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What is the difference of work $W$ and thermal energy $Q$ in thermodynamic Stirling-process (in simple form) for ideal gas?

I think that you need work to preserve this process and you bring thermal energy so that temperature $T$ rises in the process.

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Do you mean _heat $Q$_ instead of "thermal energy $Q$"? –  juanrga Dec 2 '12 at 12:23
    
I think they are similar. –  laovultai Dec 2 '12 at 12:39
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Both are different concepts. Thermal energy is usually represented as $U_\mathrm{therm}$ and $U_\mathrm{therm} \neq Q$ –  juanrga Dec 2 '12 at 13:12
    
Well maybe you can explain what are formulas for $U_{therm}$ and $Q$? Are they $U_{therm}=\Delta W$ and $Q=c m \Delta T$, where $c$ is specific heat capacity, $m$ mass of the processed object and $\Delta T$ difference of temperature? –  laovultai Dec 2 '12 at 18:00
    
The expression for thermal energy depends; for an ideal gas $U_\mathrm{therm} = (3/2) N k_\mathrm{B} T$. Work is not a state function and $\Delta W$ is meaningless. The expression for the heat $Q=C\Delta T$ is valid for special processes (e.g. constant volume or constant pressure) when the heat capacity $C=cm$ does not vary with temperature. –  juanrga Dec 4 '12 at 9:28

1 Answer 1

In a cycle you usually deal with powers $\dot L $ and $\dot Q$ both being given in $W$ (Watt) What you want from a cycle is to obtain a power source $\dot L$. In order to obtain it, you have to degrade some heat power from an hot source to a colder source. The power you can obtain from a thermodynamic cycle, applying power balance, is: $$\dot L=\dot Q_h -\dot Q_c$$ where h stands for hot and c for cold The efficiency is defined as the power you can obtain divided for the thermal flux input: $$\eta = \frac{\dot L}{\dot Q_h}$$ Replacing $\dot L$ from the first equation you get $$\eta=1-\frac{\dot Q_c}{\dot Q_h}$$ So, considering a reversible process, for which the thermal energy can be written as $Q=T\delta s$ you get the maximum efficiency (Carnot efficiency): $$\eta _{rev} =1-\frac{T_c}{T_h}$$ In the end you're right, in order to keep the same efficiency you have to keep the same temperature, but what's important is that you have to give a thermal flux to obtain a work flux.

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