introduction to biophysics - precourse · a thermometer a thermometer is an instrument that...
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Temperature
•Temperature is a property of matter that underlies the
common notions of hot and cold;
•On the macroscopic scale, temperature is the physical
property that determines the direction of heat flow
between two objects placed in thermal contact.
•If no heat flow occurs, the two objects have the same
temperature; otherwise heat flows from the hotter object
to the colder object.
•On the microscopic scale, temperature is defined as
the average energy of microscopic motions of a
single particle in the system per degree of freedom.
Temperature scales
oT = T - 273.15
T = T + 273.15
C K
K C
steam point
- 273oC 0 K
100oC 373 K
0oC 273 K ice point
absolute zero
A thermometer
A thermometer is an instrument that measures the
temperature of a system in a quantitative way
To find a substance having a property that changes
in a regular way with its temperature.
Thermometers
Mercury-in-glass (from -38.9oC to 356.7oC)
Ethyl alcohol (from -114oC to ~70oC)
Thermoelectric thermometer (from -260oC to 1800oC)
Thermal expansion
Thermal expansion of substances is a consequence
of the change in the average separation of atoms
and molecules of a substance due to increasing
amplitude of oscilations of atom and molecules at
higher temperature
Thermal expansion
Average coefficient of
volume expansion b :
TVV
/b
TVV b
ΔV– a change in volume, ΔT- a change in temperature,
V – initial volume.
Thermal expansion
Average coefficient of
linear expansion α :
TLL
/
TLL
ΔL– a change in length, ΔT- a change in temperature,
L – initial length.
in solids - b 3
Exercise:
An automobile fuel tank is filled to the brim with 55 l of gasoline at -20oC. Immediately afterwards, the vehicle is parked in a garage at 25oC. How much gasoline overflows from the tank as a result of expansion. Neglect the expansion of the tank.
ΔV = βVΔT;
Data: Tf = 25oC; Ti = -20oC; β = 9.6 x10-4 (1/oC); V = 55L
ΔV = 9.6 10-4 (1/oC) 55L (25oC -(-20oC)) = 2.38L
Exercise:
Express average summer and average
winter temperature in your country in
Celsius scale and in absolute (Kelvins)
scale of temperatures.
Ideal gas
thermodynamic variables : p, V, T
one mole
Avogadro’s number, NA = 6.022·1023
equation of state, pV = nRT
R - universal gas constant (8.315 J/mol·K), n – number of moles, T
- temperature in kelvins, P - pressure, V - volume
An ideal gas is one for which (PV)/(nT) is
constant at all pressures
Air under conditions close to normal can be treated as an ideal gas
v v
v
Exercise: How many molecules are in :
45 g of water;
245 g of sulfuric acid
at standard conditions.
N - Number of molecules; n – number of moles; MW – molecular weigth Water : MW = 2+16 = 18 g/mol
Sulfuric acid : MW = 2+32+16x4 = 98 g/mol
n=m/MW
nwater= 45g/(18g/mol) = 2.5 moles
n = 245g/(98g/mol) = 2.5 moles
N = n·NA = 2.5 x 6.022 x 1023 = 15.055 x 1023
Exercise:
Check that one mole of an ideal gas at atmospheric pressure and in 273 K (0oC) occupies the volume of 22.4 liters.
From the equation of state PV = nRT we obtain
V = nRT/P N = 1mole
P = 1 atm = 1.013x105 Pa
T = 273 K
R = 8.315 J/mol·K
V = 0.02242 J/Pa (joule/pascal)
1 J = 1Nx1m; 1 Pa = 1N/m2
V = 0.0224 m3 = 22.4 L
Ideal gas
p
V
T
m
n = m/M
equation of state
pV = nRT
pV/nT = R(const)
p’V’/nT’ = R(const)
pV/nT = p’V’/nT’
pV/T = p’V’/T’
m
V’p’
T’
A gas is heated from 27oC to 127oC while maintained at constant pressure in a vessel which volume increases. How does volume of the gas change?
P1V1 = nRT1 equation a state for the first temperature
P2V2 = nRT2 equation a state for the second temperature
comparing two equations we obtain:
P1V1/ T1 = P2V2/ T2
considering that P1 = P2
V1/ T1 = V2/ T2
V2/V1 = T2/ T1 = (127+273)K/(27+273)K = 400/300 = 1.333
final result: V2/V1 = 1.33
Exercise:
What is heat?
Heat is the net energy transferred from one object to another because of a temperature difference between these objects.
Heat is the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings.
Units of heat
SI unit of heat - joule (J)
Calorie (cal) - the amount of energy transfer
necessary to raise the temperature of 1 g of water
from 14.5 0C to 15.5 0C
Kilocalorie (kcal, Cal) - the amount of energy
transfer necessary to raise the temperature of 1 kg
of water from 14.5 0C to 15.5 0C
1 cal ≡ 4.186 J
Specific heat
How much heat is reqiured to change the
temperature of the substance ?
Q ~ m
Q ~ ΔT
Q depends on the kind of substance
Specific heat
Q = m·c· ΔT
Specific heat c of a substance is defined as
the amount of energy (of heat) Q transferred
to a unit mas of the substance reqiured to
change the temperature of the substance by
1oC.
Tm
Qc
[c] = J/(kg·K) = J/(kg·oC)
Exercise:
200 g of tea at 80o C (T1) was left in a room for few minutes and changed the temperature to 50oC(T2).
Calculate the amount of heat lost by the tea; do not consider the heat capacity of the cup; consider tea to be essentially water.
Calculate the amount of heat absorbed by the surroundings (the room); do not consider the heat capacity of the cup.
Data: m = 200g; T1= 80o C; T2= 50o C;
specific heat of water is c= 1cal/goC (or 1cal/gK)
Express the result in calories and in joules
Amount of heat transferred between the object
and the surroundings or between two objects:
Q = m·c· ΔT
Heat is absorbed (surroundings):
T2 > T1 => ΔT > 0 => Q = m·c· ΔT > 0
Heat is released – lost (tea):
T2 < T1 => ΔT < 0 => Q = m·c· ΔT < 0
Exercise:
300 g (m1) of water at 20o C (T1) was mixed with 200 g
(m2) at 50oC (T2) in a thermally isolated container.
Calculate the final temperature of water; do not consider
the heat capacity of the container.
Note: In an isolated system: Q (absorbed) = - Q (released)
300 g (m1)
20o C (T1)
200 g (m2)
50oC (T2) Q
Q1 = m1cT1 Q2 = m2cT2
T1 = Tf – T1 T2 = Tf – T2
Exercise:
300 g (m1)
20o C (T1)
200 g (m2)
50oC (T2) Q
Q1 = m1cT1
(absorbed)
Q2 = m2cT2
(released)
T1 = Tf – T1 T2 = Tf – T2
Q (absorbed) = - Q (released)
m1c(Tf – T1) = - m2c(Tf – T2)
Tf = 32oC
Phase changes (phase transitions)
Solid state
Liquid state
Gaseous state
Phase changes - processes
Melting
Condensation
Vaporization –
Evaporation or boiling
Freezing (Fusion, Solidification)
Sublimation
Phase changes (phase transitions)
Phase change involves a
change in internal energy
of a substance but not a
change in temperature
Latent heat
Energy required for phase change of
1 kg of a substance is called the
latent heat (L) or heat of
transformation
m
QL
Compare amounts of heat released to the surroundings
during freezing 1 kg of water and 1 kg of ethanol.
Latent heat of fusion for water is 333J/g.
Latent heat of fusion for ethanol is 109J/g.
Q = mL
Qw = mwLw
Qe = meLe
Qw = 1kgx333J/g = 1kgx333kJ/kg = 333kJ
Qe = 1kgx109J/g = 1kgx109kJ/kg = 109 kJ
Exercise:
Vaporization :
evaporation or boiling
Evaporation: the process whereby atoms or
molecules from the surface of a liquid gain sufficient energy to enter the gaseous state. It is the opposite process of condensation. Is is a slow process.
Boiling: is the rapid vaporization of a liquid, which typically occurs when a liquid is heated to its boiling point, the temperature at which the vapor pressure of the liquid is equal to the pressure exerted on the liquid by the surrounding atmospheric pressure.
Evaporation
Evaporation is possible in different temperatures
Evaporation is a cooling process
Rate of evaporation depends on: the temperature
of the liquid and the surroundings, the surface
area of the liquid, pressure of the air above the
liqiud surface and humidity of the air
Phase transition in water
In liquid water each molecule is hydrogen bonded to approximately 3.4 other water
Phase transition in water
water ice
An anomalous volume expansion of water near
its freezing point:
Water has minimum volume (maximum density)
at 4oC.