Problem 61 - College Physics

Question

A certain fluid has a density of \(1080 \mathrm{~kg} / \mathrm{m}^{3}\) and is observed to rise to a height of \(2.1 \mathrm{~cm}\) in a \(1.0\)-mm-diameter tube. The contact angle between the wall and the fluid is zero. Calculate the surface tension of the fluid.

Step-by-Step Solution

Answer

The surface tension of the fluid [\(\sigma\)] is calculated by using the capillary rise formula rearranged to solve for \(\sigma\) using the given parameters. \(\sigma\) is calculated to offer a solution to the exercise.

Step 1: Analyze the Given Information

Take note of the provided parameters: \(\rho = 1080 \mathrm{~kg} / \mathrm{m}^{3}\) (fluid density), \(h = 2.1 \mathrm{~cm}\) (capillary rise height), \(r = 1.0\mathrm{~mm} / 2 = 0.5 \mathrm{~mm}\) (capillary tube radius), and \(\Theta = 0\) (contact angle). Remember to convert height and radius to meters for cohesion, resulting in \(h=0.021 \mathrm{~m}\), \(r=0.0005 \mathrm{~m}\). Gravity is a constant: \( g = 9.81 \mathrm{~m/s}^{2} \).

Step 2: Rearrange the Capillary Rise Formula

Rearranging the equation \(h = \frac{2\sigma cos\Theta}{r \rho g}\) to calculate for \(\sigma\), we obtain \(\sigma = \frac{h r \rho g}{2 cos \Theta}\). As the contact angle \(\Theta = 0, cos\Theta = 1\), simplifying the formula to \(\sigma = \frac{h r \rho g}{2}\).

Step 3: Insert Values into the Formula

Substitute the given values into the revised formula: \(\sigma = \frac{h r \rho g}{2} = \frac{0.021 \mathrm{~m} \times 0.0005 \mathrm{~m} \times 1080 \mathrm{~kg} / \mathrm{m}^{3} \times 9.81 \mathrm{~m/s}^{2}}{2} \).

Step 4: Calculate Surface Tension

Using the substituted values, calculate for \(\sigma\). Ensure units are homogeneous and compatible during calculations.

Key Concepts

Fluid Density

Fluid density refers to the mass of fluid per unit volume. It is a critical parameter in fluid dynamics and is denoted by the Greek letter \( \rho \). Think of it like the weight of the fluid compacted within a specific space, helping describe how heavy or light a fluid is. In our exercise, the fluid density is given as \(1080 \mathrm{~kg} / \mathrm{m}^{3}\).
This value allows us to predict how the fluid will behave, especially in intricate scenarios like capillary action where we observe the rise or fall of liquid in a tube. While consistently necessary in calculations concerning buoyancy, pressure, and fluid flow, understanding density is vital for setting the stage for calculations related to capillary phenomena.

• Density impacts how much fluid can exert force over a specific area.
• Higher density often means more mass, influencing gravitational force at play.

When you grasp density, you unlock the potential to anticipate how fluids interact under various conditions.

Capillary Action

Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity. This behavior is due to the liquid's cohesion (molecules sticking together) and adhesion (molecules sticking to surfaces).
In the exercise, the fluid rises to a height of \(2.1 \mathrm{~cm}\) in a narrow tube. This is a perfect demonstration of capillary action, where the fluid molecules are more attracted to the tube's surface than to each other.

• This effect allows liquids to defy gravity momentarily.
• It's notably seen in plants, as water moves from roots to leaves.

The rise in the small tube is directly connected to the surface tension and the interactions between molecules both in the liquid and with the tube surface. Understanding capillary action explains numerous natural and industrial processes.

Contact Angle

The contact angle is a measure of the wettability of a liquid on a solid surface. It is the angle formed where a liquid interface meets a solid surface. In simpler terms, it tells us how well a liquid spreads or beads up on a surface.
In this exercise, the contact angle is given as zero, meaning the liquid entirely spreads on the surface, maximizing its contact with the tube wall. This situation is called perfect wetting.

• A contact angle of 0° typically results in capillary rise.
• Larger contact angles can prevent such rising in narrow spaces.

Evaluating the contact angle in problems helps predict how liquids interact with different surfaces. It is crucial in applications ranging from ink printing to waterproof coatings.

Physics Problem Solving

Physics problem solving often follows a structured approach, breaking down complex scenarios step-by-step. This helps in understanding intricate concepts by solving real-world related problems.
In our given problem, we systematically calculated the surface tension using known formulas and given values. The process involved:

• Gathering all necessary data (such as fluid density and tube dimensions).
• Converting units for consistency.
• Rearranging equations to isolate the unknown (surface tension).
• Substituting values and calculating accurately.

Problem solving in physics not only helps us find specific answers but also enhances our understanding of how fundamental principles apply in diverse scenarios. Tackling problems methodically embeds these principles deeper into our learning.