A 0.6850-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.6850-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 6.430 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

C of H20 (s): 37.7 J/(mol K)
C of H20 (l): 75.3 J/(mol K)
enthalpy of fusion of H20: 6.01 kJ/mol

27.8 J/(mol K) at 400 °C or 2.02J g¯1K¯1.

calculate how much energy must be absorbed to melt the ice cubes:

here mass of ice = 0.6850 kg = 685 g

685 g/18 g/ mol= 38.1 mole

mass of steam = 6.430 moles *18.0 g/ mole= 115.74 g

(38.1-mol x 12.40 °C x 37.7-J/mol*°C) + (38.1-mol x 6010-J/mol) + [38.1-mol x (Tf - 0 °C) x 75.3-J/mol*°C] = 6.430-mol (365 °C - Tf) x 27.8-J/mol*°C

17810.988 +228981 + 2868.93 Tf = 35245.21 – 178.754 Tf

211576.734 = 2690.176 Tf

Tf = 78.65 degree

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