How much heat (kJ) must be removed in order to change 2.91 moles of steam at 114.0 oC to water at 51.0

Question

How much heat (kJ) must be removed in order to change 2.91 moles of steam at 114.0 oC to water at 51.0 oC ?


The specific heat capacity of water vapor is 2.01 J g-1 oC -1


Use a negative sign in your answer if heat is removed.


Enter a numeric answer only, do not enter units.

Answer 1

The process occurs in 3 steps. I assume that P=1 atm and thus the boiling point of water is 100.0. The mass of water is moles*MW =2.91*18 =52.38g

Step 1. Water vapor at 114.0 C loses energy to cool down to 100.0 C
Q1=mcΔΤ= 52.38*2.01*(100-114) = -1474 J

Step 2.Water vapor at 100 C turns into liquid at the same temperature (latent heat is released). For water the latent heat of condensation is - 2272 J/g (you should have provided this value). So
Q2= -2272m = -2272*52.38 = -119007 J

Step 3. Liquid water of 100 C cools down to 51.0 C
Q3= mc'ΔΤ = 52.38*4.184*(51-100) =-10739 J

So in total Q=Q1+Q2+Q3= -131220 J= -131.22 KJ