Question 5 and 6 Question 5. 15.0 g of ice at -10 degreeC and 18.8 g of steam at

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Question 5 and 6

Question 5. 15.0 g of ice at -10 degreeC and 18.8 g of steam at 100 degreeC is added to 500 g of H2O(l) at 60 dgereeC. The process is carried out in an isolated (adiabatic) container under a constant pressure of 1 atm. Calculate the final temperature of the system and deltaS for the overall process Is the process spontaneous? The heat capacities and enthalpies are given in the following: CP(H2O(l)) = 75.3 J K^-1 mol^-1 and Cp(H2O(s) = 37.7 J K^-1 mol^-1, deltaHfusion = 6 kJ mol^-1 at 0 degree C, delta Hvaporization = 40 kJ mol^-1 at 100 degreeC . Molecular weight of water is 18 g/mole. Question 6. (Engel P5.29) One mole of H2O (l) is supercooled to -2.25 degreeC at 1 bar pressure. The freezing temperature of water at this pressure is 0.00 degreeC. The transformation H2O (l) H2O (s) is suddenly observed to occur. By calculating deltaS, delta Ssurrounding, delta Stotal, verify that this transformation is spontaneous at -2.25 degreeC. The heat capacities are given by Cp(H2O(l)) = 75.3 J K^-1 mol^-1 and C(H2O(s)) 37.7 J K^-1 mol^-1, and delta Hfusion = 6.008 kJ mol^-1 at 0.00 degreeC. Assume that the surroundings are at -2.25 degreeC. [Hint: consider the two pathways at 1 bar: (a H2O(l, -2.25 degreeC) - > H2O(s,-2.25 degreeC) and (b) H2O(l, -2.25 degreeC) - > H2O(s,0.00 degreeC) - > H2O(s,-2.25 degreeC). Because S is a state function. deltaS must be the same for both pathways.]

Explanation / Answer

When ?G is negative, a process or chemical reaction proceeds spontaneously in the forward direction.

When ?G is positive, the process proceeds spontaneously in reverse.

When ?G is zero, the process is already in equilibrium, with no net change taking place over time.

assuming all ice melts

heat gained by ice + water = heat released by steam

15/18 x 37.7 x 10 + 15/18 x 6000 + 15/18 x 75.3 x T + 500/18 x 75.3 (T - 60) = 18.8 / 18 x 40000

314.17 + 5000+ 62.75T + 2091.7T - 125500 = 41777.8

T = 75.176 degrees

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