Question

How many liters of C6H14(l), measured at 20 ∘C, must be burned to provide enough heat to warm 25.0 m3 of water from 19.8 to 38.5 ∘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.

A 1.592 g sample of a component of the light petroleum distillate called naphtha is found to yield4.880 g CO2(g) and 2.330 g H2O(l) on complete combustion.

This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula.

The compound also has the following properties:

melting point of −154 ∘C ,

boiling point of 60.3 ∘C ,

density of 0.6532 g/mL at 20 ∘C ,

specific heat of 2.25 J/(g⋅∘C) , and

ΔH∘f=−204.6kJ/mol .

Answer 1

Volume of water = 25.0 m³ = 25.0 x 10⁶ cm³   

Density of water = 1 g/cm³  

Mass of water = 1 * 25.0 x 10⁶

= 25.0 x 10⁶ g

Temperature difference = 38.5 - 19.8

= 18.7 oC

Heat absorbed by water, Q = m*c*T

= 25.0 x 10⁶ * 4.18 * 18.7

= 1.954 x 10⁹ J

Enthalpy of combustion of C6H14(l) = - 4163.2 kJ/mol

= - 4.1632 x 10⁶ J/mol

Moles of C6H14(l) burned = 1.954 x 10⁹ / 4.1632 x 10⁶

= 469.35 moles

Molar mass of C6H14 = 86.178 g/mol

Mass of C6H14(l) burned = 469.35 * 86.178

= 40447.64 g

Density of C6H14(l) = 654.8 x 10^-3 g/cm³  

Volume of C6H14(l) burned = 40447.64 / (654.8 x 10^-3)

= 61770.98 cm³  

= 61.77 L