Using a molar heat of combustion of - 890 kJ/mol, The minimum mass of methane that must be burned to warm 5.64 kg of water from 21.5°C to 77.4°C, assuming no heat losses to the environment is ab. c g.

Put your answer for ab. c in the blank.

23.78 g.

Explanation:

Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.5°C to 77.4°C using the relation:

Q = m. c.ΔT,

where, Q is the amount of heat absorbed by water (Q = ? J).

m is the mass of water (m = 5.64 kg = 5640.0 g).

c is the specific heat capacity of water (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT = 77.4 °C - 21.5 °C = 55.9 °C).

∵ Q = m. c.ΔT

∴ Q = m. c.ΔT = (5640.0 g) (4.186 J/g.°C) (55.9 °C) = 1319745.336 J ≅ 1319.745 kJ.

As mentioned in the problem the molar heat of combustion of methane is - 890.0 kJ/mol. Using cross multiplication we can get the no. of moles of methane that are needed to be burned to release 1319.745 kJ:

Combustion of 1.0 mole of methane releases → - 890.0 kJ.

Combustion of? mole of methane releases → - 1319.745 kJ.

∴ The no. of moles of methane that are needed to be burned to release 1319.745 kJ = ( - 1319.745 kJ) (1.0 mol) / ( - 890.0 kJ) = 1.482 mol.

Now, we can get the mass of methane that must be burned to warm 5.64 kg of water from 21.5°C to 77.4°C:

∴ mass = (no. of moles needed) (molar mass of methane) = (1.482 mol) (16.04 g/mol) = 23.78 g.