Question

How many liters of C6H14(l), measured at 20 ∘C, must be burned to provide enough heat to warm 28.5 m3 of water from 19.3 to 37.6 ∘C , assuming that all the heat of combustion is transferred to the water, which has a specific heat of 4.18 J/(g⋅∘C)? Recall that 1 mL=10−3 L.

Answer 1

Volume of C6H14 = 68.9 L

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The heat capacity by the first law of thermodynamics, the temperature change is linear to the heat added.

delta q = m. C. delta T

m = mass of water

C = specific heat capacity

delta T = T (final) - T(initial)

C = specific heat capacity of water = 4.186 J/g.oC

Volume of water = 28.5 m3 = 28.5 x 10^6 mL = 28.5 x 10^6 mL  

Density of water = 1 g/mL

Mass of water = 28.5 x 10^6 mL x 1 g/mL = 28.5 x 10^6 g

delta T = 37.6 - 19.3 = 18.3 oC

delta q = (28.5 x 10^6 g) x 4.18 J/g.oC x 18.3 oC = 2.18 x 10^9 J

delta Ho combustion of C6H14 = -4163 kJ/mol =   -4163 x 10^3 J/mol

No. of moles of C6H14 required for 2.18 x 10^9 J = 2.18 x 10^9/4163 x 10^3 = 523.68 mol

Molar mass of C6H14 = 86.1754 g/mol

Mass of 523.68 mol of C6H14 = 86.1754 g/mol x 523.68 mol = 45128.33 g

Density of C6H14 = 0.655 g/mL

Volume of 45128.33 g of C6H14 = 45128.33 g / 0.655 g/mL = 68898.2137 mL

Volume of C6H14 = 68898.2137 mL = 68898.2137 x 10^-3 L = 68.898 L