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 28.1°C is obtained from the

Question

combustion of methane (CH4),what volume of methane, measured at STP, must be burned?

∆H for CH4 = -891 kJ/mol

Answer 1

First, find the volume of water in the pool.

10.0 m x 4.0 m x 3.0 m = 120 m³

Next, convert this to liters, or mls.

120 m³ x 1x10⁶ ml/m³ = 120x10⁶ ml = 1.2x10⁸ ml

Now, if we assume that the density of the water is 1g/ml, we can estimate the mass of the water.

1.2x10⁸ ml x 1 g/ml = 1.2x10⁸ g

We can now use Q = mC∆T for find the energy needed to raise the temperature of this mass of water from 20.2ºC to 28.1ºC.

Q = heat = ?

m = mass = 1.2x10⁸ g

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = 7.9ºC

Solving for Q: Q = (1.2x10⁸ g)(4.184 J/g/deg)(7.9 deg) = 3.966x10⁹ J = 4.0x10⁹ J = 4.0x10⁶ kJ = 4.0x10³ MJ

For the combustion of methane, we have...

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation

∆H = -891 kJ/mol

Heat needed to raise temp of water = 4.0x10⁶ kJ

4.0x10⁶ kJ x 1 mol/891 kJ = 4.49x10³ moles CH₄ needed

At STP 1 mole of an ideal gas = 22.4 L, thus the volume of methane required would be ...

4.49x10³ L x 1 mol/22.4 L = 200 liters