Question

A 192-g aluminum calorimeter contains 606 g of water at 18° C. A 98-g piece of ice cooled to -18° C is placed in the calorimeter. (Assume that the specific heat of ice is always 2.02 kJ/kg · K.)

Answer 1

Q= m x Cp x DT

Heat needed to reach the melting point:

Q= 0.098kg x 2.02 kJ/kg.K x 18 = 3.56328 kJ

Now the fusion energy for the ice:

Q= 98g x 334 J/g = 32732 J= 32.732 kJ

Total energy= 36.29528 kJ

The temperature of the water at 18ºC will decrease:

36.29528= 606g x 4.186 J/gºC x DT x1kJ/1000J -----> DT= 14.31 ºC

Final temperature of the hot water= 18-14.31 = 3.69 ºC

The cool water need to increase the temperature until the final temperature, and the hot water will low it´s temperature a little bit:

606g x 4.186 J/gºC x (3.69 -Tf) = 98g x 4.186 J/gºC x (Tf-0)

Tf= 3.18 ºC