Answer:
5.01×10^9 J is the energy required
Explanation:
This is a calorimetry problem:
Q = m . C . ΔΤ where:
Q = heat
m = mass
ΔΤ = Final T° - Initial T°
First of all we determine the pool's volume with the measures
10 m . 4m . 3m = 120m3
As water density is 1g/mL we can determine water's mass but firstly we must convert the m3 to cm3
1mL = 1cm3
1m3 = 1x10^6 cm3
120 m3 . 1x10^6 cm3 / 1m3 = 1.2x10^8 cm3
Water density = water mass / water volume
1 g/ mL = water mass / 1.2x10^8 mL
Water mass = 1.2x10^8 g
Then, we replace the data in the formula
Q = 1.2x10^8 g .4.18 J/g°C (30°C - 20°C)
Q = 1.2x10^8 g . 4.18 J/g°C . 10°C
Q = 5016000000 joules
5.01×10^9 J
