A swimming pool, 20.0 m 12.5 m, is filled with water to a depth of 3.73 m. If the initial temperature of the water is 17.7°C, how much heat must be added to the water to raise its temperature to 28.7°C? Assume that the density of water is 1.000 g/mL.
First you want to calculate the volume of the pool, then convert to mL. The easy way to do that is to change the dimensions to cm before you start, then L x W x depth = volume. Using density convert to grams.
q = mass water x specific heat water x (Tfinal-Tinitial).
To calculate the amount of heat required to raise the temperature of the water in the swimming pool, we need to use the formula:
Q = mcΔT
Where:
Q = amount of heat required (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (in °C)
First, let's find the mass of the water in the pool. We can use the formula:
mass = density x volume
Given that the density of water is 1.000 g/mL and the volume of the pool is 20.0 m x 12.5 m x 3.73 m, we can convert the volume to liters as follows:
1 m^3 = 1000 liters
So, the volume of the pool is 20.0 m x 12.5 m x 3.73 m = 932.5 m^3 = 932,500 liters.
Next, we can find the mass of the water:
mass = 1.000 g/mL x 932,500 liters = 932,500 g = 932.5 kg
Now, let's calculate the change in temperature:
ΔT = final temperature - initial temperature = 28.7°C - 17.7°C = 11°C
Lastly, we need the specific heat capacity of water. The specific heat capacity of water is approximately 4.186 J/g°C.
Now, we can plug the values into the formula to calculate the amount of heat required:
Q = (932.5 kg) x (4.186 J/g°C) x (11°C)
Q = 43,096.54 J
Therefore, the amount of heat that must be added to the water to raise its temperature from 17.7°C to 28.7°C is approximately 43,096.54 Joules.