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

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

Apr 21, 2017

The big picture is that the water was heated from ${21}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$, and then further vaporized such that $\text{3 g}$ of water was lost as water vapor. We don't care how long this took, because heating at constant pressure (since a ${C}_{P}$ was mentioned) is a state function.

So, you have two main steps:

1. Heating the water from ${21}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$
2. Vaporizing $\text{3 g}$ of water into vapor

1) $83$ seconds is quite a long time, to go without evaporation, but we assume the mass of the water was constant as it was being heated.

Using the heat capacity of the water of ${C}_{P} = \text{4.184 J/g"^@"C}$, we have:

${q}_{P 1} = m {C}_{P} \Delta T$

$= \left(\text{250 g")("4.184 J/g"^@"C")(100^@"C" - 21^@ "C}\right)$

$=$ $\text{82634 J}$ $=$ $\text{82.634 kJ}$

2) Vaporizing the water at constant temperature and pressure (in an open system), we have:

${q}_{P 2} / n = \Delta {H}_{\text{vap" = "40.7 kJ/mol}}$

$\implies {q}_{P 2} = n \Delta {H}_{\text{vap}}$

$= 3 \cancel{\text{g H"_2"O") xx (cancel("1 mol H"_2"O"))/("18.015 g H"_2"O") xx ("40.7 kJ")/cancel("mol}}$

$=$ $\text{6.7777 kJ}$

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:

$\textcolor{g r e e n}{{q}_{P , w}} = {q}_{P 1} + {q}_{P 2} = 82.634 + 6.7777$

$=$ $\textcolor{g r e e n}{\text{89.412 kJ}}$

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 $\text{298 K}$, we ASSUME that it is the same at $\text{393 K}$:

$\Delta {H}_{C , 298}^{\circ} = - \text{2202 kJ"/"mol propane}$

This is for the reaction

$\text{C"_3"H"_8(l) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O} \left(l\right)$

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 $\Delta {H}_{C}^{\circ}$.

This means that ${q}_{P , w}$, meaning the heat flow at constant pressure into the water, having come out from the propane reaction, is equal to $- {q}_{P , \text{rxn}}$.

Therefore,

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

$\approx \Delta {H}_{C , 298}^{\circ}$

$= - \text{2202 kJ"/"mol propane" xx x " mols propane}$

$\implies x = \text{0.0406 mols propane}$,

or

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

$=$ $\textcolor{b l u e}{\text{1.79 g propane}}$,

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.