Question

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.

Answer 1

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