physics notes - a day in ducky's class

8/8/2019 physics notes - a day in ducky's class

http://slidepdf.com/reader/full/physics-notes-a-day-in-duckys-class 1/6

m p e r a t u r e a n d H e a t

ing, cooling, transitions

rnal energymentioned earlier working on collisions(when someone asked 'what happened to kinetic energy that disappeared' she said 'it was actually converted to internal energy')

internal energy is the energy of atoms and moleculesthe internal energy of a car would NOT be the motor,it would be the atoms/molecules in the materials that make it up

on the one hand you know that atoms and molecules move all the time (not at zero Kelvin but w/e) so atoms and molecules are in constant periodic motion. since they are moving, they possess a certain type of energy: motion. kinetic energy. on the other hand, they may move up and down,which means that their gravitational potential energy may vary as well.in other words, internal energy of atoms and molecules - internal energy --is a simple sum of gravitational and kinetic energy of molecules

U = internal energy = KE + PEgr of atoms and molecules

rnal energy (U) can be changed in two ways|

/ \/ \

/ \/ \

/ \/ \/ \

1.heat transfer 2.work performance-------------

i. conductionii. convectioniii. radiation

uction is a transfer of heat from one part of an object to another if they have different temperaturesex: put a metal spoon in a hot cup of coffee, and the heat travels up from the bottom

to the top of the spoon.

because metals are good conductors of heat this may happen within a few seconds.(interesting note: wood is NOT a good conductor of heat,which is good, b/c then matches don't burn you as quickly)

ection is a transfer of heat by streams of fluid (liquid or gas) from one part of a building/room/space her

ex: there is a fireplace in the room. even if you are further away, it will just take you more tifeel the streams of warm air reaching you. for another example suppose you are somewhere innew york, or the east coast, and in winter. Someone opens the door and the cold air comes in

ation is a transfer of heat by RAYS in any case, whether it's electromagnetic radiation, or ultravioleta case of heat by rays

ex: microwaves, microwaves, ultraviolet, infrared, x-ray, ?sonic-rays?



A N T I T Y O F H E A T A certain amount of heat...

For example, if you wanted to make chicken soup, you put hot water on the stove turn it on to high, and it boils.the amount of heat that it has absorbed is called the QUANTITY OF HEATwhat does it depend on? the time varies between materials, so it depends on the substance itself

Suppose that we're going to make a chicken soup; we'd start from room temperature, about 20° Celsiand we want to heat the soup to 100° so as soon as it starts boiling we'd turn it off.

so our goal here is to heat up a certain amount of water from 20°(/whatever your room temp is)to 100°we know:

water boils at 100°c; water freezes at 0°caka

0°C a phase transition from liquid to solid (ice); at 100°C it becomes vapour(also phase transi

we're going to just heat a certain amount of water between 20° to 100°, but not transition to vapour;we’re calculating the amount of internal energy OR the amount of heat that the water needs to absorb

how can we find mathematically the amount needed to gain for the water to be heated?

quantity of heat is denoted by letter 'Q'

Q = quantity of heat; it is measured in J (Joules) it can also be measured in things like calories, bif we heat a fluid OR solid between the critical points (0°c and 100°c) the formula is

heating up... Q = m * c * (t2 – t1)mass of substance (water)t2 is greater than t1

'c' is a constant: Specific Heat Capacity...it shows the amount of heat needed to heat up one kilogram of matter of a substance 1°Cit turns out that different substances..in order to be one kilogram of a different substance...in order to be heated up just one degree, need totally different amounts of energy

since t2 is greater than t one, the expression in parentheses is positive.it must ALWAYS BE POSITIVE. (t2 - t1)> 0we are not specifying which is the initial temp or final temperature, just that t 2 is biggerthis same formula can be used to find energy for COOLING temperature.we still use the same formula; we know that the subject LOSES heat,and we calculate the amount of heat LOST by substance, instead of GAINED (so the number will be

l)

in handout there is a chart.if we heat up aluminium, then all it needs is 900 J of heat

in case of copper, even less -- 387 J of heat silver: 235 (even faster)now comes water -- 4,190 Jlead 128 J

suppose that we deal with ice.s say that this ice is -5° c and we want to heat it up to bring it to a point where it starts melting (0re not going to melt it, we just want to BRING it to JUST that point --e heating it up but not bringing it to a phase transition.rder to calculate the amount of heat ice needs to reach the temperature of 0° C,same formula can be used, AS LONG AS IT REMAINS WITHIN THE SAME PHASE.

ose that we're dealing with water vapour at 115° c we want to bring it to 100° c.e not going to condense it/convert it to water, we stay with a vapour - but we want to cool it down -g it down to 100°s by cooling it down. how can we find the amount of heat LOST by the vapour in this pros the same process, as long as we're not dealing with the phase transitions)

cally this formula can be applied to solids liquids and gasses WITHIN each page

--------------------



e transitions: vaporization

e transitions take place at certain temperatures which is called atical point' whether is it a boiling point or freezing point

ng every phase transition, the system either loses or gains heat

. energy can be lost or gained at a constant temperature....remember that on the one hand a system gains heat, on the other hand, we say that there is no change inerature.e say that the substance LOSES heat, and still there is no change in temperature.ead, what happens is CHANGE IN PHASE. in order to melt, ice must absorb a certain amount of heat --hanges from a crystal structure, & in order to break the crystal structure, it needs a certain amount ofthat heat changes the structure, but it does NOT change the temperature. water vapour is...cules for water is different from that of vapour. the structure is different -- the molecular structure erent.MOLECULESS are the same (h2 0) but the STRUCTURE is different. so in order to CHANGE it -- going from vapr --as to lose some amount of heat to become water again. there is no change in temperature, but another chas place --ge of a molecular structure.. this is what phase transition is about

an't use the previous formula since it highlights change in temperature, & phase transition does not meaerature change

he formula which describes a formula for phase transition -- vaporization and condensation in particular

= Lv * m= mass of substance

vis called the HEAT OF VAPOURIZATION; it is measured in J/kg; it's a constant,

and its different for diff substances, and depending on the substance,we'll be able to find the heat of vaporization in the provided charts with each problem

hat does Lv show -- the heat of vaporization -- shows how much heat is needed for 1 kilogram of substance to transfer from LIQUID to VAPOPUR.

orization)

everse is condensation; it takes place at 100° c for water and Lv can be found at the chart

lly, let's look at melting/freezing/crystallization

ater transition takes place at 0° C, we have to do something to freeze itcan we calculate the amount of heat needed for water to lose in order to become crystals of ice?much should the water lose?

Lf * m

ed melting/crystallizatione have some ice at 0° c and we want to melt it, we have to supply a certain amount of heat to get waters of amount, by the constant, HEAT OF FUSION)

s called HEAT OF FUSIONof fusion is measured in J/kg;hows the amount of heat needed to be gained by one kilogram of ice in order to be transferred to water

ose we had pieces of ice at -5° c. how would we calculate the amount of heat needed to melt it?a two step problem -- we first have to use the FIRST formula heat it up from -5 to 0,THEN it's not water yet, it's still ice, so NOW we have to continue providing our system with heat --to find out how much, use the THIRD formula -- then you add the two amounts

mber: c, Lv and Lf are all provided in charts

use we frequently deal with water, here are the 3 constants in regards to them:

190 J/kg°s C or 4186, as another textbook has it

22.6 * 10 ^5 J /kg ........ 2 260 000

33.5 * 10 ^4 J /kg ........ 335 000



rity of problems we'll solve are based on

ERATURE/ HEAT BALANCE

problem deals with different substances or maybe the same substance but in two different amounts,h have different temperatures (for ex, 3 kg at 20° c and 6 kg of water at 35° c)approach to problems like this is the same. we assume that the amount of the lost heat is equal to the ahe gained heat.

NT OF LOST HEAT = AMOUNT OF GAINED HEAT

suppose we combine two amounts of water: 5kg at 20° and 6 at 25°. what is the temperature of the mixturee will it be? will it be 20? 25? no. where will it be? most likely somewhere between themater than 20, less than 25)... how would we apply the principle that we just wrote down?

e we have 2 amounts of water... let's call them cold and warmwater combined with cold water cools down. cold water combined with warm water heats up.

e do they meet? at what temperature? somewhere between 20° and 25°

his is the idea: that since we always deal with something HOT and COLD, the temperature....ot goes down, cold goes up, and they meet between those temperatures

et's take a look at one of those problems. it's a very simple problem but it illustrates this v. well

g of water at 70°cdded tokg of water at 20°cthe final temperature!

m=100kg 400kg t°final 100kgm=70°c --|--------|---------------------------|-

l=400kg 20°c 70°c

l=20°c it will be closer------------- to this b/c there's

al = ? more of the colder water

nt of gained heat = Q1 for the WARM WATER

mcool * c (t°final - 20°)

the warm water, how much does it lose?

mwarm * c = (70° - t°final *1)

ave to make the amount of gained heat equal to the amount of lost heat... they both contain unknown vari

mcool * c * (t°final - 20°) = mwarm * c * (70 - t°final)plug in what little we know... and solve!

400kg * 4190j/kg (t°final - 20°) = 100kg * 4190j/kg (70°- t°final)cross out "C";multiply them out

400 t°final - 80000 = 7000 - 100 t°final

+100 t°final on each side;+8000 to each side

500 t°final = 15000 t°final =15000/500 = 30° c (which is appropriate to where we guessed it would b

common mistake: students sometimes see this & go from 70° to 20°, which is wrong



lass lesson number 4, one of the problems:

m p e r a t u r e a n d H e a t

g ice cube at 0° C is dropped into 200g of water at 30° C. what is the temperature of the mixture?

= 0.03kg

= 0°c ice H2 00.2 kg ---|---------------------|-------------------------|

= 30°c 0°C tmix 30°C

= ?

t use the equation to change ice to water (gained)

Lf * mnow you have

exIce------------------v H2 0---|---------------------|-------------------------|

0°C tmix 30°C

add:

Q2 = mexIce * C (tmix° - 0°)

find out the amount of temperature LOST for the other side

Mh2 0 * C (30 - tmix)

exIce v------------------------H2 0---|---------------------|-------------------------|

0°C tmix 30°C

ine all this:

* mice ) + (mexIce * C (tmix° - 0°)) = Mh2 0 * C (30° - tmix°)

* 10^4 * 0.03kg + 0.03kg * 4190t(tmix°) = 0.2kg * 4190 * (30° - tmix°), etc

° = 15.7° C

hether we deal with vapour, gas, liquid, solid...usually just that we're going to be asked to find the final temperature.

sionally we will be asked to find like the mass... but we'll be using the same equations,

solving for a different variable. remember: it's always a matter of constructing an equation.times it's short, but more often it's long to calculate.

ESDAY NEXT WEEK: quizcovers this topic, previous topic, and waves(previous topic = ideal gas; waves = next week)



---------- side trip! -----------

e r m a l E x p a n s i o n :

e heat up a substance, it expands. why?use temperature is related to kinetic energy --up molecules move faster, further away from each other substance expands

(solid liquid gasses, doesn't matter)

he other hand if we cool it down...eratures go down molecules move slower, go closer together size of an object

(volume/linear sizes) decreases.

R is an exception; any substance that CONTAINS water is an exception

anomaly takes place in a very small temperature interval: from 0°C to 4°C.is where water behaves differently from other substances.it means is - if we have water (not ice) at 0°C and we heat it up, we go up from 0°C to 4°C-- if it were any other substance (such as olive oil)

what would happen? it would increase.any metal? it would increase/expand.

happens to water?it COMPRESSES.

we are at 10°C, we have an aluminium container filled with water at 10°C;ool the water down from 10°C to 5°C -- it's normal... from 5°C to 4°C -- it's normal.4°C to 2°C?IT GETS WEIRD AND BIGGER.

is water weird at 0°C to 4°C? It has to do with water density. What exactly happens to density?

ity of water (density: ρ, pronounced ‘rho’)

ρmax volume is at 4° C

is ρ?

m/v

ity (ρ) is directly proportional to mass, inversely proportional to volume. as density goes up, volume goes down. since density reaches its maximum density at 4° C, it reaches its minimum volume there, too

so instead of decreasing , it increases until the temperature reaches 0° C(and then there's a phase transition and it becomes ice anyway)

--------

Documents

physics notes - a day in ducky's class