5) The heating curve of H2O is shown in the figure below. Use the information pr

Question

5) The heating curve of H2O is shown in the figure below. Use the information provided to answer the following questions. Cp mater 75.3 J/mol.C CR vapor-33.6 J/mol."C As 6.01 kJ/mol AHo6.01 kJ/mol AHe 40.65 kJ/mol AHcond 40.65 kJ/mol 100.0 0.0 Energy added (kJ) a) Calculate the total heat (in kJ) required to raise the temperature of 50.0 g H:O from -15.8 C to 3.5 °C (6 pts). a) Calculate the total heat removed (in kd) when decreasing the temperature of 75.0 g H-O b) Discuss why it requires more energy to change Hy0 from a liquid to gas than a solid to liquid (3 pts).

Explanation / Answer

5) From the give curve

a) heat required to raise 50 g H2O from -15.8 oC to +3.5 oC

step 1 : H2O(s) at -15.8 oC to 0 oC

q1 = mCpdT = 50 x (37.1/18) x (0 + 15.8)/1000 = 1.6283 kJ

step 2 : H2O(s) to H2O(l) at 0 oC

q2 = mCpdT = 50 x (6.01/18) = 16.6944 kJ

step 3 : H2O(l) at 0 oC to +3.5 oC

q3 = mCpdT = 50 x (75.3/18) x (3.5 - 0)/1000 = 0.7321 kJ

Total heat required = q1 + q2 + q3 = 19.055 kJ

b) Heat to removed from 75 g H2O at 115 oC to 48 oC

step 1 : H2O(g) at 115 oC to 100 oC

q1 = 75 x (33.6/18) x (100 - 115)/1000 = -2.100 kJ

step 2 : H2O(g) to H2O(l) at 100 oC

q2 = 75 x (-40.65/18) = -169.375 kJ

step 3 : H2O(l) at 100 oC to 48 oC

q3 = 75 x (75.3/18) x (48 - 100)/1000 = -16.315 kJ

Total heat to be removed = q1 + q2 + q3 = -187.790 kJ

c) From solid to liquid the amount of intermoelcular hydrogen bonds to be broken are less than the amount of intermolecular hydrogen bonds to be broken to bring the H2O from liquid to vapor state, therefore, more heat is needed to change liquid to vapor than solid to liquid state.

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