamma \times \Delta \theta } ;] [; \mathbf{ 8.84 \times 10^{-4} = 0.255 \times 10^{3} \times 18 \times 10^(-5) \times \Delta \theta } ;]

Solving for [; \mathbf{ \Delta \theta } ;], we find: [; \mathbf{ \Delta \theta = 0.019 } ;] ° [; \mathbf{ C } ;]

using formula ΔV=V* ɑ d ΔT, we have ΔT= ΔV/(V ɑ), and ΔV=hS=h pi d^2/4. ΔT=h pi d^2/(4V ɑ)=0.05 3.14 0.15 0.15/(4 255 0.18×10^(−3))=0.01924

using formula deltaV=V* ɑ deltaT, we have deltaT=deltaV/(V ɑ), and deltaV=hS=h pi d^2/4. deltaT=h pi d^2/(4 V ɑ)=0.05 3.14 0.15 0.15/(4 255* 0.18×10^(−3))=0.01924

Looking at the units of the thermal expansion coefficient we can deduce that this coefficient multiplied by the temperature yields some sort of percentage, or fractional change of the volume.

We can find the minimum volume change needed by calculating the volume of a 0.05 mm thick slice of the thermometer.

$V = \pi \times 0.005cm \times 0.0075cm^2 = 8.83 \times 10^-7 cm^3$ (Units converted to cm to work conveniently with the given mercury container volume)

To find out what fractional change this corresponds to, we divide this change in volume by the given volume

$Fractional Change = \frac {\delta V} {V} = \frac {8.83 \times 10^-7 cm^3} {0.255 cm^3} = 3.465 \times10^-6$

Now, we finish by calculating what temperature is required for this fractional change.

$Fractional Change = Vol.Coeff \times \delta T$

$\delta T = \frac {Fractional Change} {Vol.Coeff} = \frac {3.465 \times 10^-6} {0.18 \times 10^-3 \frac {1} {K}} = 0.0193 K$

However, since a change of 1 degree Kelvin is equal to a change of 1 degree Celsius, we can conclude that a change of 0.0193 degrees Celsius is the minimum change detectable by the naked eye.

The volume of the mercury will expand $\Delta V = \alpha \Delta T V_0$ . That volume pushes the mercury inside the tube, and we can read it off as a height distance of $h = \frac{\Delta V}{A}$ . Since $h$ can minimally be $0.05~\mbox{mm}$ , we get that $\Delta T_{min} = \frac {A h}{\alpha V_0} = 0.0193^\circ \mbox{C}$ .

Simply applying expression : Co-efficient of volume expansion ( $gamma$ ) = $delta V / V * 1/ delta T$ we can solve for $delta T$ .