# Energy density and Carnot cycle

For example, diesel has about 40 MJ/kg, what does it mean? Is that if we burn it we will get 40 MJ of energy per kg, but we can not get all of it to mechanical work?

If that is the case, then why can't we get all of the energy to mechanic like in burning?

## 1 Answer

When you burn diesel the result is heat, but what we actually want is work so we need to find a way to turn heat into work. The theoretical model for this is a heat engine: (I got the image from this page, which is good background reading and I recommend it.)

The idea is that you have a hot source, heated by burning the diesel, and a cold sink, which is usually the environment around us. Heat flows from the hot source to the cold sink and on the way we siphon of some of the heat as work. Carnot's theorem tells us what the maximum possible efficiency of a heat engine is, and it turns out to be:

$$\eta = 1 - \frac{T_c}{T_h}$$

where $T_h$ is the temperature of our hot source and $T_c$ is the temperature of the cold sink. This efficiency is the maximum for a theoretically ideal engine, and all real heat engines are less efficient than this.

Heat engines usually work by heating a gas to increase its pressure, then letting the gas expand to do work. The expansion is how we convert heat to work. The gas then cools and contracts and the cycle starts again. An example of an ideal heat engine is the Carnot engine, that uses the Carnot cycle, and this has the maximum efficiency possible.

Since you mention diesel, the diesel engine uses a thermodynamic process that is a varient of the Otto cycle. The Wikipedia article gives a detailed analysis of the efficiency, which I won't reproduce here.