Question

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.

Answer 1

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$

$Q_{water} = 3599.435\,kJ$

The heat liberated by the LP gas is:

$Q_{LP} = \frac{3599.435\,kJ}{0.16}$

$Q_{LP} = 22496.469\,kJ$

A kilogram of LP gas has a minimum combustion power of $50028\,kJ$. Then, the required mass is:  

$m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }$

$m_{LP}=0.45\,kg$