The energy from 0.02 moles of butane is used to heat 328 grams of water. the temperature of the water rose from 298 k to 343 k. (the specific heat capacity of water is 4.18 j/k g.) what is the enthalpy of combustion? a. -61.7 kj b. 1,578.01 j c. 3,084,840.0 j d. 23,513,336 j

Answer : The correct option is, (C) 3,084,840.0 J

Explanation :

Formula used :

where,

= enthalpy of combustion = ?

m = mass of water = 328 g

= specific heat of water=

= final temperature = 343 K

= initial temperature = 298 K

n = moles of butane = 0.02 mole

Now put all the given values in the above formula, we get enthalpy of combustion.

Therefore, the enthalpy of combustion is, 3,084,840.0 J

icc, international association for cereal chemistry

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