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A swimming pool, 10.0 m by 4.0 m, is filled with water to a depth of 3.0 m at a temperature of 20.2°C.

If the energy needed to raise the temperature of the water to 27.3°C is obtained from the combustion of methane (CH4), what volume of methane, measured at STP,
must be burned?
AH combustion for CH4 = -891 kJ/mol
volume CH4 needed =

User Hellomello
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1 Answer

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First, we need to determine the mass of water in the pool:

mass = density x volume

density of water = 1000 kg/m³

volume = length x width x depth

volume = 10.0 m x 4.0 m x 3.0 m = 120 m³

mass = 1000 kg/m³ x 120 m³ = 120000 kg

Next, we need to calculate the heat required to raise the temperature of the water:

q = m x c x ΔT

where q is the heat energy, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.

c = 4.18 J/g°C (specific heat of water)

ΔT = 27.3°C - 20.2°C = 7.1°C

m = 120000 kg

q = 120000 kg x 4.18 J/g°C x 7.1°C = 35792400 J

Next, we need to convert the energy required to burn methane to heat energy:

-891 kJ/mol x (1 mol CH4/160 g CH4) x (1000 g/1 kg) = -5.569 kJ/g

We can now calculate the amount of methane needed:

energy = -5.569 kJ/g x mass CH4

mass CH4 = energy / (-5.569 kJ/g)

mass CH4 = 35792400 J / (-5569 J/g) = -6431.6 g

At STP, 1 mole of any gas occupies 22.4 L of volume. We can use this to convert the mass of methane to volume at STP:

1 mol CH4 = 16 g CH4

-6431.6 g CH4 x (1 mol CH4/16 g CH4) x (22.4 L/1 mol CH4) = -9074.4 L

Since we cannot have a negative volume, we can take the absolute value of the result:

|9074.4 L| = 9074 L

Therefore, approximately 9074 liters of methane gas at STP must be burned to raise the temperature of the water in the pool from 20.2°C to 27.3°C.

User Peter Severin
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