Math Problem Statement

A cylindrical bar of gold that is 1.5cm in height and .25 cm in diameter has a mass of 23.1984g, as determined on an analytical balance. An empty graduated cylinder is weighed on a triple beam balance and has a mass of 73.47 grams. After pouring a small amount of liquid into the graduated cylinder, the mass is 79.16 grams. When the gold cylinder is placed in the graduated cylinder (the liquid covers the top of the gold cylinder), the volume indicated on the graduated cylinder is 8.5 mL. Assume that the temperature of the gold bar and the liquid is 86 degrees Fahrenheit. If the density of the liquid decreases by 1.0% for each 10 degrees Celsius rise inn temperature (over the range 0 to 50 degrees Celsius), determine the density of the gold at 86 degrees Fahrenheit.

Solution

To determine the density of the gold cylinder at 86°F, we need to go through the following steps:

Step 1: Find the volume of the gold cylinder.

The formula for the volume of a cylinder is:

V=πr2hV = \pi r^2 h

Where:

  • rr is the radius (half of the diameter)
  • hh is the height

Given that:

  • The diameter of the gold cylinder is 0.25 cm, so the radius r=0.252=0.125r = \frac{0.25}{2} = 0.125 cm
  • The height h=1.5h = 1.5 cm

Substitute these values into the formula:

V=π(0.125)2(1.5)=π×0.015625×1.5=0.0733594cm3V = \pi (0.125)^2 (1.5) = \pi \times 0.015625 \times 1.5 = 0.0733594 \, \text{cm}^3

Thus, the volume of the gold cylinder is approximately:

V=0.0734cm3V = 0.0734 \, \text{cm}^3

Step 2: Calculate the density of the gold.

Density (ρ\rho) is mass divided by volume:

ρ=massvolume\rho = \frac{\text{mass}}{\text{volume}}

The mass of the gold cylinder is given as 23.198423.1984 g. The volume is 0.07340.0734 cm³ (calculated in Step 1). Therefore, the density of the gold is:

ρ=23.19840.0734=316.03g/cm3\rho = \frac{23.1984}{0.0734} = 316.03 \, \text{g/cm}^3

Thus, the density of the gold at 86°F is approximately:

ρ=316.03g/cm3\rho = 316.03 \, \text{g/cm}^3

Step 3: Confirm that temperature doesn't significantly affect gold's density.

Gold's density doesn't change significantly with temperature, and the problem doesn't indicate any major correction to the density due to temperature. Therefore, we can assume the calculated density of 316.03 g/cm³ is accurate at 86°F.

Additional Information:

For the liquid, there's a 1% decrease in density for every 10°C rise, but this pertains to the liquid and is not required for calculating the gold's density.

Would you like to dive deeper into any specific aspect or clarify any step?

Follow-up Questions:

  1. How do we determine the relationship between temperature and density for metals?
  2. What would happen to the density if the temperature were drastically lower, say at 32°F?
  3. How does the density of gold compare to other metals, like silver or copper?
  4. Can the volume of the gold cylinder be measured differently with less error?
  5. Why doesn’t temperature affect gold’s density as much as it does for liquids?

Tip: When working with physical properties like density, always make sure units are consistent, especially when converting temperatures or volumes.

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Math Problem Analysis

Mathematical Concepts

Geometry
Density
Temperature Effects on Materials

Formulas

Volume of a cylinder: V = πr²h
Density formula: ρ = mass / volume

Theorems

Volume of a Cylinder
Density Calculation

Suitable Grade Level

Grades 11-12