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Whole blood has a surface tension of \(0.0 .58 \mathrm{~N} / \mathrm{m}\) and a density of \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). To what height can whole blood rise in a capillary blood vessel that has a radius of \(2.0 \times 10^{-6} \mathrm{~m}\) if the contact angle is zero?

Short Answer

Expert verified
The whole blood can rise to an approximate height of \(0.56 \, m\) in a capillary blood vessel that has a radius of \(2.0 \times 10^{-6} \, m\) if the contact angle is zero.

Step by step solution

01

Identify values from the problem.

From the problem statement, we identify the given quantities: \(\gamma=0.058 \, N/m\), \(\rho=1050 \, kg/m^3\), \(r=2.0 \times 10^{-6} \, m\) and \(\theta=0 \, degrees\). The value of the acceleration due to gravity is a universally accepted value, \(g = 9.8 \, m/s^2\).
02

Substitute the values into Jurin's Law.

We substitute the values into the Jurin's Law formula. Considering \(\cos(0) = 1\), the equation simplifies as follows: \(h=\frac{2 \times 0.058 \times 1}{1050 \times 9.8 \times 2.0 \times 10^{-6}}\).
03

Compute the result

After substituting the values into the formula and computing, we find that \(h \approx 0.56 \, m\). This is the height that blood can rise in the capillary tube due to capillary action.

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Key Concepts

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

Surface Tension
Surface tension is a fascinating phenomenon that occurs at the surface of a liquid due to the cohesive forces among its molecules. Imagine the surface of water acting like a thin elastic sheet. This is because the molecules at the surface are pulled inwards by other molecules, creating a state of tension. Surface tension is what allows certain insects to walk on water without sinking, and it also plays a crucial role in the behavior of liquids in narrow spaces, like in our blood vessels. In mathematical terms, surface tension is represented by the Greek letter gamma (\( \gamma \)) and is measured in newtons per meter (N/m). It is this surface tension that partially facilitates the rise of liquids in capillaries, including blood in a capillary tube.
Jurin's Law
Jurin's Law provides a mathematical formula to calculate the height to which a liquid will rise or fall in a capillary tube due to the combined effects of cohesion and adhesion. This law is particularly useful in understanding capillary action, which is why it's important in medical science for understanding blood movement. Jurin's Law is expressed as follows:\[ h = \frac{2\gamma \cos \theta}{\rho gr} \]
  • \( h \) is the height the liquid rises or falls,
  • \( \gamma \) is the surface tension,
  • \( \theta \) is the contact angle,
  • \( \rho \) is the liquid's density,
  • \( g \) is the acceleration due to gravity,
  • and \( r \) is the radius of the tube.
In our problem, because the contact angle \( \theta \) is zero, the formula simplifies, allowing for straightforward calculation of blood rise in the capillary tube.
Contact Angle
The contact angle is a critical parameter in understanding how a liquid interacts with a surface. It measures the angle formed at the point where a liquid interfaces with a solid surface. Simply put, it tells you whether a liquid will wet a surface or form droplets. A contact angle of zero degrees implies complete wetting, meaning the liquid spreads entirely over the surface. In the context of our original exercise, a contact angle of zero indicates perfect wetting, allowing the blood to rise maximally in the capillary tube. If the angle were larger, the height of the rise would be reduced because the cohesion forces would exceed the adhesion to the walls of the capillary.
Density
Density is another vital concept that factors into the calculation of capillary rise as described by Jurin's Law. Density, denoted by the symbol \( \rho \), measures the mass per unit volume of a substance and is expressed in kilograms per cubic meter (kg/m³). In the problem at hand, blood's density is given as 1050 kg/m³, which is typical for human blood. The higher the density of a liquid, the less it will rise in a capillary tube because more mass needs to be supported by the surface tension. Hence, understanding density is crucial for predicting how much a liquid will rise when other factors like capillary radius and surface tension are constant.

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Most popular questions from this chapter

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0-\mathrm{L}\) bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

Take the density of blood to be \(\rho\) and the distance between the feet and the heart to be \(h_{H}\). Ignore the flow of blood, (a) Show that the difference in blood pressure between the feet and the heart is given by \(P_{F}-P_{H}=\rho g h_{j F}\) (b) Take the density of blood to be \(1.05 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\) and the distance between the heart and the feet to be \(1.20 \mathrm{~m}\). Find the difference in blood pressure between these two points. This problem indicates that pumping blood from the extremities is very difficult for the heart. The veins in the legs have valves in them that open when blood is pumped toward the heart and close when blood flows away from the heart. Also, pumping action produced by physical activities such as walking and breathing assists the heart.

A liquid \(\left(\rho=1.65 \mathrm{~g} / \mathrm{cm}^{3}\right)\) flows through two horizontal sections of tubing joined end to end. In the first section, the cross-sectional area is \(10.0 \mathrm{~cm}^{2}\), the flow speed is \(275 \mathrm{~cm} / \mathrm{s}\), and the pressure is \(1.20 \times 10^{5} \mathrm{~Pa} .\) In the second section, the cross-sectional area is \(2.50 \mathrm{~cm}^{2}\). Calculate the smaller section's (a) flow speed and (b) pressure.

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-15} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?

An iron block of volume \(0.20 \mathrm{~m}^{3}\) is suspended from a spring scale and immersed in a flask of water. Then the iron block is removed, and an aluminum block of the same volume replaces it. (a) In which case is the buoyant force the greatest, for the iron block or the aluminum block? (b) In which case does the spring scale read the largest value? (c) Use the known densities of these materials to calculate the quantities requested in parts (a) and (b). Are your calculations consistent with your previous answers to parts (a) and (b)?

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