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A thief plans to steal a cylindrical platinum medal with a radius of \(2.3 \mathrm{~cm}\) and a thickness of \(0.8 \mathrm{~cm}\) from a jewellery store. If the platinum has a density of \(21.45 \mathrm{~g} / \mathrm{cm}^{3},\) what is the mass of the medal in \(\mathrm{kg} ?\) [The volume of a cylinder is \(\left.V=\pi r^{2} h .\right]\)

Short Answer

Expert verified
The mass of the platinum medal is approximately \(0.28615 \, \mathrm{kg}\).

Step by step solution

01

Identify the given measurements

We are provided with the following measurements: - The radius of the cylindrical medal, \(r = 2.3 \, \text{cm}\) - The thickness (or height) of the medal, \(h = 0.8 \, \text{cm}\) - The density of platinum, \(d = 21.45 \, \frac{\text{g}}{\text{cm}^3}\)
02

Calculate the volume of the medal

We can find the volume of the cylindrical medal using the formula: \[V = \pi r^2 h\] Substitute the given values of radius and height into the formula: \[V = \pi (2.3)^2 (0.8)\] Now, compute the volume of the medal: \[V \approx 13.34 \, \text{cm}^3\]
03

Find the mass of the medal using the density formula

We know that the density, \(d\), is defined as the mass, \(m\), divided by the volume, \(V\). So we can find the mass using the formula: \[m = d V\] Substitute the given density and calculated volume into the formula: \[m = 21.45 (13.34)\] Now, compute the mass of the medal: \[m \approx 286.15 \, \text{g}\]
04

Convert the mass from grams to kilograms

To convert the mass of the medal from grams to kilograms, divide the mass in grams by 1,000: \[m_{\text{kg}} = \frac{286.15}{1,000}\] \[m_{\text{kg}} = 0.28615 \, \text{kg}\] So, the mass of the platinum medal is approximately \(0.28615 \, \mathrm{kg}\).

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Key Concepts

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

Volume of a Cylinder
A cylinder is a 3D shape with two parallel circular bases and a curved surface in between. To find the volume of a cylinder, the formula is used: \[ V = \pi r^2 h \] where:
  • \( V \) is the volume of the cylinder
  • \( r \) is the radius of the base circle
  • \( h \) is the height (or thickness) of the cylinder
The calculation involves squaring the radius, multiplying it by the height, and then by \( \pi \) (approximately 3.14159).
In the case of the cylindrical medal:
  • \( r = 2.3 \, \text{cm} \)
  • \( h = 0.8 \, \text{cm} \)
Plugging these values into the formula gives:\[ V = \pi (2.3)^2 (0.8) \]You multiply \( 2.3^2 \) to get 5.29, then multiply this by 0.8, and finally by \( \pi \) to calculate approximately 13.34 \( \text{cm}^3 \).
This volume tells us the amount of space inside the cylinder.
Mass Conversion
Converting mass from one unit to another is a straightforward process.
Typically, mass is expressed in grams (g) or kilograms (kg), and they are related by the factor of 1,000. That means:
  • 1 kilogram (kg) = 1,000 grams (g)
  • To convert from grams to kilograms, divide the number of grams by 1,000.
For example, if a platinum medal weighs 286.15 grams, its weight in kilograms is calculated as:\[ m_{\text{kg}} = \frac{286.15}{1,000} \]This results in approximately 0.28615 kg.
Such conversions are common in scientific calculations to ensure consistency with SI units.
Density Formula
Density is a measure of how much mass is contained in a given volume. The formula to calculate density is:\[ d = \frac{m}{V} \]where:
  • \( d \) is the density
  • \( m \) is the mass
  • \( V \) is the volume
In this exercise, you are given the density (\( 21.45 \, \text{g/cm}^3 \)) and need the mass based on the calculated volume of \( 13.34 \, \text{cm}^3 \). Re-arranging the density formula to solve for mass gives:\[ m = d \times V \]Substituting the known values:\[ m = 21.45 \times 13.34 \]The multiplication yields approximately 286.15 grams.
Understanding this relationship allows you to calculate the mass of a specific volume of material if its density is known.

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Most popular questions from this chapter

Make the following conversions: (a) \(83^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) (b) \(29^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\) (c) \(294^{\circ} \mathrm{C}\) to \(\mathrm{K}\) (d) \(832 \mathrm{~K}\) to \({ }^{\circ} \mathrm{C}\) (f) \(35^{\circ} \mathrm{F}\) to \(\mathrm{K}\). (e) \(721 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\)

Judge the following statements as true or false. If you believe a statement to be false, provide a corrected version. (a) Air and water are both elements. (b) All mixtures contain at least one element and one compound. (c) Compounds can be decomposed into two or more other substances; elements cannot. (d) Elements can exist in any of the three states of matter. (e) When yellow stains in a kitchen sink are treated with bleach water, the disappearance of the stains is due to a physical change. (f) A hypothesis is more weakly supported by experimental evidence than a theory. (g) The number 0.0033 has more significant figures than 0.033 . (h) Conversion factors used in converting units always have a numerical value of one. (i) Compounds always contain at least two different elements.

A \(32.65-g\) sample of a solid is placed in a flask. Toluene, in which the solid is insoluble, is added to the flask so that the total volume of solid and liquid together is \(50.00 \mathrm{~mL}\). The solid and toluene together weigh \(58.58 \mathrm{~g}\). The density of toluene at the temperature of the experiment is \(0.864 \mathrm{~g} / \mathrm{mL}\). What is the density of the solid?

(a) A child has a fever of \(101^{\circ} \mathrm{F}\). What is the temperature in \({ }^{\circ} \mathrm{C} ?\) (b) In a desert, the temperature can be as high as \(45^{\circ} \mathrm{C},\) what is the temperature in \({ }^{\circ} \mathrm{F} ?\) (c) During winter, the temperature of the Arctic region can drop below \(-50^{\circ} \mathrm{C}\), what is the temperature in degree Fahrenheit and in Kelvin? (d) The sublimation temperature of dry ice is \(-78.5^{\circ} \mathrm{C}\). Convert this temperature to degree Fahrenheit and Kelvin. (e) Ethanol boils at \(351 \mathrm{~K}\). Convert this temperature to degree Fahrenheit and degree Celsius.

Zirconia, an oxide of zirconium, is often used as an affordable diamond substitute. Just like diamond, it is a colorless crystal which sparkles under sunlight. Which of the following physical properties do you think would help in differentiating between diamond and Zirconia-melting point, density, or physical state?

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