An outdoor swimming pool contains 100000 kilogram of water how much energy is needed yo heat water from 15 celius to 20 celius
According to the cooling curve, what is the approximate freezing point of the substance?
A) -193°C
B) -111°C
C) 46°C
D) 70°C
To calculate the energy needed to heat water from one temperature to another, we can use the formula:
Q = mcΔT
Where:
Q = Energy (in joules)
m = Mass of water (in kilograms)
c = Specific heat capacity of water (in J/kg°C)
ΔT = Change in temperature (in °C)
Given:
Mass of water (m) = 100000 kg
Initial temperature (T1) = 15°C
Final temperature (T2) = 20°C
The specific heat capacity of water (c) is approximately 4.18 J/g°C.
First, we need to convert the mass of water from kilograms to grams:
Mass of water = 100000 kg = 100000000 g
Next, we can calculate the change in temperature (ΔT):
ΔT = T2 - T1
= 20°C - 15°C
= 5°C
Now we can substitute these values into the formula to calculate the energy (Q):
Q = mcΔT
= (100000000 g)(4.18 J/g°C)(5°C)
Calculating this:
Q = 2090000000 Joules
Therefore, approximately 2,090,000,000 Joules (or 2.09 GJ) of energy would be needed to heat the outdoor swimming pool from 15°C to 20°C.
To calculate the energy needed to heat water from one temperature to another, we can use the specific heat capacity formula:
q = m * c * ΔT
Where:
q = energy (in joules)
m = mass of water (in kilograms)
c = specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, we have the following information:
m = 100,000 kg (mass of water)
c ≈ 4,186 J/kg°C (specific heat capacity of water)
ΔT = 20°C - 15°C = 5°C (change in temperature)
Now, we can plug the values into the formula to calculate the energy needed:
q = 100,000 kg * 4,186 J/kg°C * 5°C = 2,093,000,000 joules
Therefore, approximately 2,093,000,000 joules of energy are needed to heat the water in the pool from 15°C to 20°C.