Is entropy generation in a process equal to change in entropy of universe?

I read on a source that both quantities are equal ,then why are they specified differently? Can we directly find entropy generation in any process just by calculating the entropy change of universe?

• What do you have in mind for calculating the entropy generation of the universe? What the source means is that the entropy generation in the process is contributing the the change in entropy of the universe. – Chet Miller Dec 22 '18 at 13:45
• So there was a problem which I was solving in which heat transfer to ice from surroundings was known and we had to find ds. Now ds=dq/t+s gen.But what we found was only ds=dq/t and from that ds of universe.So where does this s gen go here as it should have a value as this process is irreversible. – Jatan Amin Dec 22 '18 at 14:01
• Please share your source. – psitae Dec 22 '18 at 14:18
• quora.com/… – Jatan Amin Dec 22 '18 at 14:20
• If you can please provide the exact statement of your ice problem, I will provide the interpretation in terms of entropy generation in the system and surroundings, and total for the universe. But I want to be sure I am answering the precise question you are addressing so that I don't confuse you. – Chet Miller Dec 22 '18 at 15:48

Whenever they talk about a heat reservoir like this, they are implying that it is an "ideal" reservoir which has an infinite capacity to absorb heat or release heat, with no change in its temperature. In addition, there is no entropy generated within the reservoir, and any changes in reservoir entropy are equal to Q/T, where Q is the amount of heat received by the reservoir and T is the reservoir temperature. Furthermore, the temperature at the interface between the reservoir and any system it is brought into contact with is exactly at the reservoir temperature, in this case 373 K. So, in your problem, during the transient change from the initial to the final state of the system (the liquid water), there can be temperature gradients within the system, with lower temperatures inside the liquid water, but with the highest temperature always at the reservoir boundary. These temperature gradients are responsible for heat conduction from the reservoir to the liquid water during this irreversible change. These temperature gradients are responsible for the generation of entropy within the system.

In an irreversible change like this, the first step is to use the first law of thermodynamics to establish the initial and final states of the system and surroundings:

Initial state:

System: 1 kg of liquid water at 273 K

Surroundings: Infinite reservoir before Q is added

Final state

System: 1 kg of liquid water at 373 K

Surroundings: Infinite ideal reservoir after Q = 418.6 kJ added

The next step is to totally forget about the actual irreversible process that was responsible for the change from the initial state to the final state, and instead devise an alternate reversible path for bringing about the exact same change to the system. In doing this, we no longer need to consider that it was the surroundings that was responsible for the change in the system, and can entirely separate the system from the previous surroundings, and use a whole new set of surroundings to bring about the reversible change. In the case of the system, in order to bring it reversibly from its initial state to its final state, rather than putting it into contact with a single reservoir at a single temperature (which would again lead to an irreversible change), we instead bring it into contact with a series of reservoirs running in temperatures from 273 C to 373 in succession. In this way, there is never more than an infinitesimal difference in temperature between the system and its surroundings at any location along the path. This guarantees that the alternate path of the system is reversible. So in this case, we calculate the entropy change for the system as $$dS=\frac{dq}{T}=MC\frac{dT}{T}$$which integrates to $$\Delta S_{sys}=MC\ln{(T_f/T_i)}=(1)(4.186)\ln{(373/273)}=1.31\ kJ/K$$

Next we focus on the entropy change for the surroundings. In this case, since the surroundings is an ideal reservoir and has already experienced a reversible change, we calculate the entropy change simply by dividing the heat that had been received by the reservoir by its constant absolute temperature to determine its entropy change: $$\Delta S_{res}=-\frac{418.6}{373}=-1.12\ kJ/K$$

So the change in entropy of the universe is the sum or the change in entropy of the system and reservoir: $$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{res}=1.31-1.12=0.19\ kJ/K$$

So, in summary, during the irreversible process 1.12 kJ/K of entropy is transferred from the surroundings to the system; and 0.19 kJ/K in entropy is generated within the system. So, the change in entropy of the system is the sum of these, or 1.31 kJ/K.

For a detailed Primer on how to perform calculations of entropy changes and entropy generation in irreversible processes, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/