Question

The spirit-in-glass thermometer, invented in Florence, Italy, around 1654, consists of a tube of liquid (the spirit) containing a number of submerged glass spheres with slightly different masses (Fig. P14.76). at sufficiently low temperatures, all the spheres float, but as the temperature rises, the spheres sink one after another. The device is a crude but interesting tool for measuring temperature. Suppose the tube is filled with ethyl alcohol, whose density is 0.78945 $g/c{m}^3$ at $20.0°C$and decreases to $0.78945g/c{m}^3$ at $30.0°C$. (a) Assuming that one of the spheres has a radius of 1.000 cm and is in equilibrium halfway up the tube at $20.0°C$, determine its mass. (b) When the temperature increases to $30.0°C$, what mass must the second sphere of the same radius have to be in equilibrium at the halfway point? (c) At, $30.0°C$the first sphere has fallen to the bottom of the tube. What upward force does the bottom of the tube exert on this sphere?

Short answer

(a) The mass is $m_1-3.307g$.

(b) The mass must a second sphere of the same radius have to be in equilibrium at the halfway point is $m_2-3.271g$.

(c) The force does the bottom of the tube exert on this sphere is $3.48\times {10}^{4}N$.

Step-by-step solution

Pressure:

Fluid pressure is the pressure at a point inside a fluid due to the weight of the fluid.

The pressure in a fluid at rest is given by:

$P=\rho gh$

Where, P is the pressure of the fluid, $\rho$ is the Density of fluid, g is the gravitational constant, h is the high attained by the fluid.

(a) Determine the mass:

Consider the given data as below.

Density, $\rho =0.78097g/c{m}^3$

Radius, $r1.00cm$

Define the mass as follow.

$$\begin{gathered} m_{1}g=\rho Vg \\ =\rho \left(\frac{4}{3}\pi r^3\right)g \end{gathered}$$

$$\begin{gathered} m_1=(\rho =0.78097g/c{m}^3)\left(\frac{4}{3}\times 3.14(1.00c{m}^3)\right) \\ =3.307g \end{gathered}$$

(b) Define the mass must a second sphere of the same radius have to be in equilibrium at the halfway point:

Consider the given data as below.

The density $\rho \text{'}=0.78097g/c{m}^3$

Temperature is $30.0°C$

$$\begin{gathered} m_2=\rho \text{'}\left(\frac{4}{3}\pi R^3\right) \\ =(0.78097g/c{m}^3)\left(\frac{4}{3}\mathrm{\pi }{(1.00\mathrm{cm})}^3\right) \\ =3.271g \end{gathered}$$

Hence, the mass must a second sphere of the same radius have to be in equilibrium at the halfway point is 3.271 g

(c) Determine upward force does the bottom of the tube exert on this sphere:

When the first sphere is resting on the bottom of the tube,

$$\begin{gathered} n-B=F_{g1} \\ =m_{1}g \end{gathered}$$

Where, $n$ is the normal force and $B$is the buoyant force.

Since, the buoyant force is,

$B=\rho \text{'}Vg$

The normal force will become,

$$\begin{gathered} n=m_{1}g-\rho \text{'}Vg, \\ =\left[3.307g-(0.78097g/c{m}^3)\frac{4}{3}\mathrm{\pi }{(1.00\mathrm{cm})}^3\right](980cm/s^2) \\ =34.8g.cm/s^2 \\ =3.48\times {10}^{-4}N \end{gathered}$$

Hence, the force does the bottom of the tube exert on this sphere is $3.48\times {10}^{-4}N$.