The heat combustion of gaseous butane is -2658 kl/mol and the heat of combustion of liquid butane is:

The heat of combustion of liquid butane is: -2636 kJ/mol

Explanation:

This problem is solved by using Hess law of heats summation:

We have the heat of combustion for gaseous butane:

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)     ΔHºcomb = -2658 kJ/mol

and we want the heat of combustion for liquid butane.

What we need to do is add the heat of vaporization for liquid butane to this equation:

C₄H₁₀ (l)   ⇒  C₄H₁₀ (g)   ΔHºvap =  22.44 kJ/mol

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  ΔHºcomb = -2658 kJ/mol

C₄H₁₀ (l)   + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  

So,

ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j