A mixture 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?

Respuesta :

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

Answer: 25.4 °C