Answer:
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
