Question

i got this question wrong. the book previously said that the units of entropy are J/(mol*K). you can get mol with the grams that they gave, so why not use it?

Example: If, in a reversible process, 6.66 x 10 J of heat is used to change a 200 g block of ice to water at a temperature of 273 K, what is the change in the entropy of the system? (Note: The heat of fusion of ice = 333 J .) Solution: We know that during a phase change, the temperature is constant; in this case, 273 K. From the information given, S0.C 6.66X10° 244 Qrev AS= J K 273 K The amount of heat was exactly enough to completely melt the block of ice without changing the temperature of the resulting liquid water; therefore, Tis constant.

Answer 1

Solution: The phase change occur at constant temperature. Here change solid (ice) to liquid (water) occur at constant temperature 273K. The change in entropy As=dQ 200gx 333J/g V T 273K 244J/K is the correct answer. In certain conditions the other unit of entropy is J/(molx K) . First see the relation between mole and gram. If M is the molecular weight of gas/substance then its mass will be uM, where uis the mass of the gas/ substance in gram-molecules or mole wish to calculate the entropy For entropy calculations, we use mole as follows: Suppose change of the system due to the irreversible free expansion from an initial volume V to final we volume V. In free expansion the temperature remains constant let it be T Now from the first law of thermodynamics, T The entropy change AS S -S we have dV But dU 0 as there is no change dQ dU+dW dU+ pdV= dU +uRT in internal energy dV x uRT V dV 1 rev = uRlog. V at constant temperature. Thus we have, AS = T C directly. Let dQ be the heat Another Example: When we use molar specific heat such as V to a substance (silver) to smaller rise in temperature dT. If C, be the molar specific heat supplied at temperature T then we have dQ C,dT CodT J/molex K since the unit of C, is J/molexK T Hence AS 1S