Amixture containing 21.4g of ice (at exactly 0 c) and 75.3g of water (at 55.3 c) is placed in an insulated container. assuming no heat is lost to the surroundings, what is the final temperature of the mixture?

Energy conservation law => heat lost by the hot water = heat gained by ice

At the equilibrium the temperature of mixture is T.

heat lost by the hot water  mCsΔT = 75.3g * 1 cal /(g°C) *(55.3°C - T)

heat gained by the ice = latent fusion heat at 0°C + heat absorbed to rise temperature from 0°C to the equilibrium temperature of the mixture, T

latent fusion heat of ice per gram, at 0°C = 79.7 cal/g

latent fusion heat of ice at 0°C = 21.4g *79.7 cal/g

Heat absorbed to rise the temperature until thermal equilibrium:mCsΔT = 21.4g*1cal(g°C)*(T-0°C)

Total heat gained by ice = total heat lost by the hot water

21.4g * 79.7 cal/g + 21.4g*1cal/(g°C)*(T) = 75.3g * 1 cal /(g°C) *(55.3°C - T)

1705.58 + 21.4 T = 4164.09 - 75.3T

21.4T + 75.3T = 4164.09 - 1705.58

96.7T = 2458.51

T = 2458.51 / 96.7

T =  25.42  °C

25.4 °C

 

Solubility would be your answer.