Answer:
[tex]m_{LP}=0.45\,kg[/tex]
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
[tex]Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}][/tex]
[tex]Q_{water} = 3599.435\,kJ[/tex]
The heat liberated by the LP gas is:
[tex]Q_{LP} = \frac{3599.435\,kJ}{0.16}[/tex]
[tex]Q_{LP} = 22496.469\,kJ[/tex]
A kilogram of LP gas has a minimum combustion power of [tex]50028\,kJ[/tex]. Then, the required mass is:
[tex]m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }[/tex]
[tex]m_{LP}=0.45\,kg[/tex]
