A swimming pool, 20.0 m ✕ 12.5 m, is filled with water to a depth of 3.81 m. If the initial temperature of the water is 18.9°C, how much heat must be added to the water to raise its temperature to 30.1°C? Assume that the density of water is 1.000 g/mL.
q = mcdT
q = ?
volume = 3.81m x 20.0m x 12.5m = ? m^3 and convert to mL. Since the density is 1.00 g/mL, that will be the mass.
dT = Tfinal-Tinitial
To calculate the heat required to raise the temperature of the water, we need to use the specific heat formula:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's calculate the mass of the water:
To find the mass, we can use the formula:
mass = density × volume
Given that the density of water is 1.000 g/mL and the volume is the area of the pool multiplied by the depth:
Volume = length × width × depth
Volume = 20.0 m × 12.5 m × 3.81 m
Next, convert the volume from cubic meters (m³) to milliliters (mL) because the density is given in grams per milliliter (g/mL):
1 m³ = 1000000 mL
Now we can calculate the mass:
mass = density × volume
mass = 1.000 g/mL × (20.0 m × 12.5 m × 3.81 m) × 1000000 mL/m³
Now that we have the mass, we can calculate the heat energy (Q) using the specific heat capacity of water, which is 4.186 J/g°C:
Q = mcΔT
ΔT = final temperature - initial temperature
= 30.1°C - 18.9°C
Now we can calculate the heat energy required:
Q = mass × specific heat capacity × ΔT