Question

When 20.0g of steam at 140.0 °C condenses into 200.0g liquid water at 25.0 °C, what will be the final temperature of the water? Assume that everything takes place in a perfectly insulated enclosure from which there is no heat leak. Data: heat of vaporization for water (Hvap): 40.7 kJ/mole Heat capacity of steam: 2.09 J/g-°C Heat capacity of water: 4.184 J/g-°C Please Explain Step by step

Answer 1

Heat given out when steam moves from 140 oC to 100 oC,
Q1 = ms*Cs*delta T
       = 20 * 2.09*40
       = 1672 J

Heat required for water to go from 25 oC to 100 oC,
Q2 = mw*Cw*delta T
       = 200*4.184*(100-25)
       = 62760 J

Since Q2>Q1, steam has to convert in water,
Heat given out if whole steam converts into water,
Q3 = ns*Hvap
number of moles of steam, ns = mass / molar mass
                                                                = 20/18
                                                                = 1.11 mol
Q3 = ns*Hvap
      = 1.11 * 40.7
      = 45.222 KJ
      =45222 J

Q1+Q3 =1672 + 45222=46894 J
So, heat released is more than heat required. SO temoerature of system will be below 100 oC

Let final temperature be T oC
heat gained by steam = heat lost by water
Q1 + Q3 + ms*cw*(100-T)= mw*Cw*(T-25)
46894 + 20*4.184*(100-T) = 200*4.184*(T-25)
46894 + 8368 - 83.68*T = 836.8 *T- 20920
920.48*T = 76182
T= 82.8 oC
Answer: 82.8 oC