Sum of Temperatures in Heating??

Question

Two 20.0 g ice cubes at -20.0 C are placed into 265 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

I understand how to perform enthalpy calculations for Qp, but do not understand how to covert that to sum of T

Answer 1

heat lost by water = heat gained by ice. We will need the specific heat for ice and the ∆H_fusion for water.

I'll use the following values (you may find slightly different values)

C_ice = 2.09 J/g/deg; ∆H_fusion = 334 J/g; C_water = 4.184 J/g/deg

heat gained by ice = heat to raise 40 g from -20º to 0º + heat to melt 40 g ice at 0º + heat to raise temp of 40 g liquid water to final temperature

q = mC∆T = m∆H_fusion = (40.0 g)(2.09 J/g/deg)(20 deg) + (40.0g)(334 J/g) = 1672 J + 13360 J = 15,032 J +

q = mC∆T = (40.0 g)(4.184 J/g/deg)(Tf - 0) This is the heat to raise the temperature of the newly formed liquid

from the 40 g of melted ice.

So, in summary, for the ice, we have 15,030 J + (40.0)(4.184 J/g/deg)(Tf - 0) = heat lost by water

heat lost by water = mC∆T = (265 g)(4.184 J/g/deg)(Tf - 25.0)

Now we can equate the two processes as follows:

15,032 J + [(40.0g)(4.184 J/g/deg)(Tf-0) = -(265g)(4.184J/g/deg)(Tf-25.0)

15,030 J + 167 Tf = - 1109Tf + 27729

1276Tf = 12699

Tf = 9.95º

(Best that you check the math, but this is the approach that one should use)