Question

How many liters of C6H14(l), measured at 20 ∘C, must be burned to provide enough heat to warm 27.7 m3 of water from 19.2 to 31.4 ∘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. Express your answer to four significant figures and include the appropriate units.

Answer 1

27.7 m³ = 27700 liter = 27700000 ml density of water is 1 then  27700000 ml =  27700000 gm

specific heat = 4.18 J/g ⁰C

T = 31.4 - 19.2 = 12.2 ⁰C

q = mass specific heat T

= 277000004.1812.2

q = 1412589200 Joule = 1412589.2 KJ heat require

Combution reaction of C₆H₁₄ is

2C₆H₁₄ + 19O₂ 12CO₂  + 14H₂O - 8326.2 KJ

2 mole of hexane produce  8326.2 KJ thus 4163.1 KJ heat produced from 1 mole of hexane

then to produce 1412589.2 KJ heat require 1412589.2/4163.1 = 339.311 mole of hexane

molar mass of hexane = 86.18 gm/mol

1 mole of hexane = 86.18 gm then 339.311 mole of hexane = 339.31186.18 = 29241.82 gm

density of hexane = 0.6603gm/cm³ = 0.6603g/ml

volume = mass/density = 29241.82/0.6603 = 44285.65 ml

44285.65 ml = 44.28565 L

44.28565 Liter C₆H₁₄ require