Why is the change in entropy of a heat pump in one cycle equal to zero?

[yecko]

Homework Statement

A closed-cycle heat pump is used to move heat from a low-temp reservoir to a high-temp reservoir with fixed temperatureTH= 450 K andTC= 320 K. If the coefficient of performance of the heat pump is less thanthat of an ideal heat pump, which of the following statements is true regarding the change in the entropies? (see image attached)

Relevant Equations

S always >= 0, where S=0 only in ideal case.
for hot reservoir, delta S = negative
for cold reservoir, deta S = positive

as it is not ideal,
total change of entropy > 0
entropy loss of hot reservoir > entropy gain of cold reservoir

why would the change in entropy of the heat pump in one cycle equals to zero?
thank you

 

[Chestermiller]

entropy loss of hot reservoir > entropy gain of cold reservoir

The hot reservoir does not experience an entropy loss. It receives heat in this process, so its entropy increases. And the cold reservoir does no experience an entropy gain. It loses heat in this process, so its entropy decreases.

 

[yecko]

so the entropy loss of hot reservoir < entropy gain of cold reservoir?
but why the answer is B?
also, why entropy of the heat pump unchange?
thank you

 

[Chestermiller]

so the entropy loss of hot reservoir < entropy gain of cold reservoir?
but why the answer is B?
also, why entropy of the heat pump unchange?
thank you

No. The entropy gain of the hot reservoir is greater than the entropy loss of the cold reservoir.

The entropy of the heat pump (i.e., the working fluid) is zero for a cycle because entropy is a physical property of the working fluid (i.e., the state of the working fluid), and, in a cycle, the working fluid returns to its original state after each cycle.

 

[yecko]

No. The entropy gain of the hot reservoir is greater than the entropy loss of the cold reservoir.

why would that so?
isn't when it is not ideal, the energy loss is always larger than energy gain?

 

[Chestermiller]

With a heat pump, you are removing heat from a cold reservoir and adding a greater amount of heat to a hot reservoir. That is what a heat pump does.

 

[Delta2]

No. The entropy gain of the hot reservoir is greater than the entropy loss of the cold reservoir.

The entropy of the heat pump (i.e., the working fluid) is zero for a cycle because entropy is a physical property of the working fluid (i.e., the state of the working fluid), and, in a cycle, the working fluid returns to its original state after each cycle.

What about the hot and cold tanks, don't they return to their original state after each cycle, hence their change of entropy is also zero?

 

[Chestermiller]

What about the hot and cold tanks, don't they return to their original state after each cycle, hence their change of entropy is also zero?

No. They don’t experience a cycle. Typically, you are taking heat from outside air (the cold reservoir), adding work, and pumping heat into the house air (the hot reservoir).

 

[yecko]

No. The entropy gain of the hot reservoir is greater than the entropy loss of the cold reservoir.

But why for the magnitude of entropy change is greater for hot reservoir than cold reservoir?

 

[Chestermiller]

But why for the magnitude of entropy change is greater for hot reservoir than cold reservoir?

The sum of the two is greater than zero.

 

[yecko]

why not magnitude of entropy change is greater for cold reservoir than hot reservoir?

 

[Chestermiller]

why not magnitude of entropy change is greater for cold reservoir than hot reservoir?

Like I said, the sum of the changes has to be greater than zero.

 

[Delta2]

@Chestermiller can you restate your opinion at post #8 cause the way it is now I can't make sense, one of the reservoirs has to be the cold one, and we need to have mechanical work in the system if we going to transfer heat from the cold reservoir to the hot reservoir.

 

[Chestermiller]

@Chestermiller can you restate your opinion at post #8 cause the way it is now I can't make sense, one of the reservoirs has to be the cold one, and we need to have mechanical work in the system if we going to transfer heat from the cold reservoir to the hot reservoir.

I don't understand what you are asking.

 

[Delta2]

I don't understand what you are asking.

nvm I figure it out myself, apparently you mean that the outside air is the cold reservoir.

 

[yecko]

So do you mean with 2nd law of thermodynamics, the entropy gain always greater than entropy loss, no matter it is heat engine (hot reservoir gain>cold reservoir loss) or heat pump (hot reservoir loss<cold reservoir gain)?

 

[Chestermiller]

So do you mean with 2nd law of thermodynamics, the entropy gain always greater than entropy loss, no matter it is heat engine (hot reservoir gain>cold reservoir loss) or heat pump (hot reservoir loss<cold reservoir gain)?

Yes, for an irreversible engine, there is overall entropy generation.

 

[vela]

I don't understand what you are asking.

In post #8, you accidentally referred to the hot reservoir twice and made no mention of a cold reservoir.

 

[Chestermiller]

In post #8, you accidentally referred to the hot reservoir twice and made no mention of a cold reservoir.

Thanks. No wonder everyone was so confused. I have corrected it.