1. Feb 23, 2004

### jlmac2001

I need help finishing this problem. I did part A and B but don’t know how to do C and D. I don’t know if A and B are correct.

Problem: An ice cube (mass 30g) at 0 degrees Celsius left sitting on the kitchen table where it gradually melts. The temperature in the kitchen is 25 degrees Celsius.

A. Calculate the change in the entropy of the ice cube as it melts into water at 0 degrees Celsius. (Don’t worry about the fact that the volume changes somewhat).

C=Q/delta T Q= 30 cal/K or 126 J/K so C = 126 J/K / 0 degrees Celsius = 0 so delta S= 0 (this is the entropy)

B. Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from 0 degrees Celsius to 25 degrees Celsius.

Delta S = Sf -Si= (integral from Ti to Tf) Cv/T dT= 126 J/K (integral from 273K to 298K) 1/T dT = (126J/K) ln (298/273) = 11.0 J/K

C. Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water. (How would I do this one?)

D. Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative or zero? Is this what you would expect? (Need help with this one also, Please!)

2. Feb 23, 2004