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Mn Vt L

Mn Vt l gm 4 chuyn :

C hc: 2 bui, c Thy.

Nhit hc: 2bui. Quang hc : 2 bui, thy Khang.

Ht nhn: 1 bui, thy Khang.

Hnh thc thi: Trc nghim.

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Chuyn Nhit Hc

Chuyn Nhit hc gm 3 chng:

Chng 1: Cht kh Cht rn Cht lng.

Chng 2: Nhit ng lc hc. Chng 3: C hc cht lu.

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CHNGI

CHTKHCHTRNCHTLNG

04/10/2015

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Cht Kh

Khinh kh cu dng trong nghin cu kh tng - thu vn, thm him, du lch, qung co, ...

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Cht Kh

Cu To Cht Phn t

Chuyn ng khng

ngng Nhit cng cao,

chuyn ng cng

nhanh

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Tng Tc

Lc tng tc phn t Lc ht

Lc y

gn: lc y mnh

xa: lc ht mnh

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Rn Lng Kh

Th rn

Th lng

Th kh

Lc tng tc Rt mnh Trung gian Rt yu

Dao ng t Trung gian Hn lon

Th tch Xc nh Xc nh Khng c

Hnh dng Xc nh Khng c Khng c

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Rn Lng Kh

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Plasma

Vi ngn . Cc electronbt khi nguyn

t Nguyn t tr thnhion

dng. Nhit cng cao th s

electron bt racng nhiu. Hin tng ny c gi l

s ion ha ca cht kh.

Plasma. Tia t ngoi, tia X, tia btacc mnh chiu vo cht kh.

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BEC

BoseEinstein Condensate. Ngng t Bose Einstein. Cht kh trong nguyn t

khng cn chuyn ng mtcch hn n na m b bucphi hot ng trong mtkhng gian thng nht honton, cng vn tc, cngphng, cng hng.

Tnh siu do, c kh nngleo ngc ln thnh cc vb ra ngoi.

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Kh l tng

Kh l tng Cht im

Ch tng tc khi va

chm

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Thng S Trng Thi

Thng s trng thi p sut p

Th tch V

Nhit T

T (K) = t (0C) + 273

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ng Nhit

ng nhit Nhit T khng i

pV = hng s.

Hay p1V1= p2V2= BoyleMariotte

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ng ng Nhit

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V D: ng Nhit

Mt xilanh cha 150 cm3kh p sut 2.105

Pa. Pit-tng nn kh trong xilanh xung cn100 cm3. Tnh p sut ca kh trong xilanh lc

ny, coi nhit nh khng i. p s: 3.105 Pa

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Tm tt

V1= 150 cm3

p1= 2.105

PaV2= 100 cm

3

p2= ?

Gii

Theo nh lut Boyle Mariotte:

1 1 2 2

1 12

2

5

2

5

2

p V p Vp V

pV

2.10 . 150

p 100

p 3.10 Pa

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ng Tch

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V D: ng Tch

Mt bnh cha mt lng kh nhit 300Cv p sut 2 bar(1 bar = 105 Pa). Hi phitng nhit ln ti bao nhiu p sut

tng gp i? p s: 3330C

Gii

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Tm tt

t1= 300C = 303 K

p1= 2 barp2= 4 bar

t2= ?

Gii

Do bnh cha kn nn th tchc bo ton.

Theo nh lut Charles:

1 2

1 2

2 12

1

2

0

2

p p

T T

p TT

p

4 . 303T 606 K

2

t 606 273 333 C

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ng p

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V D: ng p

nhit 2730C, th tch ca mt lng khl 12 lt. Tnh th tch lng kh 5460Ckhi p sut kh khng i?

p s: 18 lt

Gii

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Tm tt

t1= 2730C = 546 K

V1= 12 ltt2= 5460C = 819 K

V2= ?

Gii

Theo nh lut Gay - Lussac:

1 2

1 2

2 12

1

2

V VT T

T VV

T

819 . 12V 18

546

l

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Phng trnh trngthi kh l tng

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V D: Phng trnh trng thi ca khl tng

Trong phng th nghim, ngi ta iu chc 40 cm3kh hidro p sut 750 mmHgv nhit 270C. Tnh th tch ca lng kh

trn iu kin tiu chun (p sut 760mmHgv nhit 00C).

p s: 35,921 cm3

T tt Gii

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Tm tt

V1= 40 cm3

p1= 750 mmHgT1= 270C = 300 K

p2= 760 mmHg

T2= 00C = 273 KV2= ?

Gii

Theo Phng trnh trngthi ca kh l tng:

1 1 2 2

1 2

1 2 12

2 1

2

3

2

p V p V

T T

p T VV

p T

750 . 273 . 40V

760 . 300

V 35,921 cm

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Phng trnhClapeyron - Mendeleev

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V D: Phng trnh Clapeyron -Mendeleev

Gii

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Gii

Theo phng trnhClapeyron - Mendeleev:

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

2

2

p V n RT

p V n RT

p V n T

. .p V n T

30 3103.

p 283

p 9,13 at

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Kh thc

Kh thc gn ging khl tng nhit vp sut thng thng.

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V D: Cu To Cht

Tnh cht no sau y khng phil ca phnt?

A. Chuynngkhng ngng.

B. Giacc phn tc khongcch.

C. C lc ngyn, c lc chuynng.

D. Chuynngcng nhanh th nhitcavtcng

cao.

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V D: Lc Tng Tc

Khi khong cch gia cc phn t rt nh, thgia cc phn t:

A. Chc lcht.

B. Chc lcy.C. C c lcht v lc y, nhng lc y ln hn

lcht.D. C c lcht v lcy,nhng lcynhhn

lcht.

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V D: Cc th rn, lng, kh

Tnh cht no sau y khng phil ca phnt ca vt cht th kh?

A. Chuynnghnlon.

B. Chuynngkhng ngng.

C. Chuynnghnlonv khng ngng.

D. Chuyn ng hn lon xung quanh cc v tr cn

bngcnh.

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V D: Cc Thng S Trng Thi

Trong cc i lng sau y, i lng nokhng phil thng s trng thi ca mtlng kh?

A. Thtch.B. Khilng.

C. Nhittuyti.

D. p sut.

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S N V Nhit Ca Cht Rn

Cng thc trn c suy ra t thc nghim.

S tng di ca vt rn khi nhit tnggi l s n di v nhit.

l h s n di ca vt rn, c n v l K-1.

Gi tr ca ph thuc vo cht liu ca vtrn.

o 1 t l l

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V D: S N Di

150C, mi thanh ray ca ng st di 12,5m. Hi khe h gia hai thanh ray phi c rng ti thiu bng bao nhiu cc thanh

ray khng b cong khi nhit tng ti 500C?Bit h s n di ca st l 11.10-6 K-1.

p s: 4,81 mm

Tm tt Gii

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Tm tt

t1= 150C

l= 12,5 m

t2= 500C

l= ?

Gii

rng ti thiu:

o

o 2 1

6

t

t t

11.10 . 12,5 . 50 15

4,81 mm

l l

l l

l

l

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S N Khi

S tng th tch ca vt rn khi nhit tnggi l s n khi.

l h s n khi, 3.

oV V 1 t

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V D: S N Khi

Tnh khi lng ring ca st 8000C. Bitrng khi lng ring ca st 00Cl 7,8.103

kg/m3v h s n di ca st l 11.10-6 K-1.

p s: 7,599.103kg/m3

Tm tt Gii

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Tm tt

t0= 00C

D0= 7,8.103 kg/m3

= 11.10-6K-1

t = 8000C

D = ?

Gii

Khi lng ring ca st 8000C:

00

0

00

0

30

6

3 3

m

VD V

mD V

V

V 1 tD1 3 . t

D V

D 7,8.10D

1 3 . t 1 3. 11.10 800 0

D 7,599.10 kg / m

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ng Dng

Phi tnh ton khc phc tc dng chi ca s n v nhit.

Rle nhit ng ngtin t ng.

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Hin Tng Cng B Mt Ca Cht Lng

F gi l lc cng b mt.

l h s cng mt ngoi, c n v l N/m.

H s ph thuc vo bn cht v nhit

ca cht lng: gim khi nhit tng. c bit: i vi vt c tit din trn th

F . l

2 R D l

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ng Dng

o ma. X phng.

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V D: Tch Vnh Kim Loi

Mt vng xuyn c ng knh ngoi l 44mmv ng knh trong l 40 mm. Trnglng ca vng xuyn l 45 mN. Lc bt vng

xuyn ny ra khi b mt ca glixerin 200Cl 64,3 mN. Tnh h s cng b mt caglixerin nhit ny.

p s: 73.10-3N/m

Tm tt Gii

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Tm tt

D = 44 mm

d = 40 mm

P = 45 mN

F = 64,3 mN

= ?

Gii

Sc cng mt ngoi:

FC

= FP = 64,345 = 19,3(mN).

H s cng mt ngoi:

C

3

3

3

F

D d19,3.10

3,14 44 40 .10

73.10 N / m

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Hin Tng Dnh t

Git nc nh ln bn thy tinh s b lan rng rathnh mt hnh dng bt k, v nc dnh t thytinh.

Git nc nh ln bn thy tinh ph mt lp nilons vo trn li v b dt xung do tc dng ca trnglc, v nc khng dnh t vi nilon.

B mt cht lng st thnh bnh cha n c dngmt khum lm khi thnh bnh b dnh t v cdng mt khum li khi thnh bnh khng b dnht.

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Mao Dn

ng thy tinh cng knh trong nh. Hin tng mc cht

lng bn trong cc

ng c ng knh nhlun dng cao hn,hoc h thp hn sovi b mt cht lng

bn ngoi ng gi lhin tng mao dn.

4h

gd

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V D: Mao Dn 1 ng

Tm h s cng b mt ca nc nu ngmao dn c ng knh trong l 1 mmvmc nc trong ng dng cao 32,6 mm. Bit

khi lng ring ca nc l 1000 kg/m3. Lyg = 9,8 m/s2.

p s: 79,87.10-3N/m

Tm tt Gii

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Tm tt

d = 1 mm

h = 32,6 mm

g = 9,8 m/s2

= 1000 kg/m3

= ?

Gii

H s cng mt ngoi:

3 3 3

3

4h gd

h gd

4

32,6.10 10 9,8 10

4

79,87.10 N / m

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V D: Mao Dn 2 ng

Nhng thng ng hai ng mao dn thytinh c ng knh trong ln lt l 1 mmv2 mmvo thy ngn. Hi chnh lch gia

hai mc thy ngn bn trong hai ng maodn bng bao nhiu? Cho bit h s cngb mt ca thy ngn l 0,47 N/mv khi

lng ring ca thy ngn l 13.600 kg/m3

. p s: 7 mm

Tm tt Gii

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Tm tt

d1= 1 mm

d2= 2 mm

= 0,47 N/m

= 13.600 kg/m3

h = ?

Gii

chnh lch gia hai mcthy ngn:

1 2

1 2

1 2

3 33

h h h

4 4h

gd gd

4 1 1h

g d d

4. 0,47 1 1h

10 2.1013, 6.10 9,8

h 7 mm

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S Nng Chy

Qu trnh chuyn t th rn sang th lng.

Nhit nng chy xc nh.

Nhit nng chy:

l nhit nng chy ring ph thuc vo

bn cht ca cht rn nng chy, c n vl J/kg.

mQ

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V D: S Nng Chy

Tnh nhit lng cn cung cp cho mingnhm khi lng 100 g nhit 200C, n ha lng nhit 6580C. Nhm c nhit

dung ring l 896 J/(kg.K), nhit nng chyring l 3,9.105 J/kg.

p s: 96,165 kJ

Tm tt Gii

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Tm ttm = 100 g

t1= 200C

t2= 6580C

c = 896 J/(kg.K)

= 3,9.105

J/kgQ = ?

Gii Nhit lng cn cung cp cho

ming nhm:

5

Q mc t m

Q 0,1 896 658 20 3,9.10 0,1

Q 96,165 kJ

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S Bay Hi S Ngng T

Qu trnh chuyn t th lng sang th kh b mt cht lng.

Qu trnh t th kh sang th lng gi l sngng t.

S bay hi xy ra nhit bt k v lunkm theo s ngng t.

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S Si

S chuyn t th lng sang th kh xy ra c bn trong v trn b mt cht lng.

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Tm Tt Cc Qu Trnh

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Tm Tt

Kh l tng Cc nh lut thc nghim: ng nhit, ng

tch, ng p

Phng trnh trng thi kh l tng Phng trnh Clapeyron - Mendeleev

Kh thc Cht rn Cht lng

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CHNGII

NHITNGLCHC04/10/2015

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Ni Nng

Ni nng ca vt l tng ng nng v thnng ca cc phn t cu to nn vt.

Ni nng ca mt vt ph thuc vo nhit v th tch ca vt: U = f(T, V).

bin thin ni nng.

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Cc Cch Thay i Ni Nng

Thc Hin Cng: c s bin i qua ligia ni nng v dng nng lng khc.

Truyn nhit: khng c s chuyn ho

nng lng t dng ny sang dng khcm ch c s truyn ni nng t vt nysang vt khc.

Nhit lng: Q = mct

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V D: Cn Bng Nhit

Mtbnh nhm khilng0,5 kgcha0,118 kgncnhit200C. Ngita thvo bnh mtming st khi lng 0,2 kg c nung nngti750C. Xc nhnhitcanc khi btuc s cn bng nhit. B qua s truyn nhit rami trngbn ngoi. Nhitdung ring canhml 0,92.103 J/(kg.K); ca nc l 4,18.103 J/(kg.K);castl 0,46.103 J/(kg.K). ps: 250C

Tm tt Gii

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mAl= 0,5 kg

mH2O

= 0,118 kg

mFe= 0,2 kg

t1= 200C

t2= 750

CcAl= 0,92.10

3 J/(kg.K)

cH2O= 4,18.103 J/(kg.K)

cFe= 0,46.103 J/(kg.K)tcb= ?

Nhit lng thu (nc):QH2O= mH2OcH2O.(tcbt1)

= (0,118).(4,18.103).(tcb20) Nhit lng thu (nhm):QAl = mAlcAl.(tcbt1)

= (0,5).(0,92.103).(tcb20) Nhit lng ta (st):QFe = mFecFe.(t2tcb)= (0,2).(0,46.103).(75tcb) Cn bng nhit:

Qta= Qthutcb= 25

0C

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Nguyn L I Nhit ng Lc Hc

Cng thc:U = A + Q

Qui c du: U > 0: ni nng tng; U < 0: ni nng gim.

A > 0: h nhn cng; A < 0: h thc hincng.

Q > 0: h nhn nhit; Q < 0: h truynnhit.

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Nguyn L I Nhit ng Lc Hc

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V D: Nguyn L I

Ngita cung cpcho kh trong mtxilanh nmngang nhit lng1,5 J. Kh n ra ypit-tng imton5 cmvimtlcc lnl 20 N. Tnh

binthin ninngcakh. ps: 0,5 J

Tm tt Gii

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Q = 1,5 J

S = 5 cm

F = 20 N

U = ?

Cng thc hin:

A = F.S = 20.0,05 = 1 (J)

bin thin ni nng:

U = Q + A = 1,51 = 0,5 (J)

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V D: Nguyn L I

Ngita thchincng 100 Jnn kh trongmt xilanh. Tnh bin thin ni nng ca kh,bit kh truyn ra mi trng xung quanh nhit

lng20 J. ps: 80 J

Tm tt Gii

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A = 100 J

Q = 20 J

U = ?

bin thin ni nng:

U = Q + A =20 + 100 = 80 (J)

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V D: Nguyn L I

Ngita truyncho kh trong xilanh nhitlng100 J. Kh n ra thc hin cng 70 J y pit-tngln. Tnh binthin ninngcakh.

ps: 30 J

Tm tt Gii

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Q = 100 J

A = 70 J

U = ?

bin thin ni nng:

U = Q + A = 10070 = 30 (J)

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V D: Nguyn L I

Khi truynnhitlng6.106Jcho kh trong mtxilanh hnh tr th kh n ra y pit-tng ln lmth tch ca kh tng thm 0,5 m3. Tnh bin

thin ninngcakh. Bitp sutcakh l 8.106

N/m2v coi p sutny khng itrong qu trnhkh thchincng.

ps: 2.106J

Tm tt Gii

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Q = 6.106 J

V = 0,5 m3

P = 8.106 N/m2

U = ?

Cng thc hin:A = F.h = P.S.h = P.V = 8.106.0,5 = 4.106(J)

bin thin ni nng:U = Q + A = 6.1064.106= 2.106(J)

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Thun Nghch Bt Thun Nghch

Qu trnh thunnghch l qu trnh vtttrvtrngthi ban um khng cnnscanthipcavtkhc.

Qu trnh khng thunnghch l qu trnh chc thxyra theo mtchiuxc nh,khng thtxyra theo chiungcli. Munxyra theochiungcliphicnnscan thipcavt

khc.

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Nguyn L II Nhit ng Lc Hc

Clausius: Nhitkhng thttruyntmtvtsang mtvtnng hn. Carnot: ng c nhit khng th chuyn ha

ttcnhitlngnhncthnh cng chc.

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ng C Nhit

Ngun nng cungcpnhitlng(Q1). B phn pht nggm vt trung gian nhn

nhit sinh cng (A) gi ltc nhn v cc thit bpht ng. Ngun lnh thu

nhit lng do tc nhntara (Q2).

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Vn Dng

Hiusutcangcnhit:

Hiusutccicangcnhit:

1Q

QQ

Q

|A|H

1

21

1

1T

TTQ

|A|H1

21

1

maxmax

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Vn Dng

Hiunngcamy lnh:

Hiunngccicamy lnh:

1QQ

Q

A

Q

21

22

1TT

T

21

2max

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V D: Nguyn L II

Mtngcnhitlm vicsau mtthigian thtc nhn nhntngunnng nhitlngQ1=1,5.106J, truyncho ngun lnhnhit lngQ2=

1,2.106

J. Hy tnh hiu sut thc ca ng cnhitny v so snh n vihiusutccinunhitcangunnng v ngunlnhlnltl2500Cv 300C.

ps: 20%; 42%

Tm tt Gii

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Q1= 1,5.106 J

Q2= 1,2.106J

t1= 2500C

t2= 300C

H = ?Hmax= ?

Hiu sut thc:

Hiu sut cc i:

1 2

1

6 6

6

Q Q

H Q

1, 5.10 1, 2.10H 0, 2 20%

1,5.10

1 2

max1

max

T TH

T

250 30H 0, 42 42%

250 273

V D N L II

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V D: Nguyn L II

mtngcnhit,nhitcangunnngl 5200C, cangunlnhl 200C. Hicng ccim ngcthchincnun nhntngun

nng nhit lng107

J? Cng cci l cng mngcnhitsinh ra nuhiusutcan l cci.

ps: 6,3.106J

Tm tt Gii

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Q1= 107J

t1= 5200C

t2= 200C

Amax= ?

Hiu sut cc i:

Cng cc i:

1 2

max1

max

T T

H T

520 20H 0,63 63%

520 273

maxmax

1

max max 1

7 6

max

AH

QA H .Q

A 0,63.10 6,3.10 J

V D N L II

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V D: Nguyn L II

ginhit trong phng 200C, ngi tadng mtmy lnh(trong trnghpny ngita gi l my iu ha khng kh) mi gi tiu

thcng bng5.106

J. Tnh nhitlnglyitkhng kh trong phng trong mi gi, bit rnghiunngcamy lnhl = 4.

ps: 2.107

J

Tm tt Gii

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A = 5.106J

= 4

Q2= ?

Nhit lng ly i:

2

2

6

2

7

2

Q

A

Q .A

Q 4.5.10

Q 2.10 J

V D N L II

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V D: Nguyn L II

Hiu sut thc ca mt my hi nc bngna hiu sut cc i. Nhit ca hi khi rakhil hi(ngunnng) l 2270Cv nhitca

bung ngng (ngun lnh) l 770

C. Tnh cngsutcamy hincny numigin tiuth700 kgthan c nngsuttanhitl 31.106J/kg.

ps: 904 kW

Tm tt0

Giimax 1 2H T TH

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t1= 2270C

t2= 770C

m = 700 kg

Q1= 31.106 J/kg

P = ?

1

1

1

6

AH

Q

A H.QH.QA

t t

0,15 .(31.10 )

P 3600

max 1 2

1

H2 2T

227 77

H 0,15 15%2 227 273

700 kgP P.700 904 kW

C D C T C Th

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Cc Dng Cng Trong C Th

Cng ha hc: l cng sinh ra khi tnghpcchp cht cao phn t (protid, acid nucleotid,)t cc cht c trng lng phn t thp v khithchincc phnngha hcxc nh. Cng c hc: l cng sinh ra khi dch chuyncc bphn,cquan trong c th hay ton bcthnhlcchc(co c).

C D C T C Th

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Cc Dng Cng Trong C Th

Cng thmthu: l cng sinh ra khi vnchuyncc chtkhc nhau qua mng hay qua cc hamng t vng c nng thp sang vng cnngcao hn. Cng in: l cng sinh ra khi vnchuyn cchtmang in (cc ion) trong in trng, sinhra in th sinh vt v dn truyn kch thch

trong tbo.

N L T C Th

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Nng Lng Trong C Th

Nng lng trong c th l nng lng ha hcca thc n (protid, lipid, glucid) ta ra khi b oxiha. ivithcvt,l nnglngMtTridtrtrong qu trnh quang hp. Nng lng ny cngcngvtsdngkhi nthcvt.

u tin nng lngcchuyn thnh lin ktcao nng ca nhng cht no m ch yu ladenosin triphosphat (ATP). Sau ATP phn hytrong nhng t chc tng ng ca t bo v giiphng ti ngun nng lng cn thit sinh

cng.ATP + H2O ADP + H3PO4 + (7 ~ 8,5 Kcal)

Nhit S C (Nhit C B )

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Nhit S Cp (Nhit C Bn)

Nhit s cp l lng nhit tn x trong qutrnh trao i cht, c qui nh bi tnh btthun nghch ca cc phn ng ha sinh v lsinh.

Nhit s cp t l vi cng ca qu trnhtrao i cht v t l nghch vi hiu sut cachng.

Nhit Th Cp (Nhit Hot ng)

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Nhit Th Cp (Nhit Hot ng)

Nhitthcp l lngnhit sinh ra trong ccqu trnh sinh cng khc nhau.

Nhitthcpt lvihot tnh cam, hot

ngco c.

Nguyn L I H Thng Sng

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Nguyn L I H Thng Sng

Nguyn l I nhitnglchcp dngcho hthngsngcvit:

Q = M + E + A

y l phngtrnh cbnvcn bngnhitivicthngi. Q: nhit lng sinh ra trong qu trnh ngha thcn.

M: nng lng d tr di dng ha nng(phn tATP, ADP,) trong cth.

Nguyn L I H Thng Sng

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Nguyn L I H Thng Sng

E: nnglngmtmt vo mi trngxungquanh (nnglngkhuychtn). A: cng chccththchinivimi

trngbn ngoi. Gi tr nng lng nhn c bng phngphp onhitgin tipgnnhtrng vinhitlngobngphngphp trc tip trong th

nghim(chnh xc ti1%).

V d: Cn bngnnglngcangitrong mtngy m.

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Mttrong nhngnguyn nhn cn sai lchl d cthngyn thcng sinh ra cngkhng thbngkhng c.

a vo (Kcal) Thi ra (Kcal)

Thcn

56,8g protid 237

Nhitthiqua da 1374

Khi thra 43

140g lipid 1307

Phn v nctiu 23

Bay hiqua ngh hp 181

79,9g gluxid 335

Bay hiqua da 227

Bchnh 11

Tngcng: 1879 Tngcng: 1859

H Qu: nh Lut Hess

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H Qu: nh Lut Hess

Hiungnhit(Qv v Qp) camtqu trnh hahc phc tp khng ph thuc vo cc giai ontrung gian m ch ph thuc vo trng thi u vtrngthi cuicaqu trnh.

nh lutcho ta tm chiungnhitcaccphnng. Khng thxyra trong iukinth nghim.

Khng o trc tip hiu ng nhit ca chngc.

Entropy

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Entropy

Trong nhit ng lc hc, entropy nhit nglchc (hay gingin l entropy) dS l tsgia nng lng dQ pht tn hay hp th khi

mthvt l chuyntrngthi timtnhittuytixc nhT:

T

dQdS

Bin Thin Entropy

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Bin Thin Entropy

bin thin entropy ca mt h trongqu trnh thun nghch bt k bin i ttrngthi (1) sang trngthi (2):

12 SSS

Tnh Cht Ca Entropy

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Tnh Cht Ca Entropy

S l mt hm ca trng thi (ch ph thuctrngthi uv trngthi sau).

S cxc nhsai khc mthngs.

S c tnh cngc. Entropy cahbngtngcc entropy tngphnring r.

Ngha Thng K Ca Entropy

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Ngha Thng K Ca Entropy

ctrngcho mchn lon,v trttvphn bphn tcamth.

Cc qu trnh dinbintnhin trong mth

c lpsc chiusao cho entropy S tng. Httrngthi cn bngkhi S cci(phn bngu).

Nguyn L Tng Entropy

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Nguyn L Tng Entropy

Trong mt h c lp th mi qu trnh xy rau c chiu sao cho entropy tng hoc khngiS 0.

Nhnxt: Qu trnh thunnghch: S = 0, S = const.

Qu trnh btthunnghch: S > 0, S tng.

Khi h t n trng thi cn bng S = Smaxth qu trnh btthunnghchktthc.

V D: Ni Nng

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V D: Ni Nng

Ni nng ca mt vt l:A.Tngngnngv thnngcavt.

B.Tng ng nng v th nng ca cc phn t

cutonn vt.C.Tngnhitlngv cnngm vtnhnctrong qu trnh truynnhitv thchincng.

D.Nhit lng vt nhn c trong qu trnhtruynnhit.

V D: Ni Nng

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V D: Ni Nng

Cu no sau y ni v ni nng l khngng:

A.Ninngl mtdngnnglng.

B.Ni nng c th chuyn ha thnh cc dngnnglngkhc.

C.Ninngl nhitlng.

D.Ninngcamtvtc thtngln, gimi.

V D: Nhit Lng

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V D: Nhit Lng

Cu no sau y ni v nhit lng l khngng:

A.Nhit lng l s o tng ni nng ca vttrong qu trnh truynnhit.B.Mtvtlc no cngc ninng,do lc nocngc nhitlng.C. n v ca nhit lng cng l n v ca ni

nng.D.Nhitlngkhng phil ninng.

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CHNGIII

CHCCHTLU04/10/2015

Khi Lng Ring

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Khi Lng Ring

Khilngring cachtlng(hay chtrn)lmt i lng vt l c o bng khi lngcamtnvthtch cht.

Khilngring k hiul D.

mD

V

p Sut Thy Tnh

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p Sut Thy Tnh

Cht lng c c tnh l nn ln cc vt nmtrong n. Lc m cht lng nn ln vt cphngvung gc vibmtcavt.

S

Fp

S Thay i p Sut Theo Su

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S Thay i p Sut Theo Su

p: p sutsu h. pa: p sutkh quyn.

: khilngring.

h: su.

ap p gh

Nguyn L Pascal

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Nguyn L Pascal

tngp sut ln mtcht lngcha trongbnh kn ctruynnguyn vn cho miimcachtlngv thnh bnh.

ngp p gh

My p Dng Cht Lng

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My p Dng Cht Lng

Nguyn l Pascal c p dng nhiu trong kthut v i sng, nh my p dng cht lng,my nng vtc trnglngln,phanh thylc

trong cc xe my, t

1 2

1 2

F F

S S

V D: p Sut

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V D: p Sut

Hy tnh p suttuytip su 1000 mdi mc nc bin. Cho khi lng ring cancbinl 103kg/m3v pa= 1,01.10

5N/m2.

ps: 9,9.106Pa

Tm tth = 1000 m

Gii p sut su 1000 m:

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h = 1000 m

= 103kg/m3

pa= 1,01.105N/m2

p = ?

p sut su 1000 m:

p = pa+ gh

= 1,01.105+ 103.9,8.1000

= 9,9.106(Pa)

V D: My Nng Thy Lc

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V D: My Nng Thy Lc

Mtmy nng thylccatrmsacha tdng khng kh nn ln mtpittong c bn knh5 cm. p sut c truyn sang mt pittong

khc c bn knh 15 cm. Hikh nn phitoramtlct nhtbngbao nhiu nng mt tc trnglng13000 N? p sutkh nn khi bngbao nhiu?

ps: 1444,44 N; 1,84.105Pa

Tm ttR1 = 5 cm = 0 05 m

Gii 1 1F S

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R1 5 cm 0,05 m

R2= 15 cm = 0,15 m

P = F2= 13000 N

F1= ?

p = ?

2 2

2

111 2 2 2

2 2

2

1 2

F S

RSF F . F .S R

0,05F 13000. 1444, 44 N

0,15

1 1

2

1 1

2

F Fp

S R

1444,44p 184.000 Pa

3,14.0,05

V D: Lc p Ton Phn

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V D: Lc p Ton Phn

Cangoi mtnh rng3,4 m; cao 2,1 m. Mttrn bo i qua, p sut bn ngoi gim i cn0,96 atm. Trong nh p sutvngi1 atm. Hi

lcp ton phnvo cal bao nhiu? Ch : 1atm = 1,013.105N/m2.

ps: 2,89.104N

Tm tta = 3,4 m

Gii 1 2F p.S p p S

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a 3,4 m

b = 2,1 m

p1= 0,96 atm

p2= 1 atm

F = ?

1 2

5

4

p p p

F 1 0,96 . 1,013.10 . 3, 4.2,1

F 2,89.10 N

Chuyn ng Ca Cht Lng L Tng

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Chuyn ng Ca Cht Lng L Tng

Chuynngcachtlngchia thnh hai loi: Chynnh(hay chythnh dng). Chykhngnnh(hay chycunxoy). Chuynngcachtlngl tngthamn cc

iukin: Chtlngkhng nht,tcl bqua ma st trong

lng chtlng. Schylnnhhay thnh lp,thnh dng.

Chtlngkhng chunn, tcl khilngringcachtlngkhng i.

H Thc Gia Vn Tc V Tit Din

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H Thc Gia Vn Tc V Tit Din

Hthcgiavntcv titdin:

LulngA c gi trnhnhau miim.

1

2

2

1

S

S

v

v

ASvSvSS

vv 2211

1

2

2

1

nh Lut Bernoulli

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nh Lut Bernoulli

21p v const2

Chng Minh

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Chng Minh

V D:

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V D:

Lulngnctrong mtngnmngang l 2m3/pht. Hy xc nh tc ca cht lng timtimcangc bn knh 10 cm.

ps: 1,06 m/s

Tm ttA = 2 m3/pht

Gii Tc cht lng:

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/p

r = 10 cm

v = ?

g

22A A 2

v 1,06 m / sS .r 60.3,14. 0.1

V D:

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Tit din ng mch ch ca ngi l 3 cm2,tc mu t tim ra l 30 cm/s. Tit din cami mao mch l 3.10-7 cm2; tc mu trong

mao mch l 0,05 cm/s. Hi ngi phi c baonhiu mao mch?

ps: 6 t

Tm ttS0= 3 cm

2Gii Ton b mu i qua cc mao

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0

v0= 30 cm/s

S = 3.10-7cm2

v = 0,05 cm/s

n = ?

qmch u phi qua ng mch

ch. Ta c:

0 0

0 0

9

7

S .v n.S.v

S .vn

S.v

3.30n 6.10

3.10 .0,05

V D:

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Gii Ta c:

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1 1 2 2 2

2 2 4

1 1 2 2 2

v S v S v 8 m / s

1 1p v p v p 5.10 Pa

2 2

o p Sut Tnh

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p

t mtng hnh tr h hai u, sao chomingng song song vi dng chy. Bit titdincangv cao cactchtlng,ta tnh

cp suttnhcang.

o p Sut Ton Phn

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p

Dng mtng hnh tr h hai u, mt ucunvung gc. tngsao cho mingngvung gc vidng chy. Bit titdincang

v cao cactchtlng,ta tnh cp sutton phntiimtng.

o Vn Tc Cht Lng

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g

ng Venturi c t nm ngang, gm mtphntitdinS v mtphnc titdins. Mtp khnh chU, c hai univihai ng,

cho bithiup suttnhgiahai titdin. Bithiup sutp v cc titdinS, s ta c thtnhvntc:

)sS(ps2v

22

2

o Vn Tc Cht Lng

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g

ng Venturi c tnm ngang, gm mtphn tit din S v mtphn c tit din s. Mtp k hnh ch U, c haiu ni vi hai ng ,cho bithiup suttnhgia hai tit din. Bithiu p sut p v cc

tit din S, s ta c thtnh vntc:

)sS(

ps2v

22

2

o Vn Tc My Bay

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y y

ng Pito c gn vo cnh my bay, dngkhng kh bao xung quanh ng. Vn tc khchy vung gc vi tit din S ca mtng

nhnh ch U. Nhnh kia thng qua mt bungbng p sut tnh ca mt dng khng kh bnngoi. chnh cahai mcchtlngtrongngch U cho php ta tnh c vn tc ca dng

kh tcl vntccamy bay.

kk

gh2v

o Vn Tc My Bay

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o Vn Tc My Bay

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V D:

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Mi cnh my bay c din tch l 25 m2. Bitvn tc dng khng kh pha di cnh l 50m/s, cn pha trn cnh l 65 m/s, hy xc nh

trng lng ca my bay. Gi s my bay baytheo ngnmngang vivntckhng ivlcnng my bay chdo cnh gy nn. Cho bitkhilngring cakhng kh l 1,21 kg/m3.

ps: 52181 N

Tm ttS = 25 m2

Gii Ta c:

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v1= 50 m/s

v2= 65 m/s

KK= 1,21 kg/m3

S: din tch cnhmy bay

P = ?

Trng lng my bay:

2 2

1 1 2 2

2 2

2 1 1 2

1 1p v p v

2 21

p p v v2

2 1

2 2

1 2

F p p .S.2

1F v v .S.2 52181 N2

V D:

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Mt ngi thi khng kh vi tc 15 m/sngang qua ming mt nhnh ng ch U chanc. Hichnh mcncgiahai nhnh l

bao nhiu? ps: 1,4 cm

Tm ttv = 15 m/s

Gii Ta c:

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h = ?

chnh mc nc:

2 2

1 1 2 2

2 2 2

2 1 1 2

1 1

p v p v2 2

1p p v v 136,25 N / m

2

p

0,014 m 1,4 cm

g

l

V D: p Sut

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Cu no sau ykhng ng:A.Khi xungcng su trong ncth ta chumtpsutcng ln.B.p sutcachtlngkhng phthuckhilng

ring cachtlng.C. chnh p sut ti hai v tr khc nhau trongchtlngkhng phthucp sutkh quynmtthong.

D. tngp sut ln mtbnh kn c truyninguyn vnkhpbnh.

V D: nh Lut Bernoulli

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Cu no sau ykhng ng:A. Trong mtngdng nmngang, nino c tc lnth p suttnhnh,nino c tcnhth p suttnhln.

B. nh lut Bernoulli p dng cho cht lng v cht khchynnh.C. p sut ton phn ti mt im trong ng dng nmngang th tlbcnhtvivntcdng.

D. Trong mtng dng nm ngang, ni no cc ngdng cng nmxt nhau th p suttnhcng nh.

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HTCHUYN NHIT HC

04/10/2015