### Physics SAE

```Physics Subject Area
Test
Thermodynamics
*
There are three commonly used temperature scales, Fahrenheit,
Celsius and Kelvin.
Converting Between the Kelvin and Celsius Scales
Converting Between the Fahrenheit and Celsius Scales
Thermal expansion
• Expanding solids maintain original shape
• Expanding liquids conform to the container
Linear expansion
ΔL = αLΔT
L = length
α = coefficient of liner expansion
ΔT = temperature change
Example:
The highest tower in the world is the
Poland, which has a height if 646m. How
much does its height increase between a
cold winter day when the temperature is
-35⁰C and a hot summer day when the
temperature is +35 ⁰C ?
ΔL = αLΔT
= 12x10-6/ ⁰C x 646 x 70 ⁰C
= 0.54m
• Volume expansion
ΔV= βVΔT
L = length
β = coefficient of liner expansion
ΔT = temperature change
cold
Β=3α
hot
Heat flow:
the heat current; the amount
of heat that passes by some given place on the
rod per unit time
∆
∆
= heat flow
k = proportionality constant, thermal conductivity
A = cross sectional area
ΔT = change in temperature
Δx = distance crossed, thickness of material
* convection
Heat is stored in a moving fluid and is carried from one place to another
by the motion of this fluid
The heat is carried from one place to another by electromagnetic waves
* conduction
the process of handing on energy from one thing to the next
(I) by a body at temperature T and
having a surface area A is given by the Stefan-Boltzmann law
=   4
e = emissivity (between 0 and 1)
σ = Stefan’s constant = 5.6703 x 10-8 W/m2·K
A = surface area
T = temperature
Shiny objects are not good absorbers or radiators & have emissivity close to 0
Black objects have emissivity close to 1
Latent heat ( heat of transformation) – the heat
absorbed during the change of state
=
∆

ΔQ = quantity of heat transferred
m = mass of the material
Heat of fusion - heat absorbed when changing from a
solid to a liquid
Heat of vaporization - heat absorbed when changing
from a liquid to a gas
GAS LAWS
P= pressure
V = volume
P1V1=P2V2
V1/V2=T1/T2
V = volume
T = temperature
P1/T1=P2/T2
P = pressure
T = temperature
Combined Gas Law
1 1
1
=
2 2
2
Ideal Gas Law
PV = n R T
P= pressure
V = volume
T = temperature
n = moles
R = Gas constant
= 0.08206 L-atm/mol K
*Gas Density
PV = nRT
n/V = P/RT
Molarity = n/V
Density D = m/V
Molecular Wt M = m/n
PM/RT
M= DRT/P
D=Mn/V =
*
Energy can be neither created nor
destroyed but only transformed
THE GENERAL ENERGY EQUATION
Energy In = Energy Out or U2 - U1 = Q -W
where
U1: internal energy of the system at the beginning
U2: internal energy of the system at the end
Q : net heat flow into the system
W : net work done by the system
Q = ΔU + ΔW
A closed tank has a volume of 40.0 m2 and is
filled with air at 25⁰C and 100 kPa. We want to
maintain the temperature in the tank at 25⁰C as
water is pumped into it. How much heat will have
to be removed from the air in the tank to fill it half
full?
=
1
2
2  =    1
=  1
=

2

= (100kPa) (40.0 m2)(-0.69314) = -2772.58kJ
*
• Isobaric
– the pressure of and on the working fluid is constant
– represented by horizontal lines on a graph
• Isothermal
– temperature is constant
– Temperature doesn’t change, internal energy remains constant, &
the heat absorbed by the gas = the work done by the gas
– The PV curve is a hyperbola
– there is no transfer of heat to or from the system
during the process
– Work done = decrease in internal energy & the temperature falls
as the gas expands
– -the PV curve is steeper than that of and isothermal expansion
*
Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes – sufficiently
slow processes, any intermediate state can be considered
as an equilibrium state (the macroparamers are welldefined for all intermediate states).
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (small) number of macro
parameters as for a system in equilibrium (e.g., for an ideal gas in quasiequilibrium processes, this could be T and P). By contrast, for nonequilibrium processes (e.g. turbulent flow of gas), we need a huge number
of macro parameters.
Examples of quasiequilibrium processes:
isochoric:
isobaric:
isothermal:
V = const
P = const
T = const
Q=0
For quasi-equilibrium processes, P, V, T are
well-defined – the “path” between two states
is a continuous lines in the P, V, T space.
P
T
2
1
V
*
Work
A – the
The work done by an external force on a gas
enclosed within a cylinder fitted with a piston:
piston
area
W = (PA) dx = P (Adx) = - PdV
force
x
P
The sign: if the volume is decreased, W is positive (by
compressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreases
its internal energy by doing some work on the
environment).
W 1 2   
V2
P (T , V ) dV
V1
W = - PdV - applies to any
shape of system boundary
dU = Q – PdV
The work is not necessarily associated with the volume changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent of
heat”, the system (water) was heated by stirring.
*Specific Heat
the heat absorbed during the change of state
Q = nCv ΔT
Q = amount of heat required
n = number of moles
Cv = specific heat at a constant volume
ΔT = Change in temperature
How to calculate changes
in thermal energy
Specific heat is the amount of heat required to raise the temperature of 1 kg of a
material by one degree (C or K).
C water = 4184 J / kg C
Q = m x T x Cp
Q = change in thermal energy
m = mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance
Second Law of Thermodynamics
Entropy = the transformation of energy to a more
disordered state
- can be thought of as a measure of the randomness
of a system
- related to the various modes of motion in molecules
The second law of thermodynamics: entropy of an
isolated system not in equilibrium tends to increase
over time
• No machine is 100% efficient
• Heat cannot spontaneously pass from a colder to a
hotter object
The relationship between kinetic energy and intermolecular
forces determines whether a collection of molecules will be a
solid, liquid or a gas
* Pressure results from collisions
* The # of collisions and the KE contribute to pressure
* Temperature increase KE
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