Two 20.0-g ice cubes at –13.0 °c are placed into 255 g of water at 25.0 °c. assuming no energy is transferred to or from the surroundings, calculate the final temperature, tf, of the water after all the ice melts.

Let's apply the concept of conservation of energy.

∑E = 0
Energy released by ice + Energy absorbed by water = 0
Energy released by ice = - Energy absorbed by water 
(mass of alloy)(Cp,ice(Tf - T₀) = -(mass of water)(Cp,water)(Tf - T₀)
where Cp is the specific heat capacity
*Cp for ice is 2.108 J/g·°C
*Cp for water is 4.184 J/g·°C

Substituting the values,
(2×20 g)(2.108 J/g·°C)(Tf - -13) = -(255 g)(4.184 J/g·°C)(Tf - 25)
Solving for Tf,
Tf = 22.22 °C

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