Physics: latent heat of fusion

Question

If you want to find the latent heat of fusion, you take a cup with 0.755 kg of water at 30.00 C and drop 0.0850 kg of ice initially at 0.000 C and it all reaches an equilibrium temperature of 19.00 C, what is L_f? (the specific heat of water is 4186 J/(kg C))

Answer 1

Given a final temperature of 19.00 it is clear that all of the ice is melted at equilibrium.

 

The heat removed from the water equals the heat accepted by the ice.  The heat accepted by the ice includes the heat of fusion plus the heat required to raise the ice from 0-19 degrees after melting.

 

0.755 kg * 4186 J/kgC *(30-19)  = Lf*0.0850Kg +  (19-0)*4186*0.0850kg

 

34.76kJ = Lf*0.0850Kg + 6.76KJ

 

Lf = (34.76kJ-6.76KJ)/0.085kg = 330 KJ/Kg = 330 J/g

 

For comparison, the value given in Wikipedia is 334J/g