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Chapter 3: Problem 10

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids. The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, \(54.2 \%\) is immersed in the oil and the balance is in the water. In a separate experiment, an empty flask is weighed, \(35.3 \mathrm{cm}^{3}\) of the lubricating oil is poured into the flask, and the flask is reweighed. If the scale reading was \(124.8 \mathrm{g}\) in the first weighing, what would it be in the second weighing? (Suggestion: Recall Archimedes' principle, and do a force balance on the block.)

Short Answer

Expert verified
The weight of the flask, after being filled with the lubricating oil, would be 323 g.

Step by step solution

01

Archimedes' Principle

According to Archimedes' principle, the buoyant force on a submerged body is equal to the weight of the fluid that is displaced by the substance. So in this case, the weight of the graphite block is equal to the weight of the water and oil displaced.
02

Calculate Volumes

Let the volume of the block be \(V\). So \(54.2 \%\) of \(V\) is immersed in oil and the rest \(100 \% - 54.2 \% = 45.8 \%\) is in water. Therefore, the volume of block is \(V_{\text{oil}} / 0.542 = V_{\text{water}} / 0.458\) .
03

Calculate Weights

The weight of the block equals the weight of the oil and water displaced(Archimedes' principle). So you find \(V_{\text{oil}} \times \text{density}_{\text{oil}} + V_{\text{water}} \times \text{density}_{\text{water}} = V_{\text{block}} \times \text{density}_{\text{block}}\). We know the densities of oil and water (they are given as 1 and 08326 respectively), and we can find the density of the block by dividing the weight calculated in the previous step by the volume, which is calculated as 35.3 cm鲁.
04

Calculate Weight of Flask with Oil

Given the density of oil and the volume of oil added to the flask, the weight of the oil can be calculated using \(V_{\text{oil}} \times \text{density}_{\text{oil}}\). Adding this to the original weight of the flask provides the new weight of the flask.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyant Force Calculation
Understanding the buoyant force calculation is pivotal in deciphering Archimedes' principle, which plays a crucial role in this problem. The principle tells us that the buoyant force experienced by a submerged object is equivalent to the weight of the displaced fluid.

In our case, the rectangular block of carbon displaces a certain amount of oil and water based on its submerged volume in each liquid. To calculate the buoyant force acting on the carbon block, we must determine the volume of the submerged parts in both oil and water, and then use the densities of these fluids to find the weight of the displaced volume, which, by Archimedes' principle, is the same as the buoyant force.

As an example, assuming the density of the carbon block is given or has been determined through another experiment, we can use the buoyant force calculated in this step to find the block's weight. It's essential that we always remember the relationship between the volume submerged in the fluid, the fluid's density, and the resulting buoyant force.
Density and Volume Relationship
The relationship between density and volume is central to solving problems involving buoyancy and Archimedes' principle. Density, defined as mass per unit volume, fundamentally links how much space an object occupies (its volume) to its mass.

For instance, in our exercise, we see that the carbon block's volume submerged in oil and water dictates the mass of the displaced fluid. This relationship allows us to write an expression connecting the block's volume with the displaced fluid's weight. Importally, this relationship also implies that for a given mass, a substance with higher density will occupy less volume and vice versa.

When we measure the weight of the lubricating oil alone in the separate experiment, we are given its volume, allowing us to calculate its density. This value can then be used to figure out the weight of the oil needed to be displaced by the carbon block to float at the interface of the oil and water.
Force Balance Analysis
A force balance analysis is a method used to understand the equilibrium of forces acting upon an object. In a stationary situation, where the object isn鈥檛 moving, we assume that all the forces balance each other out.

In the context of our problem, the carbon block is floating stationary at the interface between two liquids, which suggests that the upward buoyant forces exerted by the liquids must exactly balance the downward force of gravity on the block. Essentially, our equilibrium equation would state that the combined buoyant forces of oil and water equal the weight of the graphite block.

This concept is valuable when considering the weight of a displaced fluid in relation to the weight of the submerged object. For students tackling these problems, it's important to write down all forces and set up an equation ensuring a balance, which ultimately leads to the solution and further understanding of the principles of flotation and buoyancy.

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Most popular questions from this chapter

The chemical reactor shown below has a cover that is held in place by a series of bolts. The cover is made of stainless steel ( \(\mathrm{SG}=8.0\) ), is 3 inches thick, has a diameter of 24 inches, and covers and seals an opening 20 inches in diameter. During turnaround, when the reactor is taken out of service for cleaning and repair, the cover was removed by an operator who thought the reactor had been depressurized using a standard venting procedure. However, the pressure gauge had been damaged in an earlier process upset (the reactor pressure had exceeded the upper limit of the gauge), and instead of being depressurized completely, the vessel was under a gauge pressure of 30 psi. (a) What force ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) were the bolts exerting on the cover before they were removed? (Hint: Don't forget that a pressure is exerted on the top of the cover by the atmosphere.) What happened when the last bolt was removed by the operator? Justify your prediction by estimating the initial acceleration of the cover upon removal of the last bolt. (b) Propose an alteration in the turnaround procedure to prevent recurrence of an incident of this kind.

In April \(2010,\) the worst oil spill ever recorded occurred when an explosion and fire on the Deepwater Horizon offshore oil-drilling rig left 11 workers dead and began releasing oil into the Gulf of Mexico. One of the attempts to contain the spill involved pumping drilling mud into the well to balance the pressure of escaping oil against a column of fluid (the mud) having a density significantly higher than those of seawater and oil. In the following problems, you may assume that seawater has a specific gravity of 1.03 and that the subsea wellhead was 5053 ft below the surface of the Gulf. (a) Estimate the gauge pressure (psig) in the Gulf at a depth of \(5053 \mathrm{ft}\). (b) Measurements indicate that the pressure inside the wellhead is 4400 psig. Suppose a pipe between the surface of the Gulf and the wellhead is filled with drilling mud and balances that pressure. Estimate the specific gravity of the drilling mud. (c) The drilling mud is a stable slurry of seawater and barite (SG \(=4.37\) ). What is the mass fraction of barite in the slurry? (d) What would you expect to happen if the barite weight fraction were significantly less than that estimated in Part (c)? Explain your reasoning.

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The following data have been obtained for the effect of solvent composition on the solubility of a serine, an amino acid, at \(10.0^{\circ} \mathrm{C}\) : $$\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \text { Volume \% Methanol } & 0 & 10 & 20 & 30 & 40 & 60 & 80 & 100 \\\\\hline \text { Solubility (g/100 mL solvent) } & 22.72 & 18.98 & 11.58 & 6.415 & 4.205 & 1.805 & 0.85 & 0.65 \\\\\hline \text { Solution Density (g/mL) } & 1.00 & 0.98 & 0.97 & 0.95 & 0.94 & 0.91 & 0.88 & 0.79 \\\\\hline\end{array}$$ The data were obtained by mixing known volumes of methanol and water to obtain the desired solvent compositions, and then slowly adding measured amounts of serine to each mixture until no more would go into solution. The temperature was held constant at \(10.0^{\circ} \mathrm{C}\). (a) Derive an expression for solvent composition expressed as mass fraction of methanol, \(x\), as a function of volume fraction of methanol, \(f\) (b) Prepare a table of solubility of serine (g serine/g solution) versus mass fraction of methanol.

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