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. -61.7 kJ
B. 1,578.01 J
C. 3,084,840.0 J
D. 23,513,336 J

Answer 1

Enthalpy of combustion can be calculated as:

\Delta H=m\times c\times \Delta T

Where,

m=mass of the compound

Delta H=Enthalpy of the combustion

c=specific heat

Delta T=Temperature difference

Here the values are given as:

m=328 g

c=4.18 J/K g

Delta T=(343-298)K

= 45 K

Putting all the values in the equation,

\Delta H=328\times 4.18\times \45

\Delta H=61.7 kJ

As it is exothermic process, so the value will be:

\Delta H= -61.7 kJ .

So the answer is option A that is -61.7 kJ.

Answer 2

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

Explanation :

Formula used :

$\Delta H_{combustion}=\frac{q}{n}$

$\Delta H_{combustion}=\frac{m\times c\times (T_{final}-T_{initial})}{n}$

where,

$\Delta H_{combustion}$ = enthalpy of combustion = ?

m = mass of water = 328 g

$c$ = specific heat of water= $4.18J/g.K$

$T_{final}$ = final temperature = 343 K

$T_{initial}$ = 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.

$\Delta H_{combustion}=\frac{328g\times 4.18J/g.K\times (343-298)K}{0.02mole}$

$\Delta H_{combustion}=3,084,840.0J$

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