Question

# The density of water is 1000 kilograms per cubic meter, and the density of ice is about 916 kilograms per cubic meter. If 575 kilograms of water and 275 kilograms of ice is combined in a container, about how much volume would the mixture take up?

Answer

0.875 m^3

- Q: What is the total mass of the mixture? A: The total mass is 575 + 275 = 850 kg.
- Q: What is the total volume of the mixture, using the density of water as an estimate? A: If we assume the entire mixture has a density of 1000 kg/m^3 (since water is the majority component), then the total volume can be calculated as follows: 850 kg ÷ 1000 kg/m^3 = 0.85 m^3.
- Q: What is the estimated volume of the ice? A: The mass of the ice is 275 kg, and its density is about 916 kg/m^3. So the volume of the ice can be calculated as follows: 275 kg ÷ 916 kg/m^3 ≈ 0.3 m^3.
- Q: What is the estimated volume of the water? A: The mass of the water is 575 kg, and its density is 1000 kg/m^3. So the volume of the water can be calculated as follows: 575 kg ÷ 1000 kg/m^3 = 0.575 m^3.
- Q: What is the total estimated volume of the mixture, taking into account the volumes of the water and ice separately? A: The total estimated volume is the sum of the volumes of the water and ice: 0.575 m^3 + 0.3 m^3 = 0.875 m^3.