[adianadiadi]

Recently I got a question from my online student that: How the energy of Universe is constant if the entropy of it increases?

My answer is: Entropy in one way refers to the heat amount that is not available for useful work. So the amount of useless energy in the form of heat is increasing and energy of the universe is not.

So am I wrong in suggesting entropy is the heat lost to the surroundings and not useful for doing work? Or is there any other good explanation?

[Jorriss]

Qualitatively, you can think of entropy as energy 'spreading' out. As energy is spread out, it can 'go back' to do more work useful work. The energy is constant, it's just 'moved around,' so that it can't be reused.

That's handwavy but hopefully sufficient for the student.

[rabolisk]

To put it another way, as processes happen, the total energy of the universe is constant, but they cannot be reversed. This is equivalent to saying that energy goes from useful, work-producing type to being unusable. The simplest model is mixing hot water and cold water. Eventually they will converge to one uniform temperature, but this process is irreversible. Energy clearly accompanies this process (in the form of heat transferred from hot to cold), but this same energy can't be transferred back to reverse the process. Entropy has increased, although energy is still all within the system of water.

[Juan R.]

Recently I got a question from my online student that: How the energy of Universe is constant if the entropy of it increases?

My answer is: Entropy in one way refers to the heat amount that is not available for useful work. So the amount of useless energy in the form of heat is increasing and energy of the universe is not.

So am I wrong in suggesting entropy is the heat lost to the surroundings and not useful for doing work? Or is there any other good explanation?

I would not use ill-defined concepts as heat but state variables as U, V, N,...

By simplicity suppose that universe is divided into two parts: e.g. system A plus environment B. And consider only changes in energy U, then

dU = dU_A + dU_B = 0

dS = dS_A + dS_B

dS = (1/T_A)dU_A + (1/T_B)dU_B

If the universe is at equilibrium T_A =T_B = T

dS = (1/T)dU_A + (1/T)dU_B = (1/T)dU = 0

If it is not at equilibrium

dS = (1/T_A - 1/T_B) dU_A

which is different from zero until that the system reaches the equilibrium state.

The 'intuitive' reason which a constant energy does not implies a constant entropy is because entropy is energy per kelvin and temperature can vary in isolated systems.