Question

A 0.5530-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 6.310 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. (Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.)

Answer 1

mass of 6.910 moles of steam=6.910*18
=124.38
mass of ice=667gm
specific heat capacity of steam=2.1
specific heat capacity of water=4.2
specific heat capacity of steam=2
latent heat of fusion=334
latent heat of vaporisation=2256
so it is obvious that the temperature will be close to 100 degree celcius
heat energy requires for steam to come to 100 degree =124.38*2*365
=90797.4 J
heat required to released in converting ice to water= 12.4*2.1*553+553*334
=199102.12 J
heat energy required to convert steam to water =124.38*2256
=265982.4J
so net energy required to convert steam to water=265982.4+90797.4
=356779.8 J
so residual energy required=356779.8-199102.12
=157677.68 J
so the energy supplied by water from ice till temp=157677.68/553*4.2
=1197.552 K
now let the final temperature be t , so
553*4.2*9(t-1197.552)=124.38*4.2*(100-t)
ot t=117 K
this is the final temperature
hope this will suffice and u will award me points