In my physics course we have recently started covering thermodynamics.
It’s quite overwhelming but at the same time it’s really fun. So I’ll just post and explain everything I learn in class here - since it’s a good way of learning, explaining. As the late and incredible Richard Feynman once said,
If you cannot explain something in simple terms, you don’t understand it - Richard Feynman
With that said, into thermodynamics!
Thermodynamics and Theory of Heat
Thermal Efficiency
Let’s start with something which is quite simple to understand. The famous quote that,
“Energy is neither created nor destroyed, only changed from one form to another”.
With this in mind we understand that we never ‘create’ energy. It’s also a sensible conclusion that if we had machines with 100% efficiency, we wouldn’t have whole fields in physics about energy-efficiency.
Informal definition of thermal efficiency
(We will later define a more formal definition of thermal efficiency)
$$ y = kx + m $$
Real Life Examples
A usual car that runs on gas has an efficiency of $\approx 20-25\%$ ($e \approx 0.2 - 0.25$)
Thermal expansion
Imagine we have a block of some kind of material. If we apply heat to this block it’s natural that it will expand. The same logic goes if we lower the heat/temperature - it will compress instead.
Formula for Thermal Expansion
$$ L = L_0 (1 + \alpha \Delta T), $$
where $\alpha$ is called the expansion coefficient with unit(s) $[K^{-1}]$.
This is for a 1D expansion, if the expansion happens in 2D it is,
$$ L = L_0 (1 + \beta \Delta T) $$
where $\beta$ is the expansion coefficient for 2D expansion. Note, $\beta \approx 2\alpha$ is only true for most metals, not all materials.
Thermal Conductivity for Simple Geometric Objects
Say if we have a system of 2 bodies, one which gains energy/heats up from an external source, with an opening to the second body.
This will transfer the heat from the hot body $T_h$ to the colder body $T_c$ since some heat will be lost in the transfer. The opening has a so-called thermal conductivity, the ability to transfer heat. With this in mind we can derive,
$$ P = \frac{dQ}{dt} = Ak \frac{T_h - T_c}{l}, $$
where,
$$ P - \text{Effect } [W] \newline Q - \text{Heat exchange } [J] \newline A - \text{Cross-sectional Area} \newline k - \text{Thermal Conductivity } [\frac{W}{mK}] \newline l - \text{Length of the opening} $$
We can rewrite this also as, $$ P = -Ak \ \frac{dT}{dx}, $$
or more generally,
$$ P = -Ak \ \nabla T $$
Thermal Conductivity for more Complex Geometric Objects
For more complex geometric shapes, we need to consider how the thickness of the body increases/decreases. For example, for a cylinder, the formula becomes,
$$ P = -\underbrace{A}_{2\pi lr} k \ \frac{dT}{dr} = 2\pi lr \cdot k \ \frac{dT}{dr} $$
If we’re looking at a cylinder that generates heats in its core for example, being near the core (small inner radius) will obviously be warmer than being at its surface (larger inner radius)
Since $2\pi l$ is always constant — we can rewrite the equation and integrate both sides to get,
$$ \int_{T_h}^{T_c} -\frac{2\pi l\ \cdot k}{P} = \int_{R_1}^{R_2} \frac{dr}{r} \Rightarrow P = \frac{2\pi k l}{\ln(\frac{R_2}{R_1})} (T_h - T_c) $$
Where $R_1$ and $R_2$ are two inner-radii to the cylinder.
Suppose we have a system which upholds a certain temperature $T_i$ with a constant effect of $P$ — Now consider if we cut the constant effect of heat, $P$ and let the system naturally cool down due to its surrounding temperature being $T_c$.
We can describe this relation with the following, $$ dQ = -c m dT $$
Where $c$ is the conductivity coefficient and $m$ is the mass of the substance.
With a bit of rewriting and integrating we also get,
$$ \int_{T_1}^{T_2} \frac{dT}{T - T_0} = \int_0^t \frac{-Ak}{lcm} dt \Rightarrow \ln(\frac{T_2 - T_0}{T_1 - T_0}) = -\frac{Ak}{lcm}t $$
What Thermal Conductivity really means
Just so we understand the equation $Q = cm \Delta T$ — the $c$ coefficient tells us how much energy we need to put into a substance to change its temperature. For example, $\approx 4180 J$ are required to increase the temperature of 1 kg of water by 1 ‘kelvin’. Often we can use $^\circ$C as well - but this is due to the difference between two temperature being absolute. So really it’s an increase in any kind of temperature, just by 1 index.
Latent Heat
Last thing we’re about to cover is the so-called Latent heat release of required to transition from one phase to another.
The formula for how much energy is needed is,
$$ Q = mL $$
Where $L$ is the specific latent heat for a particular substance with units $[J/kg]$.
Conclusion
This was everything for the first part of the thermodynamics, next time we’ll look into how gases behave and why they are so important for thermodynamics.