A 22.5-g sample of ice at - 10.0°C is mixed with 110.0 g water at 85.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities for H₂O (l) and H₂O (s) are 4.18 J/g･°C and 2.03 J/g･°C, respectively, and the enthalpy of fusion of ice is 6.02 kJ/mol.

The answer to the question is

56.166 °C

Explanation:

To solve the question, we list out the varibles as follws

Mass of ice, mi = 22.5 g

Mass of warer, mw = 110 g

Initial temperature of the ice = - 10.0 °C

Initial temperature of the water = 85.0°C

Specific heat capacity of ice, ci = 2.03 J/g･°C

Specific heat capacity of water, cw = 4.18 J/g･°C

Enthalpy of fusion of ice = 6.02 kJ/mol.

From the first law of thermodynamics we have

Heat lost by the water = heat gained by the ice

mi*ci*ΔTi + mi*L = mw*cw*ΔTw

22.5 g * 2.03 J/g･°C * (0 - (-10)) °C + 22.5 g / (18.01528 g/mol) * 6.02 kJ/mol * 1000 + 22.5 g * 4.18 J/g･°C * T = 110.0 g * 4.18 J/g･°C * (85.0°C - T)

7975.4 + 94.05 T = 39083 - 459.8T

Making T the subject of the formula we have

553.85 T = 31107.6 or T = 56.166 °C

The final temperature of the mixture = 56.166 °C