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?
1 m^3 = 1000 L so
145 m^3 = 145,000 L = 145,000,000 mL 145,000,000 grams.
q needed = mass H2O x specific heat H2O x (Tfinal-Tinitial). Substitute and solve for q in J.
To calculate the amount of heat energy required to heat the water in the swimming pool, we need to use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy required,
m is the mass of the water (in kg),
c is the specific heat capacity of water (in J/kg°C),
ΔT is the change in temperature (in °C).
First, let's calculate the mass of water using the formula:
m = volume * density
The density of water is approximately 1000 kg/m3 (or 1 g/cm3), so:
m = 145 m3 * 1000 kg/m3
m = 145000 kg
Next, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 32°C - 20°C
ΔT = 12°C
Lastly, let's substitute the values into the formula to find the heat energy required:
Q = 145000 kg * c * 12°C
The specific heat capacity of water is approximately 4,186 J/kg°C. So:
Q = 145000 kg * 4186 J/kg°C * 12°C
Now, let's calculate the result:
Q = 727,368,000 J
Therefore, approximately 727,368,000 joules of heat energy must be supplied to heat the water in the swimming pool.