What is the largest size gallon container of mercury, SG = 13.6, if you wish to restrict the weight to 50 pounds (neglecting the weight of the container itself)?
The correct answer and explanation is:
To solve this problem, we first need to understand the relationship between the weight, volume, and specific gravity (SG) of mercury. The specific gravity of a substance is the ratio of its density to the density of water. Since the density of water is 62.4 pounds per cubic foot (lb/ft³), the density of mercury will be: Density of mercury=SG×Density of water=13.6×62.4 lb/ft3=850.64 lb/ft3\text{Density of mercury} = \text{SG} \times \text{Density of water} = 13.6 \times 62.4 \, \text{lb/ft}^3 = 850.64 \, \text{lb/ft}^3
Next, we need to relate the weight (50 pounds) to the volume of mercury. The weight of an object is calculated by multiplying the volume by the density: Weight=Density×Volume\text{Weight} = \text{Density} \times \text{Volume}
Given that the weight should be 50 pounds: 50 lb=850.64 lb/ft3×Volume50 \, \text{lb} = 850.64 \, \text{lb/ft}^3 \times \text{Volume}
Solving for the volume: Volume=50 lb850.64 lb/ft3=0.0588 ft3\text{Volume} = \frac{50 \, \text{lb}}{850.64 \, \text{lb/ft}^3} = 0.0588 \, \text{ft}^3
Now, we need to convert this volume into gallons. Since 1 cubic foot equals 7.48 gallons: Volume in gallons=0.0588 ft3×7.48 gallons/ft3=0.440 gallons\text{Volume in gallons} = 0.0588 \, \text{ft}^3 \times 7.48 \, \text{gallons/ft}^3 = 0.440 \, \text{gallons}
Thus, the largest size gallon container of mercury that would weigh 50 pounds is approximately 0.44 gallons.
This calculation neglects the weight of the container itself, focusing solely on the weight of the mercury.