Question

A 0.6410-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

Let us see how much energy is available with steam
Steam first get converted into steam of 100 C then water of 100 C then water of 0 C then ice of 0 C then ice of -12.4 C .
So this much of energy is available with us .
Total energy =
1 mole of steam = 18 grams
6.31 mole = 6.31*18 = 113.58 grams
Q = mcdT
C is specifc heat
C_steam = 2.108 kJ/kg k, C_water = 4.18 kJ/kg C , C_ice = 2.03 kJ/kgC
L_vaporisation = 2257 kJ/kg , L_fusion = 335 kJ/kg
Q = (.11358*(2.108)*(365 - 100)) + (0.11358*2257) + (.11358*(4.2*100)) + (0.113*335) + (0.113*(2.03)*(12.4))
Q = 408.335 kJ
Now we have total mass of ice = 0.6410 +0.11358 kg = 0.75458 kg
Now energy require to convert this mass into ice of 0 C
Q = mcdT = 0.7548*2.03*(0-(-12.4)) = 19 kJ
Energy require to convert this mass into 0C water
Q = mL_fusion = 0.75458*(335) = 252.78 kJ
Eenrgy left = 408.335 - (252.78+19) = 136.5507 kJ
Now energy require to convert water into 100 C water
= 0.75458*(4.2*100) = 316.9236
but energy left is less than this therefore this will not get converted into 100 C
let say final temperature is T
energy left = 136.5507 = 0.75458*4.2(T-0)
T = 43.08 C