Homework 4F.13

[Jackson Crist 1G]

It asks to solve for the entropy change for part a and part b for two different systems. The equation for this is q/T, so we need to plug in the deltaH value for the q and then the temp. This is on table 4C.1. For part a though, they change the deltaH value from table 4C.1 to a negative, but for part b, they keep it positive. Why?

[Francis Le 2K]

Since (a) involves the freezing of water, the value of H would be negative as heat must be removed from the water to freeze it. In Table 4C.1, the value of H is given for fusion (melting), and since the question asks for the change in entropy associated with freezing, you must use the negative of the enthalpy change for fusion to find the enthalpy change for freezing. For part b, they keep H positive because the question asks for the change in entropy from vaporization of ethanol, and the value of H in Table 4C.1 corresponds to vaporization.