The answers are 12) A. 13) D. Please solve these step by step and show how to ge

Question

The answers are 12) A. 13) D.

Please solve these step by step and show how to get the right answers.

Thank you.

For the next two questions, consider the case where I have 70.0 mol of ice at 273 K and 52.0 mol of water at 373 K and put them together inside a perfectly adiabatic container at a constant pressure of I atm. You may assume that steam (gas), water (liquid) and ice (solid) have a molar mass of 18.0 g/mol, have molar isobarie heat capacities of 36.0, 75.0 & 34.0 J mol .K, respectively, and that the standard molar enthalpy of fusion of water at 273 K and 1 atm is 6.00 kJ/mol, while the standard molar enthalpy of vaporization at 373 K and 1 atm is 40.0 kJ/mol. You may also assume that all heat capacities are independent of temperature. 12. What is the final temperature in K at equilibrium? (B) 278 (G) 345 13. What is the change in the entropy of the universe in J/K? (B) 195 (G) 321 A) 273 (C) 285 (D) 318 (E) 323 (F) 334 (H) 357 (I) 370 (J) 373 (A) 187 (C) 203 (B) 211 (E) 219 (F) 274 (H) 368 (1) 415 (J) 462

Explanation / Answer

Heat lost of water = heat gained by ice

Let final tempertaure = T

Heat lost by water = 52 mol x 75 J/mol K x (100-T)

Ice is already at 0 C ,Here ice will absord heat and then phase will change to water ,later some water absorbs the remaining heat.

Enthalpy of fusion of ice = 70 mol x 6 kJ/mol = 420 kJ

This is the minimum heat required for all ice to convert to water.

Now check the maximum amount of heat will water looses.

Max amount of heat water looses = 50 mol x 75 J/mol K x (100-0) = 375 kJ

This energy is less than required energy by ice at 0  oC.

So, phase change will not happen. Ice will remain at 0  oC only that is 273 K

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