Question

About one-third of the body of a physicist swimming in the Dead Sea will be above the water line. Assuming that the human body density is \(0.98 \mathrm{~g} / \mathrm{cm}^{3}\), find the density of the water in the Dead Sea. Why is it so much greater than \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ?

Short answer

The density of the Dead Sea water is approximately 1.47 g/cm³, which is much higher than the density of most other bodies of water due to high concentrations of salt and minerals in the Dead Sea.

Step-by-step solution

Understand the Problem

First, it's important to recognize that the problem requires using the properties of buoyancy. In this case, the physicist's body serves as a kind of 'floating object'. Since only two-thirds of the body is submerged, this means two-thirds of the body's volume is displacing the sea's water.

Apply Archimedes' Principle

According to Archimedes' principle, the buoyant force on the physicist's body must equal the gravitational force on the body, which allows the physicist to float. Mathematically, this can be written as: \( \rho_{Dead Sea} * V_{submerged} * g = \rho_{body} * V_{body} * g \) , where \( \rho_{body} \) and \( \rho_{Dead Sea} \) are densities of the physicist's body and Dead Sea's water respectively, g is the acceleration due to gravity, \( V_{body} \) is the volume of the physicist's body, and \( V_{submerged} \) is the volume of the submerged part of the physicist's body.

Calculate the Sea Water Density

Since the volume of the submerged body \( V_{submerged} = \frac{2}{3} V_{body} \) due to the one-third of the body being above water, we can substitute this into the previous equation, simplify it and solve for the density of the Dead Sea's water: \( \rho_{Dead Sea} * \frac{2}{3} * V_{body} * g = \rho_{body} * V_{body} * g \) This simplifies to \( \rho_{Dead Sea} = \frac{3}{2} \rho_{body} \). Substituting the given body density value \( \rho_{body} = 0.98 \, \mathrm{g/cm^3} \), the density of the Dead Sea water can be calculated as: \( \rho_{Dead Sea} = \frac{3}{2} * 0.98 \, \mathrm{g/cm^3} = 1.47 \, \mathrm{g/cm^3} \).

Explain why the Density is so High

This density is significantly higher than the density of most water bodies (1.0 g/cm³) because the Dead Sea contains a high super-saturation of salt and other minerals, which adds to the mass without significantly contributing to the volume; resulting in a higher density compared to other bodies of water.

Key concepts

Archimedes' Principle

Understanding Archimedes' principle is key to explaining why things float or sink. It tells us that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that is displaced by the object. This principle can be visualized when you take a bath and notice how the water level rises when you get in - you're displacing water, and that water is exerting a force on you.

For objects floating, like our swimming physicist, the volume of water displaced is directly related to the volume of the part of the object that's submerged. So if two-thirds of the physicist is underwater, Archimedes' principle ensures that the buoyant (upward) force is equal to the weight of the volume of water that would fill up that submerged two-thirds.

Gravitational Force

Gravitational force is the attraction between any two masses. We experience it as the force that keeps us grounded on Earth. In the exercise, it's the force acting downward on the physicist's body due to the body's own mass and Earth's gravity.

In buoyancy problems, gravity is what pulls the object down while the buoyant force tries to push it up. Our bodies, having mass, are subject to this gravitational pull, which is computed as the product of mass (or density times volume for a consistent substance) and the acceleration due to gravity (g). It is this interplay between gravitational force pulling down and buoyant force pushing up that determines whether the physicist floats or sinks in the Dead Sea.

Dead Sea Physics

The unique physics of the Dead Sea allows swimmers to float much more easily than in other bodies of water. The high levels of salt and minerals in the Dead Sea increase the water's density, creating a stronger buoyant force.

This is why our physicist, with less dense human tissues, floats with a considerable portion above the water's surface. Unlike other seas and lakes where the density is generally close to that of pure water, the Dead Sea exhibits a density that can hold a human body more effortlessly afloat, all thanks to its unusual mineral composition.

Density Calculation

Density is the measure of how much 'stuff' is packed into a given volume. It's calculated by dividing mass by volume. In our example involving the Dead Sea, we utilize the known density of the human body and the proportion that is submerged to determine the density of the sea water.

By understanding that the physicist's body has a density slightly less than that of water, we see why not all of the body is submerged. When we perform the density calculation for the Dead Sea water, considering the buoyant force and the part of the body submerged, we find that the sea water must be much denser to counteract the same weight of a lesser volume of the human body. This relationship, shaped by both the density of the object and the fluid, determines the state of flotation or submersion.