Temperature and heat

introduction to biophysics -

precourse

Temperature

thermal contact

thermal equilibrium

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

Heat transfer

heat absorbed is positive; Q > 0

heat released is negative;Q< 0

Conservation of heat energy:

In a thermally isolated system

Q absorbed = - Q released

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)

Melting

Solidification

(Sublimation)

Condensation

Vaporization

Phase changes (phase transitions)

Phase change involves a

change in internal energy

of a substance but not a

change in temperature

Phase changes (phase transitions)

How much heat is reqiured to change the

phase of the substance ?

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.

That’s all for today