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.

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superman1987 superman1987 · Answer 1 · 2019-03-12T14:23:36+01:00

Answer:

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.