How many liters of C6H14 measured at 20℃ must be burned to provide enough heat to warm 27.5 m3 of water 19.3 to 32.8℃, assuming that all the heat of. Combustion is transferred to the water which has a specific heat of4.18 J/g℃
Answers
Answer:
Okay, let's first calculate the amount of heat energy needed to raise the temperature of the water using the formula Q= mcθ, where m is the mass of the water, c is the specific heat capacity of water and θ is the change in temperature.
1.00m³= 1.00x10⁶cm³
29.5m³= 29.5x10⁶cm³
Since the density of water is 1g/cm³, the mass of 29.5x10⁶cm³ is 29.5x10⁶g.
Q= (29.5x10⁶)(4.18)(31.9-18.7)
= 1.627692000x10⁹J
ΔH°combustion of C₆H₁₄= -4163.0x10³J/mol
Density of C₆H₁₄= 654.8x10⁻³g/cm³
Number of moles of C₆H₁₄ that need to be burned
= 1.627692000x10⁹/4163.0x10³
= 390.9901513
Relative Molecular Mass(RMM) of C₆H₁₄
= 6(RAM of C)+14(RAM of H)
= 6(12)+14
= 86
1 mole of C₆H₁₄= 86.0g
390.9901513 moles of C₆H₁₄
= (86.0)(390.9901513)
= 3.362515301x10⁴g
Thus, volume of C₆H₁₄ that must be burned
= mass of C₆H₁₄/density of C₆H₁₄
= 3.362515301x10⁴/654.8x10⁻³
= 5.135179141x10⁴cm³
= (5.135179141x10⁴)x10⁻³L
= 51.35179141L
= 51.4L correct to 3sf -->since 1L= 1dm³ and 1dm³= 1000cm³= 1000mL
I hope this helps and feel free to send me an e-mail again if you have any doubts!