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Chem Help! Thermodynamics?
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 6.190 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.)
1 Answer
- billrussell42Lv 73 years ago
specific heat of water is 4.186 kJ/kgC = 4.186 J/gC
specific heat of ice is 2.06 kJ/kgC = 2.06 J/gC
specific heat of steam is 2.1 kJ/kgK = 2.1 J/gK
heat of fusion of ice is 334 kJ/kg = 334 J/g
heat of vaporization of water is 2256 kJ/kg = 2256 J/g
density of fresh water at 20C = 0.998 g/cm³
6.190 moles of water x 18g/mole = 111.4 g
assuming all the steam is converted to water and all the ice melts, both at a final temperature of T
energy to warm the ice to 0º, E1 = 2.06 J/gC x 641 g x 12.4º = 16374 J
energy to melt the ice, E1 = 334 J/g x 641 g = 214094 J
energy to warm the water to T, E3 = 4.186 J/gC x 641 g x (T–0)ºC = 2683T J
energy to cool the steam to 100º E4 = 2.1 J/gK x 111.4 g x (365–100) = 61994 J
energy to condense the steam E5 = 2256 J/g x 111.4 g = 251318 J
energy to cool that water to T, E6 = 4.186 J/gC x 111.4 g x (100–T)ºC = 46632 – 466T
E1+2+3 = E4+5+6
16374 + 214094 + 2683T = 61994 + 251318 + 46632 – 466T
2683T + 466T = 61994 + 251318 + 46632 – 16374 – 214094
3149T = 129476
T = 41.1º