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Why is Q1 always greater than Q2? And what would be an ideal heat pump (such that it is most efficient; for example an ideal heat engine is one in which all heat is converted to work and not to change the internal energy of the working substance)?

I'm learning highschool level thermodynamics, by the way. So please simplify it a bit.

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Why is Q1 always greater than Q2?

Conservation of energy. Total energy out of the pump $Q_1$ equals the total energy into the pump $W+Q_2$

And what would be an ideal heat pump (such that it is most efficient;

The Carnot cycle heat pump is the most efficient. It has a coefficient of performance (COP) of

$$COP=\frac{Q_1}{W}=\frac{T_1}{(T_{1}-T_{2})}$$

for example an ideal heat engine is one in which all heat is converted to work and not to change the internal energy of the working substance)

You are describing a reversible isothermal process, not a pump operating in a cycle. According to the second law, no heat engine operating in a cycle can convert all input heat to work. Formally it is the Kelvin Planck statement of the second law:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir.

Some heat must be rejected to a lower temperature reservoir. Likewise the Clausius statement of the second law says

No refrigeration or heat pump can operate without a net work input.

That means in your diagram work $W$ can never be zero. In other words, heat can not naturally flow from a cold body to a hot body. Work is required.

So if you do an infinitesimal small amount of work, would Q1 = Q2 + W still hold? So the most efficient refrigerator must do infinitesimal work? Is there a minimum for how much work must be done? –

No heat pump can transfer heat and do an infinitesimal amount of work. A finite amount of work is always required. The amount of work required depends on the coefficient of performance (COP) of the heat pump and the desired heat transfer to the higher temperature environment.

The theoretical minimum amount of work that must be done is for the COP of an ideal (reversible) Carnot heat pump. Rearranging the above equation in terms of the minimum theoretical work required we have

$$W_{min}=\frac{Q_{1}(T_{1}-T_{2})}{T_1}$$

Where the temperatures are in $^0$K

EXAMPLE:

$T_1$=298$^0$K (25$^0$C)

$T_2$=273$^0$K (0$^0$C)

$Q_1$ = 1000 J

$W_{min}$ = 83.9 J

In this example the Carnot COP is 1000/83.9 = 11.9

I'm no expert on heat pumps, but I've read that the typical range of COP for actual air source heat pumps is 3-4. Taking the maximum of the range (4) and using it in our example, the min work required becomes 1000/4 = 250 J. This is greater than the Carnot heat pump because the Carnot is an idealization.

You will need to learn more about the Carnot Cycle in your AP course to understand why the Carnot is greater than the actual efficiencies of heat engines and heat pumps.

Hope this helps.

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  • $\begingroup$ Q1 = W + Q2. So all of Q2 is seen in Q1? Or does that depend on work done on the working substance? If it doesn't, then what is the need for work in the first place? For a more efficient refrigerator, Q1 should be close in value to Q2 right.. so for more efficiency we should have less work done? $\endgroup$ Apr 5, 2020 at 22:45
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Apr 6, 2020 at 1:12
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    $\begingroup$ @Michael Farady I will update my answer later to answer this follow up question. Please note I am on New York time $\endgroup$
    – Bob D
    Apr 6, 2020 at 8:10
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    $\begingroup$ @MichaelFaraday I have updated my answer. Hope it helps. $\endgroup$
    – Bob D
    Apr 6, 2020 at 14:17
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    $\begingroup$ @MichaelFaraday More work is required to OVERCOME the temperature difference in order to force heat to move from low temperature to higher temperature. Note from the equation if you lower T1 the temperature difference is greater and more work is needed to transfer the same heat Q1. $\endgroup$
    – Bob D
    Apr 6, 2020 at 17:33

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