A home swimming pool contains 145 m3 of water. At the beginning of the swimming season, the water must be heated from 20 to 32°C. How many joules of heat energy must be supplied?
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To calculate the amount of heat energy required to heat the water in the pool, you can use the equation:
Q = mcΔT,
where Q is the heat energy in joules, m is the mass of the water in kilograms, c is the specific heat capacity of water in joules per kilogram per degree Celsius (J/kg°C), and ΔT is the change in temperature in degrees Celsius.
In this case, we are given the volume of water, not the mass. So, we need to determine the mass of the water first. The relationship between volume, mass, and density is given by the equation:
m = ρV,
where m is the mass, ρ is the density, and V is the volume.
The density of water is constant at around 1000 kg/m³.
Therefore, the mass of the water in the pool is:
m = ρV = 1000 kg/m³ * 145 m³ = 145000 kg.
Next, we can use the specific heat capacity of water, which is approximately 4186 J/kg°C.
Plugging in the values into the heat energy equation, we have:
Q = mcΔT = 145000 kg * 4186 J/kg°C * (32°C - 20°C).
Calculating the temperature difference:
ΔT = 32°C - 20°C = 12°C.
Now, we can calculate the heat energy required:
Q = 145000 kg * 4186 J/kg°C * 12°C.
Simplifying the equation:
Q = 725305600 J.
Therefore, approximately 725,305,600 joules (or 7.25 x 10^8 joules) of heat energy must be supplied to heat the water in the pool.