Question

Help with a phase change problem please?
If 42.0 kJ of heat is added to a 32.0g sample of liquid methant under 1 atm of pressure at atemperature of -170C, what are the final state and temperatrue of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is -161.5C. The specific heats of liquid and gaseous methane are 3.48 and 2.22 J/g*K, respectively. Heat of vaporization is 8.20 kJ/mol.

Answer 1

the total energy 42 kj is used to first raise the temp. from -170 to -161.5.
then a part of it is used to convert the state . then when it is in the gaseous state, it is used to increase the temp. from -161.5 to the final.

total heat energy=(massxspecific h.c of methanex delta temp.)+(massxheat.of.vap.)+(massxspecific.h.c of methane (gaseous)x (final temp-161.5))

Answer 2

when liquid absorbs heat Q, temp rises ΔΤ accordingto Q1=m*c1*ΔΤ1
when gas absorbs heat Q, temp rises ΔΤ accordingto Q2=m*c2*ΔΤ2
at boiling poit ab absorbing heat causes m' to evaporate according to Q'=m'*c'
so
1st we mast chek if given heat is enaf to reach boiling poit

Answer 3

we assume that all energy is absorbed in liquid state and calculate ΔΤ1 and from that Tfinal
if Tfinal <-161.5 => state liquid, temp what we have
if Tfinal>-161.5
that means that we have evaporation

Answer 4

we calc heat needed for liq to rise to -161.5
that we name Q1'
then calc Q(given)-Q1'
that is heat available for evaporation

Answer 5

now we check if that heat is enaf to evaporate all liq
Qgiven-Q1'=m'*c' solve to m'
if m' found<m given => m' gr at liq -161.5 and m-m' at gas at -161.5

Answer 6

hope you figure out rest.......

Answer 7

thanks