# Entropy change and reversible/irreversible processes

## Homework Statement

A mass m is thrown from a height of h. Find the entropy change of the mass, the surroundings and the universe. The temperature of the surroundings is T.

dS=dQ/T
ΔU=Q+W

## The Attempt at a Solution

The main thing I don't understand in this question is how we even can apply dS=dQ/T to the surroundings and the mass since all processes are irreversible. I know that entropy is a state function and you can choose a reversible process between two states and get the same entropy change as in a irreversible process, provided that all quantities we use to calculate the entropy change are state functions. Take the free expansion as an example, we can't use dS=dQ/T to calculate the entropy change since it's a irreversible process and Q isn't a state function, we have to use the central equation which only dependes on state functions. From the solution to this problem, which I've seen, it seems that dS=dQ/T was used. The entropy change of the surroundings is mgh/T, but the first law gives us Q=0 since the change in internal energy and the work the ground does is the same. I don't understand how both are true. The entropy change of the mass is zero, which I can understand if dS=dQ/T can be applied somehow.

• 