If the price of electrical energy is $0.10 /KWh, what is the cost of using electrical energy to heat the water in a swimming pool (12.0space m times 9.00 space m times 1.5space m) from 15 ^oC to

The first step is to calculate the volume of the swimming pool: Volume = (12.0 m) x (9.00 m) x (1.5 m) = 162 m^3 The next step is to calculate the amount of heat needed to raise the temperature of the water from 15°C to 27°C. We can use the specific heat capacity of water to do this: Q = mc?T where Q is the amount of heat needed, m is the mass of water, c is the specific heat capacity of water, and ?T is the change in temperature. We can assume that the density of water is 1 kg/liter, so the mass of water in the pool is: Mass = Volume x Density = 162 x 1000 = 162,000 kg The specific heat capacity of water is 4.18 J/g°C, or 4,180 J/kg°C. Therefore, the amount of heat needed to raise the temperature of the water is: Q = (162,000 kg) x (4,180 J/kg°C) x (27°C - 15°C) = 9,087,360,000 J We can convert this to kilowatt-hours (kWh) by dividing by the conversion factor of 3,600,000 J/kWh: Energy = 9,087,360,000 J / 3,600,000 J/kWh = 2,524 kWh Finally, we can calculate the cost of using electrical energy to heat the water, given the price of $0.10/kWh: Cost = Energy x Price = 2,524 kWh x $0.10/kWh = $252.40 Therefore, the cost of using electrical energy to heat the water in the swimming pool from 15°C to 27°C is $252.40.