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

2 Answers 2


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.


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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