Question

A 20.0-g sample of ice at 210.08C is mixed with 100.0 g water at 80.08C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H 2 O(s) and H 2 O(l) are 2.03 and 4.18 J/g ? 8C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

Answer 1

Mass of ice m1 = 20 g

Temperature T1 = - 10 C

Melting point of Ice Tm = 0 C

Mass of water m2 = 100 g

Temperature T2 = 80 C

Heat capacity of ice Cps = 2.03 J/gC

Heat capacity of water Cpw = 4.18 J/gC

H fusion =( 6.02 kJ/mol )x (1mol/18g) = 334 J/g

Heat given by water = heat absorbed by ice

Heat given by water = heat absorbed by ice to bring the temperature to 0 C + heat absorbed to change the ice phase to liquid + heat absorbed to change the temperature from 0 C to T c

m2 Cpw ( T2 - T) = m1 x Cps ( Tm - T1) + m1 x H fusion + m1 x Cpw x ( T - Tm)

100 x 4.18 x (80- T) = 20 x 2.03 x ( 0 + 10) + 20 x 334 + 20 x 4.18 x (T - 0)

418 (80 - T) = 7086 + 83.6 T

33440 - 418 T = 7086 + 83.6 T

26354 = 501.6 T

T = 52.54 C