Question

A butane burner is used to heat water. The Mr of butane is 58.
● the enthalpy change of combustion of butane is –2877 kJ mol–1.
● 250 g of water is heated from 12 °C to 100 °C.
● The burner transfers 47% of the heat released from the burning fuel to the water.
Assume that the butane undergoes complete combustion and none of the water evaporates.

What is the minimum mass of butane that must be burnt?
A 0.068 g
B 1.85 g
C 3.94 g
D 4.48 g
(the answer is C) But i want the method so please help.

Answer 1

Answer:

The correct answer is option C.

Explanation:

Heat absorbed by water :Q

Mass of water = m = 250 g

Change in temperature of water = ΔT = 100 °C - 12 °C = 88 °C

Heat capacity of water = c = 4.186 J/g°C

Q = mc × ΔT

$Q=250 g\times 4.186 J/g^oC\times 88^oC=92,092 J$

Heat produced by the burner = Q'

The burner transfers 47% of the heat released from the burning fuel to the water.

$47\%=\frac{Q}{Q'}\times 100$

$Q'=\frac{92,092 J}{47}\times 100=195,940.42 J$

Q = 195,940.42 J = 195.940 kJ (1 kJ = 1000 J)

Heat produced by the burner = Q' = -195.940 kJ

Negative sign means that energy is released.

The enthalpy change of combustion of butane :

= $\Delta H_{comb}=-2877 kJ/mol$

Mass of butane burnt = m

Moles of butane burnt ,n= $\frac{m}{58 g/mol}$

$Q'=n\times \Delta H_{comb}$

$-195.940 kJ =\frac{m}{58 g/mol}\times -2877 kJ/mol$

$m=\frac{-195.940 kJ\times 58 g/mol}{-2877 kJ/mol}$

m = 3.94 g