If a 55-gallon drum is completely full of water, what would be the weight of all that water on Earth? Answer in units of lbf.
The correct answer and explanation is:
To calculate the weight of water in a 55-gallon drum on Earth, we first need to know the weight of the water in pounds-force (lbf).
- Convert gallons to liters:
1 gallon is approximately equal to 3.78541 liters. Therefore, the volume of water in the drum is: 55 gallons×3.78541 liters/gallon=208.19755 liters55 \, \text{gallons} \times 3.78541 \, \text{liters/gallon} = 208.19755 \, \text{liters} - Convert liters to kilograms:
The density of water is approximately 1 kilogram per liter, so the mass of the water in the drum is: 208.19755 liters×1 kg/liter=208.19755 kg208.19755 \, \text{liters} \times 1 \, \text{kg/liter} = 208.19755 \, \text{kg} - Convert kilograms to pounds:
To convert kilograms to pounds, multiply by the conversion factor of 2.20462 pounds per kilogram: 208.19755 kg×2.20462 lbs/kg=458.18 lbs208.19755 \, \text{kg} \times 2.20462 \, \text{lbs/kg} = 458.18 \, \text{lbs} - Convert pounds to pounds-force (lbf):
On Earth, the weight of an object is measured in pounds-force, and since 1 pound-force equals 1 pound of mass under standard gravity (9.8 m/s²), the weight of the water will be the same as its mass in pounds. Thus: Weight of water=458.18 lbf\text{Weight of water} = 458.18 \, \text{lbf}
Therefore, the weight of the water in the 55-gallon drum is approximately 458.18 lbf.
This calculation assumes standard gravity at Earth’s surface and does not account for any minor variations that may exist due to location or atmospheric conditions.