Question

Swimming Pool A swimming pool measuring 20.0 m × 12.5 m is filled with water to a depth of 3.75 m. If the initial temperature is 18.4°C, how much heat must be added to the water to raise its temperature to 29.0°C? Assume that the density of water is 1.000 g/mL.

Short answer

To raise the temperature of the swimming pool from 18.4°C to 29.0°C, approximately \(3.93 × 10^{10} J\) (Joules) of heat must be added.

Step-by-step solution

Find the volume of water

To find the volume of water in the swimming pool, multiply its length, width, and depth: Volume = length × width × depth Let's plug in the given values: Volume = \(20.0 m\) × \(12.5 m\) × \(3.75 m\)

Calculate the volume

Now, we can calculate the volume of the water: Volume = \(20.0 m\) × \(12.5 m\) × \(3.75 m\) = \(937.5 m^3\)

Convert the volume to liters

Next, we need to convert the volume from cubic meters to liters, since the density of water is given in grams per milliliter (g/mL), and there are 1000 liters in a cubic meter: Volume in liters = \(937.5 m^3\) × \(1000 \frac{L}{m^3}\) = \(937,500 L\)

Calculate the mass of water

Now, we can find the mass of the water using its density: Mass = volume × density Given the density of water is \(1.000 \frac{g}{mL}\), let's first convert it to grams per liter: Density = \(1.000 \frac{g}{mL}\) × \(1000 \frac{mL}{L}\) = \(1000 \frac{g}{L}\) Now, we can calculate the mass of the water: Mass = \(937,500 L\) × \(1000 \frac{g}{L}\) = \(937,500,000 g\)

Determine the temperature change

We need to find the temperature change to be able to calculate the amount of heat: ΔT = T(final) - T(initial) Let's plug in the given initial and final temperatures: ΔT = \(29.0°C\) - \(18.4°C\) = \(10.6°C\)

Calculate the heat using the specific heat formula

Finally, we can calculate the amount of heat needed to raise the temperature of the water using the specific heat formula: Q = mcΔT where Q is the heat, m is the mass, c is the specific heat (which is \(4.186 J/g°C\) for water), and ΔT is the temperature change. Let's plug in the values: Q = \(937,500,000 g\) × \(4.186 \frac{J}{g°C}\) × \(10.6°C\)

Calculate the heat

Now, we can calculate the total amount of heat needed to raise the temperature of the water: Q = \(937,500,000 g\) × \(4.186 \frac{J}{g°C}\) × \(10.6°C\) ≈ \(3.93 × 10^{10} J\) So, to raise the temperature of the swimming pool from 18.4°C to 29.0°C, approximately \(3.93 × 10^{10} J\) (Joules) of heat must be added.

Key concepts

Specific Heat Capacity

Specific heat capacity is an important characteristic of a substance; it helps us understand how much heat it takes to change its temperature. Specifically, it tells us the amount of heat required to raise the temperature of 1 gram of a substance by 1°C.

For water, the specific heat capacity is quite high at 4.186 J/g°C. This means water can absorb a lot of heat without rapidly changing temperature. This property is why water is so effective at storing heat and is used in cooling and heating systems.

In the context of the swimming pool exercise, knowing the specific heat capacity allows us to calculate the total heat, or energy, needed to raise the water’s temperature using the formula:

• Q = mcΔT

where

• Q is the total heat energy,
• m is the mass of the water,
• c is the specific heat capacity of water, and
• ΔT is the change in temperature.

Temperature Change

Understanding temperature change is crucial in heat transfer problems because it represents the difference in temperature that a substance undergoes.

In the exercise, we want to heat the swimming pool from an initial temperature of 18.4°C to a final temperature of 29.0°C. The change in temperature ( ΔT) is found by subtracting the initial temperature from the final temperature:

• ΔT = T(final) - T(initial)

Plugging in the given values:

• ΔT = 29.0°C - 18.4°C = 10.6°C

This temperature change is then used in the specific heat capacity formula to determine how much heat is required to achieve this change in the pool's temperature.

Understanding this concept ensures you can not only handle similar exercises involving heat transfer but also appreciate how much energy input is needed to effect a desired change in conditions.

Mass of Water

Mass is a fundamental property of matter that influences how much heat is required to change its temperature. To calculate the mass of water in a system, you often multiply the volume of water by its density.

In this swimming pool example, we first determined the volume of water by multiplying the pool's dimensions. Since water has a density of 1.000 g/mL, or 1000 g/L, we can easily convert between units, from volume to mass, when dealing with large quantities like in a swimming pool.

The mass of water is calculated as:

• Mass = volume × density

Using the exercise values:

• Volume = 937,500 L (from the conversion from cubic meters)
• Mass = 937,500 L × 1000 g/L = 937,500,000 g

Knowing the mass is essential for determining the heat required to increase the water's temperature. This concept is necessary for solving heat transfer problems, as greater mass generally means more heat is needed to achieve a change in temperature.