Answer:
7.46 g.
Explanation:
- Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.0°C to 70.0°C using the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by water (Q = ??? J).
m is the mass of water <u><em>(m: we will determine).</em></u>
c is the specific heat capacity of water (c = 4.186 J/g.°C).
ΔT is the temperature difference (final T - initial T) (ΔT = 70.0 °C - 21.0 °C = 49.0 °C).
- To determine the mass of 1.76 L of water we can use the relation:
mass = density x volume.
density of water = 1000 g/L & V = 1.76 L.
∴ mass = density x volume = (1000 g/L)(1.76 L) = 1760.0 g.
∵ Q = m.c.ΔT
<em>∴ Q = m.c.ΔT </em>= (1760.0 g)(4.186 J/g.°C)(49.0 °C) = 360483.2 J ≅ 360.4832 kJ.
- As mentioned in the problem the molar heat of combustion of hexane is - 4163.0 kJ/mol.
<em>Using cross multiplication we can get the no. of moles of hexane that are needed to be burned to release 360.4832 kJ:</em>
Combustion of 1.0 mole of methane releases → - 4163.0 kJ.
Combustion of ??? mole of methane releases → - 360.4832 kJ.
∴ The no. of moles of hexane that are needed to be burned to release 360.4832 kJ = (- 360.4832 kJ)(1.0 mol)/(- 4163.0 kJ) = 0.0866 mol.
- Now, we can get the mass of hexane that must be burned to warm 1.76 L of water from 21.0°C to 70.0°C:
<em>∴ mass = (no. of moles needed)(molar mass of hexane)</em> = (0.0866 mol)(86.18 g/mol) = <em>7.46 g.</em>
