If the price of electrical enegy is $0.10 per kilowatt *hour, find the cost of using electrical energy to heat the water in a 12.0m * 9.00m * 1.5m swimming pool from 15 degrees C to 27 degrees C.

So the change in temperature is 285.15 , the volume is 162m^3, the specific heat capacity (c) is 4.186 [J/kg*degrees C]. Please and Thanks.

You're not going to get an answer here, I've tried this site, and everytime, I get the wrong answer or no one responds.

Well, let's dive into the calculation, shall we?

First, we need to calculate the mass of the water in the pool. We can use the formula:

mass = volume * density

The density of water is approximately 1000 kg/m³. So, the mass of the water in the pool is:

mass = 162 m³ * 1000 kg/m³ = 162,000 kg

Now, let's calculate the energy required to heat the water. We can use the formula:

energy = mass * specific heat capacity * change in temperature

The specific heat capacity is given as 4.186 J/kg*°C, and the change in temperature is 12 degrees Celsius (27°C - 15°C = 12°C). So, the energy required is:

energy = 162,000 kg * 4.186 J/kg*°C * 12°C
energy = 8,208,336 J

Finally, we need to convert the energy from joules to kilowatt-hours (kWh). We can use the conversion factor:

1 kWh = 3.6 × 10⁶ J

energy_kWh = energy / (3.6 × 10⁶ J/kWh)
energy_kWh = 8,208,336 J / (3.6 × 10⁶ J/kWh)
energy_kWh ≈ 2.28 kWh

Now, to find the cost, we multiply the energy in kWh by the cost per kWh:

cost = energy_kWh * $0.10/kWh
cost ≈ 2.28 kWh * $0.10/kWh

And voila! The cost of using electrical energy to heat the water in the pool is approximately the same as the cost of convincing a fish to ride a bicycle. Just kidding! The actual cost is about $0.23.

Please remember that this is just a rough estimate and there may be other factors to consider, such as efficiency losses.

To find the cost of using electrical energy to heat the water in the swimming pool, we need to follow these steps:

Step 1: Calculate the mass of the water in the pool.
The volume of the pool is given as 162 m^3. Since the density of water is approximately 1000 kg/m^3, we can find the mass of the water using the formula: mass = volume * density.
mass = 162 m^3 * 1000 kg/m^3 = 162,000 kg.

Step 2: Calculate the energy required to heat the water.
The specific heat capacity (c) of water is given as 4.186 J/kg°C. The change in temperature (ΔT) is 27°C - 15°C = 12°C = 12 K (since 1°C = 1 K).
The energy required (Q) to heat the water can be calculated using the formula: Q = mass * c * ΔT.
Q = 162,000 kg * 4.186 J/kg°C * 12 K = 8,364,672 J.

Step 3: Convert the energy to kilowatt-hours.
Since the cost of electrical energy is given in dollars per kilowatt-hour, we need to convert the energy obtained in Step 2 to kilowatt-hours.
1 kilowatt-hour (kWh) = 3,600,000 J.
Energy in kilowatt-hours = 8,364,672 J / 3,600,000 J/kWh = 2.323 kWh.

Step 4: Calculate the cost of the electrical energy.
The cost of electrical energy is $0.10 per kilowatt-hour.
Cost = Energy in kilowatt-hours * cost per kilowatt-hour.
Cost = 2.323 kWh * $0.10/kWh = $0.2323.

Therefore, the cost of using electrical energy to heat the water in the swimming pool is approximately $0.2323.

To find the cost of using electrical energy to heat the water in the swimming pool, we first need to calculate the amount of energy required to heat the water.

The formula to calculate the energy required is given by:
Energy (Q) = mass (m) * specific heat capacity (c) * change in temperature (ΔT)

To find the mass of the water, we can use the formula:
Mass (m) = density (ρ) * volume (V)

The density of water is approximately 1000 kg/m^3.

Let's calculate the mass of water in the swimming pool:
mass (m) = 1000 kg/m^3 * 162 m^3
mass (m) = 162,000 kg

Now, let's calculate the energy required to heat the water:
Energy (Q) = 162,000 kg * 4.186 J/kg∙°C * 285.15 °C
Energy (Q) ≈ 192,822,570 J

Next, let's convert the energy from joules to kilowatt-hours:
1 kilowatt-hour (kWh) = 3,600,000 joules (J)

Energy (Q) ≈ 192,822,570 J / 3,600,000 J/kWh
Energy (Q) ≈ 53.56 kWh

Now, we can calculate the cost of using electrical energy:
Cost = Energy (Q) * Price per kilowatt-hour

Given the price of electrical energy is $0.10 per kilowatt-hour:
Cost ≈ 53.56 kWh * $0.10/kWh
Cost ≈ $5.36

Therefore, the cost of using electrical energy to heat the water in the swimming pool from 15°C to 27°C would be approximately $5.36.

The change in temperature is 12 deg C if it rises from 15 to 27 degrees.

1. Find the mass of water from volume of water times its density.
2. Find the heat in Joule needed from Q=mct where m is mass of water, c is specific heat capacity and t is rise in temperature.

Then the cost.
3. One kilowatt hour is
1000W x 60(min) X 60(sec) Joules.
4. Then find how many of these you need to heat that water.
5. Each one costs $0.1