Question

A waterbed filled with water has the dimensions \(8.0 \mathrm{ft} \times 7.0 \mathrm{ft} \times 0.75 \mathrm{ft}\). Taking the density of water to be \(1.00 \mathrm{~g} / \mathrm{cm}^{3},\) how many kilograms of water are required to fill the waterbed?

Short answer

Answer: 1194.364 kg

Step-by-step solution

Convert Dimensions from Feet to Centimeters

To convert the dimensions of the waterbed from feet to centimeters, we'll use the conversion factor 1 foot = 30.48 centimeters: Length = 8.0 ft × 30.48 cm/ft = 243.84 cm Width = 7.0 ft × 30.48 cm/ft = 213.36 cm Height = 0.75 ft × 30.48 cm/ft = 22.86 cm

Calculate the Volume of the Waterbed

Next, we'll calculate the volume of the waterbed by multiplying the length, width, and height that we found in step 1. The formula for volume is: Volume = Length × Width × Height Volume = 243.84 cm × 213.36 cm × 22.86 cm = 1,194,364.066 cm³

Find the Mass of the Water using Density

Now we'll find the mass of the water using the given density (1.00 g/cm³) and the volume we just calculated. The formula for mass is: Mass = Volume × Density Mass = 1,194,364.066 cm³ × 1.00 g/cm³ = 1,194,364.066 g

Convert the Mass from Grams to Kilograms

Lastly, we'll convert the mass of the water from grams to kilograms by using the conversion factor 1 kg = 1000 g. The conversion formula is: Mass (kg) = Mass (g) / 1000 Mass (kg) = 1,194,364.066 g / 1000 = 1194.364 kg So, it takes 1194.364 kilograms of water to fill the waterbed.