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What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings. Express your answer using two significant figures.

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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]

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