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Let's say I have:

1: one mole of extremely cold ideal gas
2: unlimited amount of ideal gas at temperature 300 K
3: one ideal heat engine

Can I generate for example 1 MWh of mechanical energy using those three things?

Alternative formulation: When temperature of the cold gas approaches 0 K, what does the amount of generated energy approach?

(there is no other heat sink than one mole of cold ideal gas)

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2 Answers

Edit: I just saw John Rennie post his answer, but he seems to assume that all the heat flow $\delta Q$, which comes from the hot bath, translates to the $\delta Q$, which eventually heats up the cold gas via $\delta Q=C dT$. I'm interested in clearing that up.

enter image description here

I think this is the way to go:

We have the efficiency $$\eta:=\frac{\delta W}{\delta Q_{\text{bath}}}=1-\frac{T_{\text{low}}}{T_{bath}},$$ and $$\delta Q_{\text{low}}=C\ dT,$$ and of course $$dU=\delta Q_{\text{bath}}-(\delta Q_{\text{low}}+\delta W)=0.$$ $$\Longrightarrow \left(\frac{\delta Q_{\text{bath}}}{\delta W}-1\right)\delta W=\delta Q_{\text{low}}$$ $$\Longrightarrow \delta W=C\ dT\times \left(\frac{T_{\text{bath}}}{T_{\text{low}}}-1\right)$$

If you use your machine it open up the heat channel and translates $\delta Q_{\text{bath}}$ to work as long as the initially cold gas hasn't warmed up to bath temperature. In case you start out with $T_{\text{low}}=T_{\text{bath}}$, then nothing will happen.

$$\Delta W=\int_{\epsilon>0}^{T_{\text{bath}}}C\ T_{bath}\left(\frac{1}{T}-\frac{1}{T_{bath}}\right)dT=C\ T_{\text{bath}}\times \left(\text{log}\left({\frac{T_{\text{bath}}}{\epsilon}}\right)-\frac{T_{\text{bath}}-\epsilon}{T_{\text{bath}}}\right).$$

Here the first factor $C\ T_{\text{bath}}$ would be the energy from the direct heating of the cold gas.

Edit: To point that out more clearly, the term $-C\ T_{\text{bath}}\ \text{log}(\epsilon)$ in this ideal model is arbitrary big. Whenever you half the small minimal temperature, you get another constant amount of work $W$ out of the machine.

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Oops yes, that's what comes of doing things in a hurry. If your mole of gas was really at absolute zero (impossible of course) all the heat flow will go into work and you'd get an infinite amout of work out. I'll hurriedly delete my answer before someone downvotes it! –  John Rennie Jan 27 '12 at 16:21
    
but you have not given numbers. My instinctive reply would be that no, you cannot get 1MWh out of this set up.1 Megawatt is 3.6*10^9 joule/h unitconversion.org/power/…;. –  anna v Jan 27 '12 at 16:44
    
@anna v: If I didn't make a calculation mistake, then the expression says that depending on $C$ and $\epsilon$, you can get any amount of work from an infinite heat reservoir this way. The rate $dT=\frac{dQ_{\text{low}}}{C}=\frac{\delta Q_{\text{bath}}-\delta W}{C}$ means that for very big $C$, the cold gas will take forever to warm up, while for very small $T_{\text{low}}$, the efficiency of converting $Q_{\text{bath}}$ to work is quasi $1:1$. In any case, in practice this will never work, since $C$, or rather $C(T)$ and also $\eta(T)$ are more complicated than the expressions for ideal gas. –  NiftyKitty95 Jan 27 '12 at 17:17
    
@anna v: The main point being that the third law of thermodynamics says something about $C(T)$ at low temperatures, namely that it should be very very small and certainly not constant. –  NiftyKitty95 Jan 27 '12 at 17:21
    
but,but,but, the question says that there is only one mole of the extremely cold gas, that will reach the temperature of the bath and that will be it. –  anna v Jan 27 '12 at 19:53
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(Nick Kidman was a bit faster but I post my coffee break answer anyway)

I doubt that you can extract 1 MWh from your experiment but let's see: We start with an ideal engine that runs with the Carnot efficiency: $$\eta=1-\frac{T_{cold}}{T_{warm}}$$ So if you have one mole of extremely cold gas the initial efficiency of the ideal engine will be very high, still you will warm up the cold reservoir. For one mole of an ideal gas the specific heat at constant volume is $$C_v = \frac{3}{2}R \approx 12 \frac{J}{K}$$ The energy you can extract from your machine is $$dE = dQ_{warm} \cdot \eta(T_{cold})$$The heat going into your cold gas is $$dQ_{cold} = \frac{T_{cold}}{T_{warm}} dQ_{warm}$$ With $C= dQ/T$ we can calculate the total energy $$E = \int C\left(\frac{T_{warm}}{T_{cold}} - 1 \right) dT_{cold}$$With a mole of very cold gas at 1mK and a room temperature bath at 300K Wolfram Alpha thinks we are able to extract 41kJ out of this machine and as we can see the efficiency drops rapidly when the cold gas warms up: enter image description here

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