If 100 g of Ice at 10 c is placed in 200 g of water at 800 c

If 10.0 g of Ice at -10 c is placed in 200 g of water at 80.0 c in an insulated container, what will be the temperature of the system when equilibrium is established? Â (sp Heat of H2O(s) = 2.09j/g . c (sp heat of H2O (l) = 4.18 J/g.c) Heat of Fus. of H2O(s)=333j/g

Solution

let the final temperature be x

heat absorbed in solid state = 2.09j/g * 10g * (0-(-10))C = 209 joules

heat of fusion = 333 * 10g = 3330joulse

heat absorbed by ice in liquid state = 4.18 j/g * 10g * x C = 41.8x

total heat absorbed by ice = sum of the above 3 heats = 3330 + 209 +41.8x = 3539 +41.8x

heat given by water = 4.18 j/g * 200g * (80-x) C =66880 - 836x

at equilibrium these two heats are equal

therefore 66880 - 836x = 3539 + 41.8x

solving the above equation we get x = 72.15 C

final temperature = 72.15C