5 The Second Law and Entropy Free Online FlashCards

What does the second law of thermodynamics add to the first law?

The second law determines which energy transformations can occur naturally and limits how much heat can be converted into useful work.

Which way does heat flow spontaneously?

Heat flows spontaneously from a warmer object to a cooler one. Reversing that transfer requires work, as in a refrigerator.

What happens during a heat-engine cycle?

A heat engine absorbs heat from a hot reservoir, produces work, and rejects some heat to a colder reservoir.

How do a refrigerator and a heat pump differ in focus?

A refrigerator focuses on removing heat from a cold region; a heat pump uses the same kind of cycle but focuses on delivering heat to a warm region.

Why can a refrigerator’s COP exceed 1?

A COP can exceed 1 because it measures heat moved per unit of work supplied; the device moves heat rather than converting work entirely into heat.

Can a system’s entropy decrease?

Yes. A system’s entropy may decrease if entropy increases elsewhere by at least as much, leaving the total entropy change nonnegative.

How does entropy generation affect energy’s usefulness?

Irreversible processes generate entropy, making energy transferred as heat less fully available for conversion into useful work. Energy is still conserved.

What is a heat engine’s net work per cycle?

The net work is the heat absorbed from the hot reservoir minus the heat rejected to the cold reservoir: $W_{out}=Q_H-Q_C$.

How is a heat engine’s thermal efficiency calculated?

Thermal efficiency is the fraction of input heat converted to work: $\eta=\frac{W_{out}}{Q_H}=1-\frac{Q_C}{Q_H}$.

How are heat-pump and refrigerator COPs related for the same cycle?

For the same cycle, the heat-pump COP is one greater than the refrigerator COP: $\mathrm{COP}_{HP}=\mathrm{COP}_R+1$.

How is entropy change calculated using a reversible path?

Entropy change between two states can be calculated along a reversible path: $\Delta S=\int_{\text{initial}}^{\text{final}}\frac{\delta Q_{\mathrm{rev}}}{T}$, where $T$ is absolute temperature in kelvins.

What sets the maximum efficiency of an engine between two reservoirs?

The Carnot limit is $\eta_{max}=1-\frac{T_C}{T_H}$. A real engine between the same reservoirs cannot exceed it; temperatures must be in kelvins.