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Thermodynamic efficiency can be expressed as the ratio of Work done(W) to Energy invested (Q).

Thermodynamic efficiency= W/Q

How can one measure work done by a biological system?

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    $\begingroup$ Shouldn't this be W/Q? If Q/W then "thermodynamic efficiency" is very high when you do little work but put in a lot of energy. Doesn't seem very efficient. $\endgroup$
    – Conner
    Commented Aug 16, 2012 at 20:38

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Measuring the work done by a biological system seems pretty impossible. Imagine how many different ways one cell of your body uses energy (ATP). You can't really measure all the work done by every cell on a macro scale. Metabolic efficiency has been defined as... "health". That seems just a little ambiguous. That's why we use things like averages to determine if energy use is normal or not, such as in metabolic age.

In short, work is a more tangible term in discrete physics examples, but there is so much complicated energy use in biological systems that total systemic work can't be easily defined.

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Just like with work in non biological systems, one needs to consider the path that the system takes. The definition of work for a process depends on a path. As well, different types of potential energy can be equilibrated for different systems.

So what's your system? Let's say a plant is in a sealed system, with a movable piston. It starts with a small amount of CO2 gas and a lot dissolved in solution. Then the plant is placed in a light source, which can enter the closed system and drive photosynthesis. As the plant makes oxygen it comes out of solution and increases the volume/pressure of gas in the system. As a result, the photosynthesis could do work against the piston. Then, since the energy inputted from the light source can also be readily estimated, the thermodynamic efficiency of "driving a cylinder using photosynthetically derived oxygen gas."

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