Problem 66
Question
A titanium bicycle frame displaces \(0.314 \mathrm{~L}\) of water and has a mass of 1.41 kg. What is the density of the titanium in \(g / \mathrm{cm}^{3} ?\)
Step-by-Step Solution
Verified Answer
The density of the titanium is approximately 4.49 g/cm³.
1Step 1: Convert the Volume to cubic centimeters (cm³)
Since the density formula requires consistency of units and the volume is initially given in liters, convert the volume of water displaced by the titanium bicycle frame into cubic centimeters. Use the conversion where 1 liter equals 1000 cubic centimeters.
2Step 2: Convert the Mass to grams (g)
Similarly, for consistency in the density calculation, convert the mass of the titanium bicycle frame from kilograms to grams. Use the conversion where 1 kilogram equals 1000 grams.
3Step 3: Calculate Density
Use the density formula, which is density (\rho) equals mass (m) divided by volume (V): \[\rho = \frac{m}{V}\] Substitute the mass and volume in grams and cubic centimeters respectively into the formula, and solve for density.
Key Concepts
Volume ConversionMass ConversionDensity Formula
Volume Conversion
Understanding how to convert different units of volume is essential in many scientific calculations, especially when determining the density of a substance. In the given problem, the volume of water displaced by the titanium bicycle frame is initially measured in liters, but for the density calculation, we need to express the volume in cubic centimeters (cm³).
To convert from liters to cubic centimeters, it's important to note that:
\[0.314 \text{ L} \times 1,000 \frac{\text{cm}^3}{\text{L}} = 314 \text{ cm}^3\]
This step sets the stage for a more accurate density calculation, as we will discuss next.
To convert from liters to cubic centimeters, it's important to note that:
- 1 liter is equivalent to 1,000 cubic centimeters (1 L = 1,000 cm³).
\[0.314 \text{ L} \times 1,000 \frac{\text{cm}^3}{\text{L}} = 314 \text{ cm}^3\]
This step sets the stage for a more accurate density calculation, as we will discuss next.
Mass Conversion
Mass conversion is just as crucial as volume conversion when you're working with the density formula. Since the mass of the bicycle frame is given in kilograms (kg), and the density is more commonly expressed in grams per cubic centimeters (g/cm³), we must convert the mass to grams.
The conversion factor between kilograms and grams is:
\[1.41 \text{ kg} \times 1,000 \frac{\text{g}}{\text{kg}} = 1,410 \text{ g}\]
The exercise emphasizes consistent units; hence, understanding and applying this conversion ensures we can correctly compute the density.
The conversion factor between kilograms and grams is:
- 1 kilogram equals 1,000 grams (1 kg = 1,000 g).
\[1.41 \text{ kg} \times 1,000 \frac{\text{g}}{\text{kg}} = 1,410 \text{ g}\]
The exercise emphasizes consistent units; hence, understanding and applying this conversion ensures we can correctly compute the density.
Density Formula
The crux of the original exercise lies in applying the density formula. Density, often denoted by the Greek letter rho (\(\rho\)), is a measure of mass per unit volume. The formula for calculating density is:
\[\rho = \frac{m}{V}\]
where \(m\) represents mass in grams (g), and \(V\) represents volume in cubic centimeters (cm³).
Using the converted values:
\[\rho = \frac{1,410~\text{g}}{314~\text{cm}^3}\]
Solving this, we arrive at the density of the titanium used in the bicycle frame. The ability to manipulate and use this formula with the correct units is pivotal to accurate and meaningful results in density calculations across various scientific fields.
\[\rho = \frac{m}{V}\]
where \(m\) represents mass in grams (g), and \(V\) represents volume in cubic centimeters (cm³).
Using the converted values:
- Mass, \(m\), of the titanium frame = 1,410 g
- Volume, \(V\), of the displaced water = 314 cm³
\[\rho = \frac{1,410~\text{g}}{314~\text{cm}^3}\]
Solving this, we arrive at the density of the titanium used in the bicycle frame. The ability to manipulate and use this formula with the correct units is pivotal to accurate and meaningful results in density calculations across various scientific fields.
Other exercises in this chapter
Problem 63
How many \(1-\mathrm{cm}\) squares would it take to construct a square that is \(1 \mathrm{~m}\) on each side? MISSED THIS? Read Section 1.6
View solution Problem 64
How many \(1-\mathrm{cm}\) cubes would it take to construct a cube that is \(4 \mathrm{~cm}\) on edge?
View solution Problem 67
Glycerol is a syrupy liquid often used in cosmetics and soaps. A \(3.25 \mathrm{~L}\) sample of pure glycerol has a mass of \(4.10 \times 10^{3} \mathrm{~g}\).
View solution Problem 68
A supposedly gold nugget displaces \(19.3 \mathrm{~mL}\) of water and has a mass of 371 g. Could the nugget be made of gold?
View solution