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Blood. (a) Mass of blood. The human body typically contains 5 L of blood of density 1060 \(\mathrm{kg} / \mathrm{m}^{3} .\) How many kilograms of blood are in the body? (b) The average blood pressure is \(13,000 \mathrm{Pa}\) at the heart. What average force does the blood exert on each square centimeter of the heart? (c) Red blood cells. Red blood cells have a specific gravity of 5.0 and a diameter of about 7.5\(\mu \mathrm{m}\) . If they are spherical in shape (which is not quite true), what is the mass of such a cell?

Short Answer

Expert verified
(a) 5.3 kg (b) 1.3 N (c) ≈ 2.8 × 10^{-14} kg

Step by step solution

01

Calculate Mas of Blood

Given the density of blood is \(1060 \, \mathrm{kg/m^3}\) and the volume is \(5 \, \mathrm{L}\). We first convert the volume to cubic meters: \(5 \, \mathrm{L} = 0.005 \, \mathrm{m^3}\). Use the formula \( \text{mass} = \text{density} \times \text{volume} \). Thus, the mass of blood is \(1060 \, \mathrm{kg/m^3} \times 0.005 \, \mathrm{m^3} = 5.3 \, \mathrm{kg}\).
02

Calculate Force on Heart

Use the given average blood pressure \(13000 \, \mathrm{Pa}\). Pressure is defined as force per unit area. Thus, force \(F = P \times A\). To find force per square centimeter, convert 1 square centimeter to square meters: \(1 \, \mathrm{cm^2} = 0.0001 \, \mathrm{m^2}\). Therefore, the force is \(13000 \, \mathrm{Pa} \times 0.0001 \, \mathrm{m^2} = 1.3 \, \mathrm{N}\).
03

Determine Mass of Red Blood Cell

The specific gravity of a red blood cell is given as 5.0, meaning its density is \(5 \times 1000 \, \mathrm{kg/m^3} = 5000 \, \mathrm{kg/m^3}\), since water has a density of \(1000 \, \mathrm{kg/m^3}\). The volume of a spherical cell is calculated using the formula \(V = \frac{4}{3}\pi r^3\), where radius \(r = \frac{7.5 \times 10^{-6}}{2}\) meters. Thus, \(V = \frac{4}{3} \pi (3.75 \times 10^{-6})^3 \, \mathrm{m^3}\). Then, the mass of the red blood cell is \(\text{density} \times \text{volume} = 5000 \, \mathrm{kg/m^3} \times V\). On calculation, this gives mass \( \approx 2.8 \times 10^{-14} \, \mathrm{kg}\).

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Key Concepts

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

Density Calculation
Density plays a crucial role in determining the mass of substances when their volume is known. The basic relationship is expressed by the formula:
  • \[\text{mass} = \text{density} \times \text{volume}\]
Here, density is the amount of mass per unit volume of a substance. It is usually expressed in kilograms per cubic meter (\(\mathrm{kg/m^3}\)). For example, the density of blood is typically \(1060 \, \mathrm{kg/m^3}\). To find the mass, you need to convert the volume into cubic meters, particularly when it is given in liters since \(1 \, \mathrm{L} = 0.001 \, \mathrm{m^3}\). Once the volume is in the correct units, multiply by the density to get the mass. In this exercise, a 5-liter volume of blood corresponds to a mass of 5.3 kilograms.
Pressure and Force
Understanding pressure involves relating it to how much force is applied over a certain area. The formula for pressure is:
  • \[F = P \times A\]
where pressure \(P\) is in pascals (\(\mathrm{Pa}\)), force \(F\) is in newtons (\(\mathrm{N}\)), and area \(A\) is in square meters (\(\mathrm{m^2}\)). If you're given the area in square centimeters, remember to convert it to square meters:
  • \(1 \, \mathrm{cm^2} = 0.0001 \, \mathrm{m^2}\)
In this exercise, the average blood pressure is \(13,000 \, \mathrm{Pa}\). Applying this over an area of \(1 \, \mathrm{cm^2}\), the resulting force is calculated as \(1.3 \, \mathrm{N}\). This illustrates how the application of pressure at the heart converts into force, which explains why specific force values are crucial in medical contexts.
Specific Gravity
Specific gravity is a way to express density relative to another standard, typically water. It is a dimensionless number and is given by:
  • \[\text{Specific gravity} = \frac{\text{density of substance}}{\text{density of water}}\]
The density of water is \(1000 \, \mathrm{kg/m^3}\). Therefore, if a substance has a specific gravity of 5, its density is \(5 \times 1000 = 5000 \, \mathrm{kg/m^3}\). For red blood cells in the exercise, this derived density is critical to calculate their mass when paired with volume calculations. Specific gravity thus provides a direct mathematical way to comprehend the density of different materials relative to water.
Volume of a Sphere
When dealing with spherical objects, you need to compute the volume first if you want to find mass. The formula for the volume of a sphere is:
  • \[V = \frac{4}{3}\pi r^3\]
Here, \(r\) is the radius of the sphere. The radius is half of the diameter. For example, if the diameter of a red blood cell is given as \(7.5\, \mu\mathrm{m}\), the corresponding radius is \(3.75 \times 10^{-6} \, \mathrm{m}\). Substitute the radius value into the formula to calculate the volume. Once you have the volume, you can multiply by the cell’s density to get its mass. This approach helps when approximating objects as spheres, providing useful results in biology.
Red Blood Cell Mass
Red blood cells are tiny but can still have their mass calculated with the right methods. Once the volume of the cell has been determined using the sphere volume formula, the mass is computed as:
  • \[\text{mass} = \text{density} \times \text{volume}\]
With a specific gravity of 5, the red blood cell density is \(5000 \, \mathrm{kg/m^3}\). After calculating its volume using the formula for a sphere, you multiply the density by this volume. The resulting mass of a single cell is extraordinarily small, approximately \(2.8 \times 10^{-14} \, \mathrm{kg}\). This calculation is a remarkable display of how even microscopic elements maintain predictable, calculable physical properties.

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Most popular questions from this chapter

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At area, \(1,\) the cross-sectional area of the pipe is 0.070 \(\mathrm{m}^{2}\) and the magnitude of the fluid velocity is 3.50 \(\mathrm{m} / \mathrm{s} .\) What is the fluid speed at points in the pipe where the cross-sectional area is \((a) 0.105 \mathrm{m}^{2},\) (b) 0.047 \(\mathrm{m}^{2}\) ?

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3} .\) The balloon fabric (the envelope) weighs 900 \(\mathrm{N} .\) The basket with gear and full propane tanks weighs 1700 \(\mathrm{N}\) . If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

A cube 5.0 \(\mathrm{cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 \(\mathrm{cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.50 \(\mathrm{N}\) (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

An ore sample weighs 17.50 \(\mathrm{N}\) in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 \(\mathrm{N}\) . Find the total volume and the density of the sample.

How big is a million dollars? At the time this problem was written, the price of gold was about \(\$ 1239\) per ounce, while that of platinum was about \(\$ 1508\) an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 \(\mathrm{g} / \mathrm{cm}^{3}\) and that of platinum is 21.4 \(\mathrm{g} / \mathrm{cm}^{3} .\) (a) If you find a spherical gold nugget worth 1.00 million dollars, what would be its diameter? (b) How much would a platinum nugget of this size be worth?

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