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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 cm³.
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’t 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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