Question

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. 1,578.01 J B. 3,084,840.0 J C. 23,513,336 J

Answer 1

c

Step-by-step explanation:

C4H10 + 13/2 O2 = 4CO2 + 5H2O

0.02 = 328*4.18*45

1= x

making x the subject of the formula you'll get 3084840.0J

Answer 2

B) The enthalpy of combustion = 3,084,840 J

Step-by-step explanation:

Given:

Moles of butane = 0.02

Mass of water, m = 328 g

Initial temperature T1 = 298 K

Final temperature T2 = 343 K

Specific heat of water, c = 4.18 J/g-K

To determine:

Enthalpy of combustion

Step-by-step explanation:

Heat lost during combustion of butane = heat gained by water

Heat gained (q) by water is given as:

q = mc\Delta T = mc(T2-T1)

substituting for m, c, T2 and
`q = 328g*4.18J/g-K*(343-298)K = 61697.8 \ J`T1

`Enthalpy \ of \ combustion = (q)/(moles\ of\ butane) \\\\= (61697.8)/(0.02) = 3,084,890\ J`