In: Chemistry
If 30g of water at 10C is added to 180g of ice at -110C in an insulated cup, what is the final temperature when equilbrium is reached? How much liquid is in the cup when equilibrium is reached? What is the total entropy change for the reaction?
Note: dH fusion for water = 6.025kJ/mol at 0C, Cp,m for H2O(l) = 75.48 and Cp,m for H2O(s) = 37.68
Molar heat capacity = Specific heat x Molar mass.
Specific heat of water = 75.48 / 18 = 4.19
Specific heat of ice = 37.68 / 18 = 2.09
= 30 s (10 - t) (water) = 180 s (t+110) (ice)
= 30 x 4.19 ( 10 - t ) = 180 x 2.09 ( t + 110)
= 125.7 ( 10 - t ) = 376.2 ( t + 110 )
= 1257 - 125.7 t = 376.2 t + 41383
= 1257 - 41383 = 376.2 t + 125.7 t
= - 40125 = 501.9 t
= t = - 79.9 C.
= 30 g of water would have changed into ice at equilibrium temperature. A part of ice would have changed into water, as temperature of the ice has increased to - 79.9 C.
At - 110 C ------- There is 180 g of ice
At - 79.9 C ------- There is (180 x -79.9) / -110 = 130 g of ice.
Total entropy change of the reaction:
Change in entropy = dH fusion / T
( In the above situation ice is melting, hence its a fusion reaction)
= Here T is the absolute temperature at which ice melts = 273 K
= Change in entropy :
=dH fusion = 6.025 kJ/mol = 6025 J/mol.
Change in entropy = 6025 / 273 = 22.1 J/K/mol.