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

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