The spirit-in-glass thermometer, invented in Florence,...

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. $\mathrm{P} 14.41$ ). 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 $0.78945 \mathrm{~g} / \mathrm{cm}^{3}$ at $20.0^{\circ} \mathrm{C}$ and decreases to $0.78097 \mathrm{~g} / \mathrm{cm}^{3}$ at $30.0^{\circ} \mathrm{C}$. (a) Assuming that one of the spheres has a radius of $1.000 \mathrm{~cm}$ and is in equilibrium halfway up the tube at $20.0^{\circ} \mathrm{C},$ determine its mass. (b) When the temperature increases to $30.0^{\circ} \mathrm{C}$, what mass must a second sphere of the (c) At $30.0^{\circ} \mathrm{C}$, the first sphere has fallen to the bottom of the tube. What upward force does the bottom of the tube exert on this sphere?

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. $\mathrm{P} 14.41$ ). 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 $0.78945 \mathrm{~g} / \mathrm{cm}^{3}$ at $20.0^{\circ} \mathrm{C}$
and decreases to $0.78097 \mathrm{~g} / \mathrm{cm}^{3}$ at $30.0^{\circ} \mathrm{C}$. (a) Assuming that one of the spheres has a radius of $1.000 \mathrm{~cm}$ and is in equilibrium halfway up the tube at $20.0^{\circ} \mathrm{C},$ determine its mass.
(b) When the temperature increases to $30.0^{\circ} \mathrm{C}$, what mass must a second sphere of the (c) At $30.0^{\circ} \mathrm{C}$, the first sphere has fallen to the bottom of the tube. What upward force does the bottom of the tube exert on this sphere?

Key Concepts

Archimedes’ Principle

This principle states that any object wholly or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. It is fundamental in determining whether an object will float, sink, or remain in equilibrium within a fluid, and is used to relate the submerged volume of an object to the displaced fluid’s weight.

Equilibrium of Floating Bodies

The concept of equilibrium involves balancing forces acting on an object. For a floating or partially submerged object, equilibrium occurs when the downward gravitational force (its weight) is exactly balanced by the upward buoyant force generated by the displaced fluid. This balance is used to calculate properties such as the mass of the object when its submerged fraction or displaced volume is known.

Density Variations with Temperature

Density is a measure of mass per unit volume and can change with temperature, especially in liquids. As temperature increases, the density of fluids typically decreases. This variation directly affects the buoyant force on objects submerged in the fluid since the weight of the displaced fluid (and hence the buoyant force) depends on its density.

Normal (Reaction) Force

When an object is in contact with the bottom of a container, the bottom exerts an upward reaction force, known as the normal force, to support the object. This force is part of the overall force balance that determines the state of motion or rest of the object. In cases where buoyancy is not sufficient to counteract the object’s weight entirely, the additional upward normal force prevents further sinking.

Transcript

00:01 Hi there, so for this problem, for part of this problem, we are asked about, assuming that one of the spheres has a radius that is one centimeter and is in equilibrium halfway up at a degree at a degree at a temperature of 20 celsius degrees, we need to determine its mass.

00:23 So since the upward point force is balanced by the weight of the sphere, we will have that the mass m1 times the accelerators.

00:32 Due to gravity, which is equal, we can write the mass as the density times the volume, this time is acceleration due to gravity.

00:40 Now the volume of a sphere, and we know that is defined as 4 divided by 3 times pi times the radius to the 3, and this time is acceleration due to gravity.

00:49 Now what we need to do is to solve for, well, we can cancel the acceleration due to gravity with the acceleration due to gravity, and then we have the mass for this.

01:00 So the mass for this n1 is equal to the density that we are given for this, which is equal to a value of 0 .78, 945, this in units of grams per cubic centimeter.

01:19 And this times 4 divided by 3 times pi times the radius, and the radius is 1 centimeter, and that to the 3.

01:28 So using our calculator, we obtained that the mass m1 is equal to 3 .37 grams.

01:44 So that's a solution for part a of this problem.

01:48 Now for part b, we are asked about when the temperature increases to a value of 30 celsius degrees, why mass amounts a second sphere? so in this case, following the same procedure as we did for part a, but in this case, we now have a density, we're going to call the density prime that is equal to 0 .7807, and this in grams per cubic centimeter, and this at a temperature of 30 celsius degrees.

02:32 So we need to find that the mass m2, again, is just the only thing that changes is the density.

02:38 So that will be the density times 4 divided by 3.

02:42 Pi times the radius is just the same as before.

02:45 So we substitute the values in here.

02:48 So the density is 0 .78, 095 grams per cubic centimeter.

02:54 This times 4 divided by 3 times pi times the radius.

02:58 Again, that is 1 centimeter.

03:00 End up to the 3.

03:01 So from this we obtain that now the mass is equal to 3 .271 grams.

03:10 So that's a solution for part b of this problem...