Further explanation of entropy using language descriptions and more clarifications

Muzzie’s advice: You always need a change in temperature to have a change of heat, but not vice versa.

You can still have an entropy with an isothermal (constant temperature) process. You just put S= (∆Heat)/T, where T is constant. You might think this is a contradiction, and it actually is. However, what happens if you drop an ice cube into the water off the coast of Brazil? Well, the ice cube will melt, and its temperature will change causing a heat gain. However, the ocean also produces a change of heat. The ocean loses heat. Therefore the ocean must undergo a change in temperature to cause the change in heat. However, the change is so small you say it has the same temperature before and after. Trick question.

When the temperature is actually constant (oceans excluded), the heat must be constant. Okay, but this is only for an open system. You can actually keep the same temperature and produce heat if the state of matter of the object changes by volume and pressure changes in a (closed system). You must be careful to apply and remove heat from a system while varying the volume and pressure in the correct amounts. But does the temperature still change by a negligible amount here? According to the math it doesn’t.

ΔU=0=Q-W

The above is for an isothermal process. The total heat produced must equal the amount of work produced. This is a completely efficient process that is in essence impossible, even in the highest temperature nuclear power process. So, yes, you still have a temperature change, but a very very small one.

Finding the work gives: Q= W=nRTlnVfVi

Since the temperature is a constant, we can apply the ideal gas law PV=nRT and find the initial and final pressures of the process. So in a cylinder with a piston with no heat loss you can have a theoretical equation with constant temperature.