Two 20.0 g ice cubes at −11.0 ∘C are placed into 205 g of water at 25.0 ∘C.

Question

Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.

heat capacity of H2O(s) 37.7 J/(mol⋅K)

heat capacity of H2O(l) 75.3 J/(mol⋅K)

enthalpy of fusion of H2O. 6.01 kJ/mol

Answer 1

since no energy is gained/lost to the surroudings we know that all the heat gained by the ice is going to be equal to the heat lost by the water. The ice is going to gain heat in three processes; one process will involve the heat gained to go from -11 to 0, the second process will be the heat gained to melt the ice, and the third process will be the heat gained to go from 0C to the final temp. The heat lost by the water will just be one process because there is no phase change, it will just be the heat lost from going from the initial temp (25) to the final temp. We also know that the final temperature of the ice and water will be the same

q(ice to 0 C) + q(ice melting) + q(water at 0C to final temp)= q(water from 25 C to final temp))

the heat gained by the ice for the temperature change of -11 to 0 C :

q=mcdT

m=20g= 1.11mol. c= 37.7 dT= 0-(-11)=11

q= 1.11(37.7)(11)= 46.027 J= 0.046027 kJ

the heat gained by the ice melting:

q=nHf

n=moles= 1.11 Hf= 6.01 kJ

q=1.11(6.01) q=6.67 kJ

heat gained to go from 0C to final temp

q=1.11(75.3)(Tf-0)

all of these three q's will be equal to the heat lost by the water:

-q=mcdT

m=205g= 11.376 mol c= 75.3 dT= (Tf-25)

-q=11.376 (75.3)(Tf-25)

putting all of these equations together we have:

46.027 J + 6670kJ + (1.11)(75.3)(Tf)= -(11.376)(75.3)(Tf-25)

solving for Tf gives you 15.63C