The specific weight of water at ordinary pressure and temperature is 62.4 lb/ft3 (9.81 kN/m3). The specific gravity of mercury is 13.55. Compute the density of water and the specific weight and density of mercury
The Correct Answer and Explanation is :
Problem Breakdown and Solution
Given:
- The specific weight of water, γwater=62.4 lb/ft3\gamma_{\text{water}} = 62.4 \, \text{lb/ft}^3 (which is equivalent to 9.81 kN/m39.81 \, \text{kN/m}^3).
- The specific gravity of mercury, SGHg=13.55SG_{\text{Hg}} = 13.55.
Part 1: Density of Water
The specific weight (γ\gamma) and density (ρ\rho) are related by the equation: γ=ρ⋅g\gamma = \rho \cdot g
where γ\gamma is the specific weight, ρ\rho is the density, and gg is the acceleration due to gravity. We will use the value of gg as 32.2 ft/s232.2 \, \text{ft/s}^2 (standard in imperial units).
The density of water in imperial units can be calculated by rearranging the equation: ρwater=γwaterg=62.4 lb/ft332.2 ft/s2\rho_{\text{water}} = \frac{\gamma_{\text{water}}}{g} = \frac{62.4 \, \text{lb/ft}^3}{32.2 \, \text{ft/s}^2}
This gives the density of water in lbm/ft3\text{lbm/ft}^3, which is: ρwater=1.937 lbm/ft3\rho_{\text{water}} = 1.937 \, \text{lbm/ft}^3
For SI units, the specific weight of water is 9.81 kN/m39.81 \, \text{kN/m}^3, and using the standard gravitational acceleration g=9.81 m/s2g = 9.81 \, \text{m/s}^2, the density of water is: ρwater=9.81 kN/m3÷9.81 m/s2=1 kg/m3\rho_{\text{water}} = 9.81 \, \text{kN/m}^3 \div 9.81 \, \text{m/s}^2 = 1 \, \text{kg/m}^3
Thus, the density of water is 62.4 lbm/ft362.4 \, \text{lbm/ft}^3 or 1000 kg/m31000 \, \text{kg/m}^3 in standard units.
Part 2: Density and Specific Weight of Mercury
Mercury’s specific gravity, SGHg=13.55SG_{\text{Hg}} = 13.55, is the ratio of mercury’s density to water’s density. Since we know the density of water is 62.4 lb/ft362.4 \, \text{lb/ft}^3, we can calculate the density of mercury using: ρHg=SGHg×ρwater=13.55×62.4 lbm/ft3\rho_{\text{Hg}} = SG_{\text{Hg}} \times \rho_{\text{water}} = 13.55 \times 62.4 \, \text{lbm/ft}^3
Thus, ρHg=844.32 lbm/ft3\rho_{\text{Hg}} = 844.32 \, \text{lbm/ft}^3
The specific weight of mercury in imperial units is: γHg=ρHg×g=844.32 lbm/ft3×32.2 ft/s2=27272.3 lb/ft3\gamma_{\text{Hg}} = \rho_{\text{Hg}} \times g = 844.32 \, \text{lbm/ft}^3 \times 32.2 \, \text{ft/s}^2 = 27272.3 \, \text{lb/ft}^3
For SI units, using SGHg=13.55SG_{\text{Hg}} = 13.55 and the density of water as 1000 kg/m31000 \, \text{kg/m}^3, we find: ρHg=13.55×1000=13,550 kg/m3\rho_{\text{Hg}} = 13.55 \times 1000 = 13,550 \, \text{kg/m}^3
The specific weight of mercury in SI units is: γHg=ρHg×g=13,550 kg/m3×9.81 m/s2=133,535.5 N/m3=133.5 kN/m3\gamma_{\text{Hg}} = \rho_{\text{Hg}} \times g = 13,550 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 = 133,535.5 \, \text{N/m}^3 = 133.5 \, \text{kN/m}^3
Final Answers:
- Density of Water:
- In imperial units: 1.937 lbm/ft31.937 \, \text{lbm/ft}^3
- In SI units: 1000 kg/m31000 \, \text{kg/m}^3
- Density of Mercury:
- In imperial units: 844.32 lbm/ft3844.32 \, \text{lbm/ft}^3
- In SI units: 13,550 kg/m313,550 \, \text{kg/m}^3
- Specific Weight of Mercury:
- In imperial units: 27272.3 lb/ft327272.3 \, \text{lb/ft}^3
- In SI units: 133.5 kN/m3133.5 \, \text{kN/m}^3
Explanation:
To compute these values, the relationship between specific weight and density is essential. Specific gravity is a dimensionless quantity representing the ratio of the density of a substance to the density of water. By knowing the specific gravity of mercury, we can calculate both its density and specific weight by using the density of water as a reference. Additionally, converting between imperial and SI units is straightforward once we establish the known conversion factors (e.g., 1 lb = 0.4536 kg, 1 ft = 0.3048 m, etc.).