# Question 086d5

Jan 25, 2016

Here's what I got.

#### Explanation:

The idea here is that you need to figure out how much heat will be needed to

• go from liquid methane at $- {170}^{\circ} \text{C}$ to liquid methane at its boiling point of $- {161.5}^{\circ} \text{C}$

• go from liquid methane at its boiling point to vapor at its boiling point

• heat the resulting vapor from its boiling point of $- {161.5}^{\circ} \text{C}$ to the final temperature

The two equations that you will use are

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed/lost
$m$ - the mass of the sample
$c$ - the specific heat of methane
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

and

color(blue)(q = n * DeltaH_"fus"^@)" ", where

$q$ - heat absorbed
$n$ - the number of moles of methane
$\Delta {H}_{\text{fus}}^{\circ}$ - the standard molar heat of veporization of methane

The standard heat of vaporization of methane is listed as being equal to

$\Delta {H}_{\text{vap"^@ = "8.17 kJ/mol}}$

https://en.wikipedia.org/wiki/Methane_%28data_page%29

So, you're starting with a $\text{32.0-g}$ sample of liquid methane at $- {170}^{\circ} \text{C}$. Your first step will be to determine how much heat is needed to heat this sample to its boiling point.

Use the first equation to get

${q}_{1} = 32.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 3.48"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [-161.5 - (-170)]color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{1} = \text{946.56 J}$

So, out of the given $\text{42.0 kJ}$, a total of $\text{0.94656 kJ}$ will be used to heat the liquid from $- {170}^{\circ} \text{C}$ to $- {161.5}^{\circ} \text{C}$.

Next, use the second equation to figure out how much heat is needed to vaporize the liquid. Notice that you need to use the number of moles of methane, so use the compound's molar mass to go from grams to moles

32.0color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "1.995 moles CH"_4

This means that you have

${q}_{2} = 1.995 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 8.17"kJ"/color(red)(cancel(color(black)("mol}}}}$

${q}_{2} = \text{16.3 kJ}$

So, adding $\text{16.3 kJ}$ of heat will make sure that all the liquid methane gets converted to vapor at its boiling point of $- {161.5}^{\circ} \text{C}$. The heat you consumed to get to this point will be equal to

${q}_{\text{consumed}} = {q}_{1} + {q}_{2}$

${q}_{\text{consumed" = "0.94656 kJ" + "16.3 kJ" = "17.25 kJ}}$

The remaining amount of heat will thus be used up to heat the vapor from $- {161.5}^{\circ} \text{C}$ to a final temperature.

${q}_{3} = {q}_{\text{total" - q_"consumed}}$

${q}_{3} = \text{42.0 kJ" - "17.25 kJ" = "24.75 kJ}$

Now use the first equation again to find the change in temperature, $\Delta T$, when this much heat is added to vapor methane at its boiling point.

q_3 = m * c_"vapor" * DeltaT_"vapor" implies DeltaT_"vapor" = q_3/(m * c_"vapor")

Plug in your values to get - do not forget to convert the heat from kilojoules to joules

$\Delta {T}_{\text{vapor" = (24.75 * 10^(3)color(red)(cancel(color(black)("J"))))/(32.0color(red)(cancel(color(black)("g"))) * 2.22color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 348.4^@"C}}$

Since you know that

$\Delta T = {T}_{\text{final" - T_"initial}}$

you can say that

${T}_{\text{final" = T_"initial" + DeltaT_"vapor}}$

${T}_{\text{final" = -161.5^@"C" + 348.4^@"C" = 186.9^@"C}}$

I'll leave the answer rounded to three sig figs.

T_"final" = color(green)(187^@"C")#

So, adding $\text{42.0 kJ}$ of heat to $\text{32.0 g}$ of liquid methane at $- {170}^{\circ} \text{C}$ will get you $\text{32.0 g}$ of methane vapor at ${187}^{\circ} \text{C}$.