0.5750-kg ice cube at -12.40 degree C is placed inside a chamber of steam at 365
ID: 1311366 • Letter: 0
Question
0.5750-kg ice cube at -12.40 degree C is placed inside a chamber of steam at 365.0 degree C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 6.070 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.) Since the system is isolated, then energy is conserved. This means that whatever thermal energy is absorbed by the ice must have been radiated by the steam. Therefore The absorbtion of thermal energy by the ice occurs in three stages: (1) the temperature of the ice is raised to its critical temperature (0 C), (2) the thermal energy melts the ice, and (3) the water is raised from 0 degree C to Tf. Mathematically, this means where mice and cice are the mass and specific heat of the ice, cwater is the specific heat of liquid water, Tc, 1 is the critical temperature of ice, Tice, is the initial temperature of the ice, and Lf is the latent heat of fusion of water. Using similar nomenclature, Where Lv is the latent heat of vaporization and Tc2 is the critical temperature of steam. By setting the two energy changes equal to each other, you should be able to solve for the equilibrium temperature Tf. C ice = 2034 J / (kg degree C) c water = 4186 .J / (kg - degree C) c steam = 2029 J / (kg degree c)Explanation / Answer
let equilibrium temperature be T
Heat lost bysteam = Heat gained by ice
6.07*(0.018)*2029*(365 - 100) +6.07*(0.018)* 2256*1000 + +6.07*(0.018)*4186*(100 - T)
= 0.5750*2034*12.4 + 0.575*334*1000 + 0.575*4186*(T - 0)
( 0.575*4186 + 6.07*(0.018)*4186 )*T =
6.07*(0.018)*2029*(365 - 100) +6.07*(0.018)* 2256*1000 +6.07*(0.018)*4186*100 - ( 0.5750*2034*12.4 + 0.575*334*1000 )
T = ( 6.07*(0.018)*2029*(365 - 100) +6.07*(0.018)* 2256*1000 +6.07*(0.018)*4186*100 - ( 0.5750*2034*12.4 + 0.575*334*1000 )) / ( 0.575*4186 + 6.07*(0.018)*4186 )
= 50.42
Equilibrium temperature = 50.42 degree celsuis
Answer is 50 .42
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