A 0.6630-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.
ID: 1626298 • Letter: A
Question
A 0.6630-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 6.790 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.)
Explanation / Answer
mass of the 6.79 molesof the steam is = 6.79 X 18 = 122.22
mass of the ice = 667 g
specific heat capacity of the steam = 2.1
specific heat capacity of the water = 4.2
latent heat of fusion = 334
latent heat of vaporisation = 2256
the heat energy is required for the steam to close 100oC
= 122.22 X 2 X 365 = 89220.6 J
the heat energy is required to convert ice to water = 12.4 X 2.1 X 667 + 667 X 334 = 240146.68 J
the heat energy is required to convert steam to water = 122.22 X 2256 = 275728.32 J
the net energy is required to convert steam to water = 89220.6 + 275728.32 = 364948.92 J
the energy suppiled by the water is = 89220.6 / 667 X 4.2
= 31.84 K
the final tempetature is
667 X 4.2 X ( t - 31.84 ) = 122.22 X 4.2 X (100 - t)
2801.4 t - 89196.57 = 51332.4 - 513.324 t
3314.72 t = 140528.97
t = 42.39 K
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.