If the price of electrical energy is $0.10 per kilo-watt*hour, what is the cost of using electrical energy to heat the water in a swimming pool (12.0m*9.00m*1.5m) from 15 to 27 degree C?
Can anyone please give me some hints to do it? Thanks a lot!
the heat capacity of water is 4190 Joules/kg deg C
1 meter^3 of water is about 1000 kg
so Joules to heat = 4190 J/kg C*1000 *volume in m^3 * change in temp
a Joule is a watt second
To find the cost of using electrical energy to heat the water in a swimming pool, you first need to determine the amount of energy required to raise the temperature of the water. Here are the steps to calculate the cost:
1. Calculate the volume of the swimming pool:
- The volume of the pool can be calculated by multiplying its length, width, and depth.
- In this case, the pool dimensions are given as 12m x 9m x 1.5m, so the volume would be 12m x 9m x 1.5m = 162 cubic meters.
2. Determine the specific heat capacity of water:
- The specific heat capacity of water is the amount of energy required to raise the temperature of a given amount of water by one degree Celsius.
- The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius.
3. Convert the volume of water to kilograms:
- Since the specific heat capacity of water is given in terms of grams, the volume of water (in cubic meters) needs to be converted to kilograms.
- 1 cubic meter of water is equal to 1000 kilograms, so in this case, the volume of water would be 162 cubic meters x 1000 = 162,000 kilograms.
4. Calculate the change in temperature:
- Subtract the initial temperature from the final temperature to determine the change in temperature.
- In this case, the change in temperature is 27 degrees Celsius - 15 degrees Celsius = 12 degrees Celsius.
5. Calculate the amount of energy required to heat the water:
- To calculate the energy required, multiply the mass of the water (in kilograms) by the specific heat capacity of water (in Joules per gram per degree Celsius) and by the change in temperature (in degrees Celsius).
- In this case, the energy required would be 162,000 kilograms x 4.18 Joules/g°C x 12°C = 8,163,360 Joules.
6. Convert the energy to kilowatt-hours:
- Since the cost of electrical energy is given in dollars per kilowatt-hour, the energy required needs to be converted from Joules to kilowatt-hours.
- 1 Joule is equal to 2.778 x 10^-7 kilowatt-hours, so the energy required would be 8,163,360 Joules x 2.778 x 10^-7 kWh/J = 2.27 kilowatt-hours.
7. Calculate the cost:
- Finally, multiply the energy required (in kilowatt-hours) by the cost of electrical energy (in dollars per kilowatt-hour).
- In this case, the cost would be 2.27 kWh x $0.10/kWh = $0.23.
Therefore, the cost of using electrical energy to heat the water in the swimming pool from 15 to 27 degrees Celsius would be $0.23.