posted by Shawn on .
A 2.00-kg block of ice is at STP (0¢XC, 1 atm) while it melts completely to water. What is its change in entropy? (For ice, Lf = 3.34 „e 105 J/kg)
The entropy change, which is an increase in this case, equals the heat of fusion of water divided by the absolute temperature, 273 K.
I do not understand your
¢X and „e symbols. You should precede exponents with a ^
The c with the line through it is a degree symbol, and the e is a ^ symble. I am sorry the computer put it in that way.
The answer is 2450 J/K but I can't figure out how to get that.
You have to multiply by the number of kg also. (heat of fusion)/T gives you the entropy per unit mass
3.34*10^5 J/kg * 2.00 kg /273 K = 2450 J/K, to three significant figures