Calculate the density of mercury in lbm/ft^3 from a tabulated specific gravity, and calculate the volume in ft^3 occupied by 215 kg of mercury.
The Correct Answer and Explanation is:
Answer:
- Density of mercury in lbm/ft³:
Specific gravity of mercury ≈ 13.6
Density of water = 62.4 lbm/ft³
⇒ Density of mercury = 13.6 × 62.4 = 848.64 lbm/ft³ - Volume occupied by 215 kg of mercury:
215 kg × 2.20462 lbm/kg = 473.993 lbm
Volume = mass / density = 473.993 lbm ÷ 848.64 lbm/ft³ ≈ 0.5585 ft³
Textbook-Style Explanation
To calculate the density of mercury in imperial units (pounds mass per cubic foot, or lbm/ft³), we start with its specific gravity (SG). Specific gravity is defined as the ratio of the density of a substance to the density of water at a reference temperature, typically 4°C. Mercury has a specific gravity of approximately 13.6, which means it is 13.6 times denser than water.
In the imperial system, the density of water is approximately 62.4 lbm/ft³. Since specific gravity is a unitless ratio, we can calculate the density of mercury by simply multiplying this value by the specific gravity:Density of mercury=SG×Density of water=13.6×62.4=848.64 lbm/ft3\text{Density of mercury} = \text{SG} \times \text{Density of water} = 13.6 \times 62.4 = 848.64 \, \text{lbm/ft}^3Density of mercury=SG×Density of water=13.6×62.4=848.64lbm/ft3
Next, to determine the volume occupied by a given mass of mercury, we need to convert the mass from kilograms to pounds mass. Using the conversion factor:1 kg=2.20462 lbm1 \, \text{kg} = 2.20462 \, \text{lbm}1kg=2.20462lbm215 kg×2.20462=473.993 lbm215 \, \text{kg} \times 2.20462 = 473.993 \, \text{lbm}215kg×2.20462=473.993lbm
Now we apply the definition of density:Density=MassVolume⇒Volume=MassDensity=473.993848.64≈0.5585 ft3\text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{473.993}{848.64} \approx 0.5585 \, \text{ft}^3Density=VolumeMass⇒Volume=DensityMass=848.64473.993≈0.5585ft3
Thus, 215 kg of mercury occupies approximately 0.5585 ft³. This calculation is commonly seen in fluid mechanics and thermodynamics to convert between metric and imperial units or to determine how much space a given mass of a liquid will occupy based on its known properties.
