Chemistry Calorimetry problem below please explain. The steps you take to get the answer. I will AWARD BEST ANSWER?

Calculate total heat required to raise the temperature of the pool:
q = m c (T2-T1)
The terms in [ ] get to the mass of H2O, assuming a constant density of 1 g/mL
q = [70.0 m^3 X 1000 L/m^3 X 1000 g/L] X 4.184 J/gC X (22.3 - 2.66C = 5.75 X10^9 J = 5.75X10^6 kJ

Moles and mass of methane:
5.75X10^6 kJ / 803 kJ/mol X 16.01 g/mol = 1.15X105 g = 115 kg
This is the mass required if the efficiency of heat transfer was 100%. Because it only 65% efficient,
115 / 0.65 = 176 kg methane would be required.

Mass of methane = m kg = 1000m g
Molar mass of methane (CH₄) = (12.0 + 1.0×4) g/mol = 16.0 g/mol
No. of moles of methane = (1000m g) / (16.0 g/mol) = 62.5m mol
Heat released by burning m kg of methane = (62.5m mol) × (803 × 10³ J mol⁻¹)

Density of water = 1.00 g/cm³
Mass of water, m = (70.0 × 10⁶ cm³) × (1.00 g/cm³) = 7.00 × 10⁷ g
Heat absorbed by water = m c ΔT = (7.00 × 10⁷ g) × (4.184 J/g °C) × [(22.3 - 2.66) °C]

(Heat released by burning m kg of methane in J) × 65% = (Heat absorbed by water in J)
62.5m × (803 × 10³) × 65% = (7.00 × 10⁷) × 4.184 × (22.3 - 2.66)
m = (7.00 × 10⁷) × 4.184 × (22.3 - 2.66) / [62.5 × (803 × 10³) × 65%]
m = 176

Mass of methane burned = 176 kg

Methane = CH4
Mass of one mole = 12 + 4 = 16 grams = 0.016 kg

Q = mass * specific heat * (Tf – Ti)

The specific heat of water is 4,186 J/(kg * ˚C). The density of water is 1000 kg/m^3,
Mass = 70,000 kg
Tf – Ti = 22.3 – 2.66 = 19.64˚C

Q = 70,000 * 4,186 * 19.64 = 5.7549128 * 10^9 J

Heat of combustion = 803,000 J/ mole
Since the natural gas fired heater has an efficient of 65% only 65% of this will be used to increase the water’s temperature.

Heat of combustion = 521,950 J/mole
To determine the number of moles of methane, divide the number of joules by the heat of combustion.

n = 5.7549128 * 10^9 ÷ 521,950
This is approximately 11,026 moles.

Mass = 0.016 * (5.7549128 * 10^9 ÷ 521,950)
This is approximately 176.4 kg.