Question #086d5
1 Answer
Here's what I got.
Explanation:
!! LONG ANSWER !!
The idea here is that you need to figure out how much heat will be needed to

go from liquid methane at
#170^@"C"# to liquid methane at its boiling point of#161.5^@"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^@"C"# to the final temperature
The two equations that you will use are
#color(blue)(q = m * c * DeltaT)" "# , where
and
#color(blue)(q = n * DeltaH_"fus"^@)" "# , where
The standard heat of vaporization of methane is listed as being equal to
#DeltaH_"vap"^@ = "8.17 kJ/mol"#
https://en.wikipedia.org/wiki/Methane_%28data_page%29
So, you're starting with a
Use the first equation to get
#q_1 = 32.0 color(red)(cancel(color(black)("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 = "946.56 J"#
So, out of the given
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 = 1.995color(red)(cancel(color(black)("moles"))) * 8.17"kJ"/color(red)(cancel(color(black)("mol")))#
#q_2 = "16.3 kJ"#
So, adding
#q_"consumed" = q_1 + q_2#
#q_"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
#q_3 = q_"total"  q_"consumed"#
#q_3 = "42.0 kJ"  "17.25 kJ" = "24.75 kJ"#
Now use the first equation again to find the change in temperature,
#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
#DeltaT_"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
#DeltaT = T_"final"  T_"initial"#
you can say that
#T_"final" = T_"initial" + DeltaT_"vapor"#
#T_"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