Question

A mercury thermometer has a cylindrical capillary tube with an internal diameter of 0.2 mm. If the volume of the thermometer and that of the bulb are not affected by temperature, what volume must the bulb have if a sensitivity of 2mm/°C is to be obtained? Assume operation near 24°C. Assume the stem volume is negligible compared with the bulb internal volume. Differential expansion coefficient of Hg=1.82x10^-4 ml/(ml°C).

Answer 1

To solve this problem we will proceed to calculate the specific volume from the area of the cylinder and the sensitivity. Later we will calculate the volumetric coefficient of thermal expansion and finally we will be able to calculate the volume through the relation of the two terms mentioned above. Our values are

$\text{Sensitivity}= 2mm/\°C$

$\text{Internal diameter } d= 0.2mm$

$\text{Differential expansion of Hg } \lambda_L = 1.82*10^{-4}/\°C$

Let's start by calculating the specific volume which is given by

$v = \pi (\frac{d}{2})^2 \gamma$

Here,

d = Diameter

$\gamma$ = Sensitivity

Replacing our values we have

$v = (\frac{\pi}{4})(0.2mm)^2(2mm/\°C)$

$v = 0.0628mm^3 /\°C$

Now we will obtain the value of the volumetric coefficient of thermal expansion of mercury through the differential expansion coefficient of Hg whic is three times, then

$\lambda_V = 3\lambda_L$

$\lambda_V = 3(1.82*10^{-4}/\°C)$

$\lambda_V = 5.46*10^{-4}/\°C$

Finally the relation to calculate the volume the bulb must is

$\text{Specific volume} = \text{Bulb Volume} \times \text{Volumetric Coefficient}$

$v = v_B \times \lambda_V$

$v_B = \frac{v}{\lambda_V}$

$v_B = \frac{0.0628mm^3/\°C}{5.46*10^{-4}/\°C}$

$v_B = 115mm^3$

Therefore the volume that the bulb must have is $115mm^3$

Answer 2

The volume of the bulb has 115 mm³ if a sensitivity of 2 mm/ °C must be obtained.

The specific volume for an area of a cylinder can be determined provided the diameter of the cylinder and the sensitivity are known.

Using the formula for calculating the specific volume for an area of a cylinder, we have:

$\mathbf{V = \pi (\dfrac{d}{2})^2\gamma}$

where;

• sensivity = $\mathbf{\gamma}$

By substituting the values from the parameters given, we have:

$\mathbf{V = \pi (\dfrac{0.2 \ mm}{2})^2\times 2 \ mm/^0 C}$

V = 0.0628 mm³/ °C

However, the volumetric coefficient of thermal expansion for a mercury thermometer by using the differential expansion coefficient is expressed as:

• $\mathbf{\lambda _V= 3 \lambda _L}$
• $\mathbf{\lambda _V= 3(1.82 \times 10^{-4} \ /^0 C)}$
• $\mathbf{\lambda _V= 5.46 \times 10^{-4} /^0 C}$

Now, the volume of the bulb can be computed by using the relation:

• Specific volume (V) = Volume of the bulb × Volumetric coefficient

Making the Volume of the bulb the subject of the formula, we have

$\mathbf{V_B = \dfrac{0.0628 \ mm^3/^0 C}{5.46 \times 10^{-4}/ ^0 \ C}}$

$\mathbf{V_B =115\ mm^3}$

Learn more about the specific volume of a cylinder here:

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