# How much propane is required to power the stove that heats and vaporizes this water? What was the efficiency?

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Suppose you have a propane stove and you want to cook #"250 g"# of water. First, you want to heat it from #21^@ "C"# to #100^@ "C"# over the course of #83# seconds. Then, you want to vaporize #"3 g"# of the water (#DeltaH_(vap) = "40.7 kJ/mol"# ). If the enthalpy of combustion of propane at #"298 K"# is #-"2202 kJ/mol"# , calculate the mass of propane needed. If #"3.7 g"# of propane was available and the calculated mass is used up, what is the efficiency of the propane stove?

Suppose you have a propane stove and you want to cook

##### 1 Answer

The big picture is that the water was heated from

So, you have two main steps:

- Heating the water from
#21^@ "C"# to#100^@ "C"# - Vaporizing
#"3 g"# of water into vapor

Using the heat capacity of the *water* of

#q_(P1) = mC_PDeltaT#

#= ("250 g")("4.184 J/g"^@"C")(100^@"C" - 21^@ "C")#

#=# #"82634 J"# #=# #"82.634 kJ"#

*constant temperature and pressure* (in an open system), we have:

#q_(P2)/n = DeltaH_"vap" = "40.7 kJ/mol"#

#=> q_(P2) = nDeltaH_"vap"#

#= 3 cancel("g H"_2"O") xx (cancel("1 mol H"_2"O"))/("18.015 g H"_2"O") xx ("40.7 kJ")/cancel("mol")#

#=# #"6.7777 kJ"#

Because it was an *open* system, water was able to be lost as water vapor (we still have conservation of mass!). Combining the two steps gives the total heat flow involved:

#color(green)(q_(P,w)) = q_(P1)+q_(P2) = 82.634 + 6.7777#

#=# #color(green)("89.412 kJ")#

This is *positive* because **heat was put into** the heating and the vaporization, *relative* to the system.

Knowing the **enthalpy of combustion** of propane at

#DeltaH_(C,298)^@ = -"2202 kJ"/"mol propane"#

This is for the reaction

#"C"_3"H"_8(l) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(l)#

Thus, in order to figure out how much propane is needed, equate this to the heat flow we just found. The heat had to flow **out of the propane combustion reaction**, so it is exothermic *with respect to the propane reaction itself*, i.e. negative

This means that

Therefore,

#q_(P,"rxn") = -"89.412 kJ"#

#~~ DeltaH_(C,298)^@#

#= -"2202 kJ"/"mol propane" xx x " mols propane"#

#=> x = "0.0406 mols propane"# ,

or

#0.0406 cancel("mols propane") xx "44.096 g propane"/cancel"mol propane"#

#=# #color(blue)("1.79 g propane")# ,

was needed to perform this heating and vaporization.

The **efficiency** is simply the mass of propane used up divided by the mass of propane put in. It's how much mass was actually needed compared to the total mass we thought we needed.

#color(blue)("% efficiency") = (1.79 cancel"g propane")/(3.7 cancel("g propane")) xx 100%#

#= 48._(4)%#

#=> color(blue)(48%)# to two sig figs.