Thermodynamics

Thermodynamic Experiments §

Name of Scientist	Experiment	Importance
Cornelis Drebbel	Used salt to cool air	May have been the inventor of air conditioning
Robert Boyle	Weighed a barrel of water and let it freeze, then weighed it again. It weighed exactly the same.	Figured out that cold is not a substance
Fahrenheit	Calibrated a temperature scale for a thermometer	Developed the Fahrenheit scale
Michael Faraday	Put chlorine under pressure and cooled it, then cooled in an ice bath	Figured out that energy can transform liquid into gas- ‘produced cold’
Count Rumford	Water could be boiled in a closed box for hours, the supply of frictional heat was inexhaustible	Heat isn’t a substance (again)
Guillome Amonton	Heated up a tube of Mercury and saw that it went up and down	Discovered a temp/pressure correlation & absolute zero
Frederick Tutor	Takes the same amount of energy to melt a lot of ice as it takes to heat up a lot of water	Helped develop artificial refrigeration
Sadit Carnot	Analyzed how a steam engine turned heat into work	Greater temp difference -> More work produced
James Joule	Measured temperature increase when movement occurred, energy advent	Figured out that heat turns into mechanical work

Laws of Thermodynamics §

First Law §

Energy and matter can change forms, but they cannot be created or destroyed.

Second Law §

Heat flows from hot to cold and the entropy (disorder) of a system tends to increase.

Entropy §

Entropy is the measure of “disorder” in a system, and often refers to energy that is no longer able to be used to perform work. For example, the engine of a car burns gasoline changing it from usable energy in the gas into mechanical work (the movement of the car) and exhaust, which cannot be used to create work.

Expansion and Heat §

Since heat adds energy to a system (solid, liquid or gas), hotter molecules move faster, thereby expanding in overall size. Different materials expand at different rates!

Heat Transfer §

Conduction §

Conduction can be undergone by solids and liquids. The energy of the vibrating molecules of one substance transfers to another when they touch.

Convection §

Convection can be undergone by liquids and gases. Hot gas molecules rise, cold molecules sink. As they move, energy and gas molecules are transferred.

Radiation §

Transfer through electromagnetic radiation— you can see this in microwaves and infrared heat lambs. Molecules absorb radiation at different rates. Gases in our atmosphere absorb radiation that keeps us warm, but humans can make them faster than nature can remove them, causing global warming.

Albedo §

Albedo is the amount of energy reflected by a surface without being absorbed. An object with high albedo reflects the majority of the radiation while an object with low albedo reflects only a small amount of incoming radiation but absorbs the majority of the radiation. Snow has an albedo of around 0.8.

Thermal Expansion §

When a solid object heats up, the molecules move faster, pushing each other apart, through expansion. This expansion can be calculated by using the linear thermal expansion formula:

• Type of material ($\alpha$), measured in $\frac{1}{K}$ (or $K^{-1}$)
• Length ($L_i$)
• Change in Temperature ($\Delta T$)

$$\begin{array}{rl}\text{change in length}& =\text{coefficient of thermal expansion}\times \text{initial length}\times \text{temperature chance}\\ & \\ \Delta L& =\alpha L_i\Delta T\end{array}$$

Consider that the Eiffel Tower is made of iron, and is 301.00 meters tall on a day that the temperature is 22.0 C. How much does the height decrease when the temperature cools to 0.0 C?

$$\begin{array}{rl}{T}_i& =22^{\circ }C\\ T_f& =0^{\circ }C\\ \Delta T& =-22^{\circ }C\\ L_i& =301.00\text{ meters}\\ \alpha & =1.2\times 10^{-5}{\text{ K}}^{-1}\end{array}$$

$$\begin{array}{rl}\Delta L& =\alpha L_i\Delta T\\ & =(1.2\times 10^{-5})(301)(-22)\\ & =-0.079464\text{ meters OR }-7.9\text{ cm}\end{array}$$

So, the height decreases by $-7.9$ centimeters.

Heat Capacity and Energy §

Energy and heat are typically measured in Joules (J) or calories (cal), but they can also be measured in derived units of this, like:

• kJ ($1000$ J) & mJ ($1\times 10^6$ J)
• Calories / kcal (1000 cal in a kcal, 1000 calories in a Calorie)

Heat Transfer §

The heat capacity of aluminium is $0.900{\frac{J}{g}}^{\circ }C$, whilst the heat capacity of water is $4.18{\frac{J}{g}}^{\circ }C$ . Different materials absorb heat differently.

Specific Heat Capacity §

Specific heat capacity is the heat capacity per gram, in ${\frac{J}{g}}^{\circ }C$. We can use this and the heat capacity formula to calculate how much energy is needed to increase the temperature of a certain substance to a certain point.

We have two substances:

• 1000g of water, with a specific heat capacity of 4.18 ${\frac{J}{g}}^{\circ }C$
• 1000g of gold, with a specific heat capacity of 0.129 ${\frac{J}{g}}^{\circ }C$ How much energy do we need to apply to each of them to increase their temperature by 20 ${\frac{J}{g}}^{\circ }C$?

First, we can set up the variables for water.

$$\begin{array}{rl}m& =1000\text{ g}\\ \Delta T& =20{}^{\circ }C\\ c& =4.18\frac{J}{g}{}^{\circ }C\end{array}$$

We’re looking to find heat, $Q$, in Joules. Because heat capacity is measured in Joules over grams, we do not have to convert the mass into kilograms as we have done before.

$$\begin{array}{rl}Q& =mc\Delta T\\ & =(1000)(4.18)(20)\\ & =83600\text{ J}\end{array}$$

Now, when we go to set this up for gold, we are expecting a much lower value because gold has a lower specific heat capacity— it doesn’t take as much energy to heat up or cool gold as it does to heat up or cool water. So, for gold, our variables are:

$$\begin{array}{rl}m& =1000\text{ g}\\ \Delta T& =20{}^{\circ }C\\ c& =0.129\frac{J}{g}{}^{\circ }C\end{array}$$

Again, 0.129 is a lot lower than 4.18— because it is under 0, when we multiply $m\Delta T$ by it, we will get a smaller number out.

$$\begin{array}{rl}Q& =mc\Delta T\\ & =(1000)(0.129)(20)\\ & =2580\text{ J}\end{array}$$

Conservation of Energy §

The principle of conservation of energy still applies here. The heat that we gain will be equal to the heat that something else lost.

Let us say that we have 1000g of gold. If the temperature of that gold decreases by $80^{\circ }C$, by how much will the temperature of 100g of water increase when the good is placed in the water? Consider that $c_{H_{2}0}=4.18{\frac{J}{g}}^{\circ }C$ and $c_{Au}=0.129{\frac{J}{g}}^{\circ }C$.

$$\begin{array}{rl}{m}_{Au}=1000g,\Delta T_{Au}& =80^{\circ }C,c_{Au}=0.129{\frac{J}{g}}^{\circ }C\\ m_{H_{2}0}=1000g,\Delta T_{H_{2}0}& =?{?}^{\circ }C,c_{H_{2}0}=4.18{\frac{J}{g}}^{\circ }C\end{array}$$

$$\begin{array}{rl}{Q}_{Au}& =mc\Delta T\\ Q_{Au}& =(1000)(80)(0.129)\\ Q_{Au}& =10320\text{ J}\\ & \end{array}$$