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Chapter 14: Problem 45

Suppose you have a coffee mug with a circular crosssection and vertical sides (uniform radius). What is its inside radius if it holds \(375 \mathrm{g}\) of coffee when filled to a depth of \(7.50 \mathrm{cm} ?\) Assume coffee has the same density as water.

Short Answer

Expert verified
The inside radius of the coffee mug is approximately \(3.054 \: cm\).

Step by step solution

01

Identifying the given parameters and formula

We're given the mass of coffee (m) is 375 grams and the depth of coffee (h) is 7.5 cm. The density of coffee (蟻) is the same as water, which is 1000 kg/m鲁 or 1 g/cm鲁. We'll be using the volume formula for a cylinder: \(V = 蟺r^2h\).
02

Calculate the volume of coffee

Since we know the mass and density of the coffee, we can calculate the volume (V) using the formula: \(V = \frac{m}{蟻}\) Substitute the given values: \(V = \frac{375}{1}\) \(V = 375\) cubic centimeters
03

Solve the volume formula for the radius

Now, substitute the value of V and h in the volume formula of the cylinder and solve for r: \(375 = 蟺r^2(7.5)\) To isolate the term with r, divide both sides by \(7.5蟺\): \(r^2 = \frac{375}{7.5蟺}\)
04

Solve for the radius

To get the value of the radius, take the square root of the result obtained in step 3: \(r = \sqrt{\frac{375}{7.5蟺}}\) \(r 鈮� 3.054\) cm The inside radius of the coffee mug is approximately 3.054 cm.

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Key Concepts

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

Cylinder Volume
Understanding cylinder volume is crucial when dealing with objects like coffee mugs, which are essentially cylinders with circular bases. The volume of a cylinder is determined using the formula:
  • \[V = \pi r^2h\]
Here, \(V\) stands for volume, \(r\) is the radius of the base, and \(h\) is the height or depth of the cylinder.
To understand this formula, visualize the base of the cylinder as a circle with radius \(r\). The area of that circle is \(\pi r^2\). When you multiply this area by the height \(h\), you get the full volume of the cylinder.
This volume calculation is essential to determine how much liquid the cylinder can hold.
Density of Water
Density is a key concept when converting between mass and volume. The density of water is a standard reference in many scientific calculations because it is very consistent.
  • For water, the density is 1000 kg/m鲁 or 1 g/cm鲁. This means that every cubic centimeter of water weighs 1 gram.
In practical terms, knowing the density of water allows us to easily find the volume of a liquid when its mass is given.
In this exercise, since coffee is assumed to have the same density as water, we use it to accurately convert the mass of the coffee to its volume.
Using the density formula \[V = \frac{m}{\rho}\] where \(m\) is mass and \(\rho\) (rho) represents density, helps us transition from mass to volume effectively.
Radius Calculation
Finding the radius is often necessary when you know the volume and height of a cylinder beforehand. In such cases, you need to rearrange the cylinder volume formula to solve for the radius \(r\):
  • Start with \[V = \pi r^2 h\]
  • Solve for \(r^2\) by dividing both sides by \(\pi h\):\[r^2 = \frac{V}{\pi h}\]
  • Take the square root to find \(r\):\[r = \sqrt{\frac{V}{\pi h}}\]
By substituting the volume \(375 \text{ cm}^3\) and height \(7.5 \text{ cm}\) into the equation, you get the radius.
Calculations like these are routinely used in practical problems where the dimensions of containers or structural elements need to be determined.
Mass to Volume Conversion
Converting mass to volume is often necessary in scenarios involving liquids, especially when you know the density. This conversion is straightforward because the mass and volume of a substance are directly related by its density.
  • Use the formula \[V = \frac{m}{\rho}\]
Where \(V\) is volume, \(m\) is mass, and \(\rho\) is density.
In this particular exercise, coffee has a density equivalent to water, simplifying our calculations.
The known mass of the coffee is used to determine its volume directly, by dividing the mass by the density.
This concept is widely used in cooking, laboratory settings, and industries where precise liquid measurements are essential.

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Most popular questions from this chapter

A simple compass can be made by placing a small bar magnet on a cork floating in water. (a) What fraction of a plain cork will be submerged when floating in water? (b) If the cork has a mass of \(10.0 \mathrm{g}\) and a 20.0 - \(\mathrm{g}\) magnet is placed on it, what fraction of the cork will be submerged? (c) Will the bar magnet and cork float in ethyl alcohol?

Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which the fluid flows.)

Why are gases easier to compress than liquids and solids?

Fluid originally flows through a tube at a rate of \(100 \mathrm{cm}^{3} / \mathrm{s} .\) To illustrate the sensitivity of flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of \(1.50 .\) (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50 .

(a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take if you could divert a moderate size river, flowing at \(5000 \mathrm{m}^{3} / \mathrm{s}\) into the pool?
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