Nathaniel Cooper Ph.D.
September 10, 2018
Now on Blackboard:
Purdue University has a small nuclear reactor in the basement of the EE building. It is used to make chemical isotopes for research. It is a 10.0 Kw reactor that operates at 1 kW. It is the size of a microwave oven. It sits in the bottom of a 17.0 ft deep cylindrical pool of water that has a radius of 4.00 feet. The water is at room temperature of 20.0oc. How much time would it take for the reactor, operating at capacity to bring the water surrounding it to the boiling point?
\(Q = mc \Delta T\)
What do we have to calculate before we calculate time?
Volume of a cylinder: \(V = \pi r^2*h\)
r = \(4 ft \times \frac{0.3048 m}{1 ft} = 1.22 m\)
h = \(17 ft \times \frac{0.3048 m}{1 ft} = 5.18 m\)
\(V = \pi 1.22m^2*5.18m = 24.2m^3\)
\(m=\rho V = \frac{1000 kg}{m^3}*24.2m^3 = 24200 kg\)
\(Q = ?\)
\(m = 24200 kg\)
\(c = \frac{4184 J}{kg*^o C}\)
\(\Delta T = 100^oC - 20^oC = 80^oC\)
\(Q = mc\Delta T\)
\(Q = 24200 kg \times \frac{4184 J}{kg*^o C} \times 80^oC\)
\(Q = 8.1\times 10^9 J\)
\(P = \frac{Q}{t} \rightarrow t = \frac{Q}{P}\)
\(P = 10 kW = 10000 W = 10000 \frac{J}{s}\)
\(t = \frac{8.1\times 10^9 J}{10000 \frac{J}{s}} = 8.1\times 10^5 s\)
\(1 d = 86400 s \rightarrow 8.1\times 10^5 s\times\frac{1d}{86400s}\)
\(t= \boxed{9.38 \space d}\)
1 mol = \(6.02\times 10^{23}\) of something
Used for counting atoms/molecules
molar heat capacity: the amount of energy needed to change 1 mol of a substance by 1 kelvin
Often used for gases, as moles of a gas are determined by the volume and pressure of a gas.
\(Q = nC\Delta T\)
similar to Specific Heat Capacity, but n is the number of moles and C is the molar heat capacity.
Latent heat is the energy it takes to cause a phase change.
Phase of a substance is determined by the strength of bonds between atoms/molecules.
Changing Phase represents a change in the Potential Energy of the system.
Temperature measures average Kinetic Energy, which is not changing.
Therefore temperature does not change during a phase change.
\(Q = mL\)
Where L is the Latent Heat of the phase change:
\(L_f\) for Latent Heat of Fusion (freezing/melting)
\(L_v\) for Latent Heat of Vaporization (boiling/condensing)
Units are Joules per kg.
If the boiling point of the cooling pool has been reached, how much time before the water boils away?
Recall:
\(m = 24200 kg\)
\(P = 10000 \frac{J}{s}\)
\(L_v = 2256 \frac{J}{kg}\) for water
\(Q = 24200 kg * 2256 \frac{kJ}{kg} = 5.46 \times 10^{10} J\)
\(t = \frac{5.46 \times 10^{10} J}{1.00 \times 10^4 \frac{J}{s}} = 5.46 \times 10^6 s\)
\(\boxed{63.1 \space days}\)
As a liquid vaporizes, the vapor carries heat away.
Evaporation is the process where liquid near the surface becomes vapor before the boiling point is reached.
Recall that temperature is based on Average KE. Some molecules may have above-average KE and may be able to break the inter-molecular forces keeping them in the liquid.
Such molecules become vapor.
This is how sweating works and why you feel cold when leaving a pool.
Conduction
Convection
Radiation
Heat transferred due to physical contact.
The molecules of the higher temperature object collide with the molecules of the lower temperature object.
Internal KE in the high temperature object goes down, and internal KE in the lower temperature object goes up, until they have the same average KE.
\(H = \frac{dQ}{dt} = kA\frac{T_H - T_c}{L}\)
H is heat current, the rate the heat is transferred. Units: Watts.
k is the coefficient of thermal conductivity a measure of how readily a material transmits heat. Units: \(\frac{W}{m*K}\)
A is the cross sectional area of contact. Units: m2
\(\frac{T_H - T_c}{L}\) is the magnitude of the Thermal Gradient. How temperature changes per unit length. Units: \(\frac{K}{m}\)
Convection is the heat transfer process due to a flowing fluid.
If the fluid flows due to it own thermal gradient this is called Natural Convection.
If the fluid is forced to move (e.g., a pump or a fan is used) this is Forced Convection.
Flowing fluids tend to be a better heat transfer mechanism than static fluids.
Heat current is directly proportional to surface area of contact. This is why CPU heat sinks have fins.
A fluid’s viscosity produces a insulation layer near the surface of contact for natural convection. Forced convection reduces the thickness of this layer which increases rate of heat transfer. This is the reason for ‘Wind Chill Factor’
\(H \propto T^{\frac{5}{4}}\)
All atoms and molecules are constantly moving and colliding or vibrating about inter-molecular bonds.
All atoms consist of electrons, protons and neutrons. Which are negative, positive, and neutral changes respectively.
If you accelerate a charged particle it creates an electromagnetic wave: light.
Radiation is the process of heat being transferred via this light.
\(H = Ae\sigma T^4\)
We can see that the Sun’s Heat Current per area varies by about 1 W/m2 per 5.5 years. Assume in 1995 the average temperature of the Earth was 278.75 K. The Sun’s Irradiance was 1365.5 W/m2. In 2000, when the Sun’s Irradiance was about 1366.5 W/m2, what would you expect the Earth’s average Temperature to be?
Since the Sun’s Irradiance (I) is W/m2 the formula becomes:
\(I = e\sigma T^4\), We divide both sides by area and call \(\frac{H}{A} = I\)
We can solve by ratio:
\[ \frac{I_f}{I_i} = \frac{e\sigma T^4_f}{e\sigma T^4_i} \]
\[ T_f = (T^4_i\frac{I_f}{I_i})^{\frac{1}{4}} = T_i (\frac{I_f}{I_i})^{\frac{1}{4}} \]
\[ T_f = 287.75 K (\frac{1366.5 W/m^2}{1365.5 W/m^2})^{\frac{1}{4}} = 287.80 K \]
When we talk about thermal processes we are often talking about measurements made on a large scale (macroscopic), but the objects mediating the thermal process are atoms and molecules which are microscopic.
How exactly these tiny atoms can cause measurable effect on large systems comprised 2 out of 5 of Einstein’s earliest published works: His ‘miracle year’ paper on Brownian Motion and his doctoral thesis on the size of atoms (more or less an estimation of Avogadro’s Number, \(6.02\times 10^{23}\), in which he was off by a little bit).
At the atomic scale we ought to be able to measure momentum, kinetic energy, and number of particles. What we can measure easily in a lab is pressure, temperature, and volume or mass.
The State of a material, i.e. Solid, Liquid, Gas are determined by factors such as pressure, temperature, and volume.
Such variables are therefore called Variables of State.
Equations that relate these variables together are called Equations of State.
The Ideal Gas Law relates the relationship between Temperature, Pressure, and Volume of a gas in a container.
This law assumes the gas is mono-atomic and the atoms’ size is negligible and that atoms cannot attract each other.
\[ pV = nRT \] - P is pressure of the gas in Pa, V is volume in m3, T is temperature in KELVIN.
n is the number of mols in the gas.
R is the Ideal Gas Constant: \(8.314 \frac{J}{mol*K}\)
It can get you within a few percent of a non-ideal gas: Great for estimations.
Molar mass is the mass of a pure substance containing one mole of molecules of that substance.
In the ideal gas equation, of we can calculate the number of moles, we can determine the mass:
\[ m_t = nM \] -m_t is total mass (kg)
n is number of mols
M is molar mass.
Assume air is an ideal gas. You have a container of compressed air at a temperature of 293 K and volume of 0.10 m3 and pressure of 500 kPa. How many mols of air do you have? If the molar mass of air is 29 g/mol, what is the mass of the air in the container?
\[ n = \frac{pV}{RT} = \frac{500000 \frac{N}{m^2} * 0.1 m^3}{ 293 K * 8.314 \frac{J}{mol*K}} = 20.5 \space mols \] \[ m_t = nM = 20.5*29\frac{g}{mol} = 595 g = 0.595 kg \]
This is after the gas is allowed to cool to room temperature. When it was first filled the temperature and pressure were higher.
At the age of 35, former high school physics teacher Johannes Van der Waals defended a Ph.D. thesis in which he explained how real gasses depart from the ideal gas equation.
Atoms will tend to attract each other if they get close enough, like when they collide.
Atoms are not point particles and have volume.
These results in corrections to the Ideal Gas Law:
\[ (p + \frac{an^2}{V^2})(V-nb) = nRT \]
a is a measure of attraction between gas molecules
b is the volume of the gas molecules.
a and b are different for every gas and are measured empirically.
Why pV diagrams are useful
\[ \frac{N}{m^2}\times m^3 = N*m = J \]
\[ W = \int pdV \]
For a variable volume process (isothermal, isobaric, or adiabatic) work done is the Area under the PV curve.
This work done is from the Internal Energy (total mechanical energy of the atoms/molecules) of the System.
Intramolecular forces affect the chemical identity of a substance, e.g, the ionic bond in NaCl
Intermolecular forces allow solids to maintain their shape and liquids to be contained in an open container, chemical identity is not affected.
As stated internal energy is the mechanical energy of the atoms and molecules in the system.
Heat may cause a phase change or chemical reaction if the heat increases kinetic energy enough to overcome the potential energy of the bonds:
Recall that a mol is a counting number like a dozen (12) or a gross (\(12^2\)).
Specifically:
One mole is the amount of substance that containes as many elementary entities as there are atoms in 0.012 kg (12 g) of Carbon-12
This is equal to Avogadro’s Number: \(N_A = 6.02214129(27) \times 10^{23}\)
Elementary entities is a molecule for chemical compounds and an atom of a pure substance.
The scale is calibrated to C-12 because the protons and neutrons within a nucleus ‘give up’ some of their mass to the nuclear bonds (Nuclear Strong and Weak forces) via \(E = mc^2\). The amount ‘given up’ changes from isotope to isotope.
For an ideal gas: \[ K_t = \frac{3}{2} nRT \]
\[ k = \frac{R}{N_A} = \frac{8.138 \frac{J}{mol*K}}{6.022\times10^{23} \frac{particle}{mol}} = 1.381\times10^{-23} \frac{J}{particle *K} \]
\[ \frac{1}{2}mv^2_{av} = \frac{3}{2} kT \]
\[ v_{rms} = \sqrt{v_{av}^2} = \sqrt{3\frac{k}{m}T} = \sqrt{3\frac{R}{M}T} \]
When a helium balloon pops and releases it’s helium to atmosphere, what is the rms speed of the helium? Assume the helium is 297 K, mass per molecule is \(6.68\times10^{-27} kg\).
Earth’s Escape Speed is \(1.12\times10^4 \frac{m}{s}\), how does rms velocity compare to this?
\[ \sqrt(\frac{3*1.381\times10^{-23}*297}{6.68\times10^{-27}}) = \]
So the helium’s root mean squared speed is about 0.1 Earth’s escape speed. Helium with above average KE can escape Earth’s gravity!
It may be of interest to know how far a molecule goes on average before colliding with another molecule.
The Statistical basis for this is called a Random Walk, and is used to model how objects often drift or diffuse from a starting point (stock prices can be modeled this way.)
The Random Walk distance between collisions of a molecule in a gas is:
\[ \lambda = \frac{V}{\sqrt{2} *\space4\pi r^2 *N} = \frac{kT}{\sqrt{2} *\space4\pi r^2 *p} \]
-\(\lambda\) is the mean free path (m), V is volume of the container (m^3), N is number of molecules, r is radius of the molecules (m).
The heat capacities for gases are based on how many degrees of freedom the gas has. That is in how many ways the gas can move.
Monoatomic Gases (e.g., the Noble Gases) have 3 degrees of freedom (x,y,z)
For an isobaric process, heat capacity \(C_V\): \[ C_V = \frac{3}{2} R \]
-Diatomic gases (e.g., \(O_2\)) have 5 degrees of freedom (x,y,z), 2 rotational modes about the center of mass.
For an isobaric process, heat capacity \(C_V\): \[ C_V = \frac{5}{2} R \]
For an isobaric process, heat capacity \(C_V\): \[ C_V = \frac{7}{2} R \]
For a crystalline solid of a pure substance an atom bond in the crystal will have 2 degrees of freedom: x,y,z
The motion will be vibration along the bonds will be due to Hooke’s Law: \(U = \frac{3}{2}kT\).
Kinetic Energy is also \(U = \frac{3}{2}kT\), like in a gas so total energy is: \(E = 3kT\)
Heat Capacity becomes:
\[ C = 3R \]
Since Temperature is an mean KE, what the the statistical distribution of molecular speed look like?
This is determined by the Maxwell-Boltzmann Distribution:
\[ f(v) = 4\pi (\frac{m}{2\pi kT})^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2kT}} \]
Compare to the Normal Distribution (aka, The Bell Curve)
\[ f(v) = \frac{1}{\sqrt{2\sigma^2}} e^{\frac{-(v-\mu)^2}{2\sigma^2}} \]
