Math Problem Statement

The seawater has an approximate density of 1.04 g/cm³. We want to fill a swimming pool with the following dimensions: 2 meters deep, 15 meters wide, and 25 meters long. How much water will it hold, and how many kilograms of seawater will be needed?

Solution

The problem in the image asks for the volume of water needed to fill a swimming pool and how many kilograms of seawater would be required based on the given density. Let's break down the problem:

Information given:

  • Density of seawater: 1.04g/cm31.04 \, \text{g/cm}^3
  • Pool dimensions:
    • Depth: 2meters2 \, \text{meters}
    • Width: 15meters15 \, \text{meters}
    • Length: 25meters25 \, \text{meters}

a) Volume of water the pool can hold:

The volume VV of a rectangular pool is calculated by multiplying its dimensions: V=length×width×depthV = \text{length} \times \text{width} \times \text{depth} Substituting the values: V=25m×15m×2m=750m3V = 25 \, \text{m} \times 15 \, \text{m} \times 2 \, \text{m} = 750 \, \text{m}^3 So, the volume of water required to fill the pool is 750 cubic meters.

b) Kilograms of seawater needed:

We need to convert the volume from cubic meters to cubic centimeters to use the density given in g/cm3\text{g/cm}^3.

  • 1m3=1,000,000cm31 \, \text{m}^3 = 1,000,000 \, \text{cm}^3

So: 750m3=750×1,000,000cm3=750,000,000cm3750 \, \text{m}^3 = 750 \times 1,000,000 \, \text{cm}^3 = 750,000,000 \, \text{cm}^3

Now, using the density of seawater (1.04g/cm31.04 \, \text{g/cm}^3), we can find the mass: Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume} Mass=1.04g/cm3×750,000,000cm3=780,000,000g\text{Mass} = 1.04 \, \text{g/cm}^3 \times 750,000,000 \, \text{cm}^3 = 780,000,000 \, \text{g} Since 1kg=1000g1 \, \text{kg} = 1000 \, \text{g}, we convert to kilograms: Mass=780,000,000g1000=780,000kg\text{Mass} = \frac{780,000,000 \, \text{g}}{1000} = 780,000 \, \text{kg}

Thus, 780,000 kilograms of seawater are needed to fill the pool.

Would you like more details on any step or have further questions?


Here are 5 related questions to think about:

  1. How would the volume change if the pool depth were doubled?
  2. What is the total weight of seawater in pounds instead of kilograms?
  3. If the pool had an irregular shape, how would you calculate the volume?
  4. What would happen to the required mass if the density of the water changed to 1.03g/cm31.03 \, \text{g/cm}^3?
  5. How would evaporation affect the amount of water required to keep the pool full?

Tip: Always remember to keep track of unit conversions, especially when working with densities and volumes!

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Math Problem Analysis

Mathematical Concepts

Volume Calculation
Unit Conversion
Density and Mass Relationship

Formulas

Volume of a rectangular pool: V = length × width × depth
Mass = Density × Volume
1 m³ = 1,000,000 cm³

Theorems

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Suitable Grade Level

Grades 9-12