WILL RATE!! A 0.6410-kg ice cube at -12.40°C is placed inside a chamber of steam
ID: 1479667 • Letter: W
Question
WILL RATE!! 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 6.070 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.)
Explanation / Answer
To warm the ice to 0 0C
Q1 = mice Cice (del T) { eq.1 }
Cice = specific heat constant of ice = 2.087 J/(g 0C
Q1 = (641 g) (2.087 J/(g 0C) [(0) - (-12.4)] 0C
Q1 = 16588.3 J
To melt the ice, we have
Q2 = mice (del L) { eq.2 }
Q2 = (641 g) (333.6 J/g)
Q2 = 213837.6 J
To cool the steam to 100 0C, we have
Q3 = (6.07 mol) (36.03 J/(mol·0C) (365 0C - 100 0C)
Q3 = 57956.1 J
To condense the steam, we have
Q4 = (6.07 mol) (40660 J/mol)
Q4 = 246806.2 J
Heat lost by the steam is given as :
Qsteam = Q3 + Q4 = (57956.1 J) + (246806.2 J)
Qsteam = 304762.3 J
Heat gained by the ice is given as :
Qice = Q1 + Q2 = (16588.3 J) + (213837.6 J)
Qice = 230425.9 J
The excess amount of heat is given as :
Q = Qsteam - Qice = (304762.3 J) - (230425.9 J)
Q = 74336.4 J
Heat produced into the water is given as :
Q = mice Cw (del T) { eq.3 }
(74336.4 J) = (641 g) (4.184 J/(g·0C) (del T)
(del T) = (74336.4 J) / (2681.9 J/0C)
(del T) = 27.7 0C
If the chamber initially contained 6.07 moles of steam (water) molecules before the ice is added, then the final temperature of the puddle once it settled to equilibrium which will be given as :
[(641 g) + (109.2 g)] (del T) = (641 g) (100 0C) + (109.2 g) (27.7 0C)
(del T) = (64100 + 3024.8) g 0C / (750.2 g)
(del T) = 89.4 0C
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