Fahmidda Mabaadi'da Madow
Haddii aan ku qasno biyo kulul iyo kuwo qabow weel furan (tusaale ahaan, baaldi), kulaylku wuxuu ka gudbaa biyaha kulul una gudbaa biyaha qabow. Maadaama weelku furan yahay, qaar ka mid ah kulaylku wuxuu u gudbaa hawada. Weelku sidoo kale wuu kululaadaa. Kulaylka ay sii daayaan biyaha kulul ma aha oo kaliya biyaha qabow laakiin sidoo kale hawada iyo weelka. Xaaladdan, baaldigu waa nidaam aan go'doon ahayn.
Marka aan biyaha kulul iyo kuwa qabow ku qasno heerkulbeeg xiran, kulaylku wuxuu ka gudbaa biyaha kulul una gudbaa biyaha qabow. heerkulbeeggu waa nidaam go'doon ah , sidaa darteed kuleylku uma gudbo hawada ama heerkulbeeggu. Kulaylka ay sii daayaan biyaha kulul waxaa si fudud u nuuga biyaha qabow ilaa isku darku gaaro dheelitirka kulaylka.
Mabda'a Black wuxuu sheegayaa in nidaamka xiran ee go'doonsan, kulaylka uu sii daayo shay heerkul sare leh = kulaylka uu nuugo shay heerkul hooseeya.
Qaacidada aasaasiga ah ee Madow
Qlepas =Qnuugid
Sharaxaad: Qlepas = kulayl ay sii daayaan shay heerkul sare leh, Qnuugid = kulayl ay nuugaan walxaha heerkulka hooseeya.
Su'aalaha tusaalaha ah ee mabda'a Black
1. Bir kulul oo miisaankeedu yahay 1 kg iyo heerkulkeedu yahay 100 oC waxaa lagu shubaa weel ay ku jiraan biyo miisaankoodu yahay 2 kg iyo heerkulkiisu yahay 20 oC. Waa maxay heerkulka ugu dambeeya ee isku darka? Kulayl gaar ah oo bir ah = 450 J/kg Co, kulayl gaar ah oo biyo ah = 4200 J/kg Co.
Dood:
Waa la garanayaa :
bir m = 1 kg, Tbesi = 100 oC, m biyo = 2 kg, Thawada = 20 oC, c birta = 450 J/kg Co, c biyo = 4200 J/kg Co.
La weydiiyay :
Isku darka T?
Jawab :
Birtu waxay leedahay heerkul ka sarreeya biyaha sidaa darteed birtu waxay sii deysaa kulaylka, biyuhuna waxay nuugaan kulaylka.
(m)(c)(delta T) = (m)(c)(delta T)
(1)(450)(100-T) = (2)(4200)(T-20)
(450)(100-T) = (8400)(T-20)
45000 – 450T = 8400 T – 168000
45000 + 168000 = 8400 T + 450 T
213000 = 8850 T
T = 213000: 8850
T = 24 oC
Suuye Heerkulka kama dambaysta ah ee isku darka birta kulul iyo biyaha qabow marka labaduba ay ku jiraan dheelitirka kulaylka waa 24 oC.
2. Imtixaanka Qaranka ee Fiisigiska SMA/MA 2015/2016 Su'aal Lambarka 20
75 garaam oo biyo ah heerkul ah 0 o C ayaa lagu qasi karaa 50 garaam oo biyo ah si heerkulka isku darka uu u noqdo 40 o C, sidaas darteed heerkulka bilowga ah ee 50 garaam oo biyo ah waa...
A. 125 o C
B. 120 o C
C. 110 o C
D. 100 o C
E. 80 o C
Dood
Waa la ogyahay in:
Cufka biyaha 1 (m1 ) = 75 garaam
Heerkulka bilowga ah ee biyaha 1 (T 1 ) = 0 o C
Cufka biyaha 2 (m2 ) = 50 garaam
Heerkulka kama dambaysta ah ee isku darka (T) = 40 o C
Cutubyada cufku waa isku mid sidaa darteed looma baahna in la beddelo.
La Rabay: Heerkulka bilowga ah ee biyaha 2 ( T2 )
Jawaab:
Mabda'a Black wuxuu sheegayaa in nidaamka xiran ee go'doonsan, kulaylka uu sii daayo shay heerkul sare leh uu la mid yahay kulaylka uu nuugo shay heerkul hoose leh.
Q la sii daayay = Q la nuugo
m 1 c ( Δ T 1 ) = m 2 c ( Δ T 2 )
Kulaylka gaarka ah ee biyuhu waa isku mid sidaa darteed c waa laga saarayaa isla'egta.
m 1 ( Δ T 1 ) = m 2 ( Δ T 2 )
(75)(40 – 0) = (50)(T 2 – 40)
(75)(40) = (50)(T 2 – 40)
3000 = 50 T 2 – 2000
3000 + 2000 = 50 T 2
5000 = 50 T 2
T 2 = 100
Jawaabta saxda ah waa D.
3. Imtixaanka Qaranka ee Fiisigiska SMA/MA 2015/2016 Su'aal Lambarka 20
Bir 200 garaam ah ayaa lagu kululeeyaa ilaa 120 o C ka dibna lagu shubaa 100 garaam oo biyo ah heerkul ah 30 o C. Ka dib dheelitirka, heerkulka isku darka wuxuu noqonayaa 60 o C. Haddii kulaylka gaarka ah ee biyuhu yahay 4200 J.kg -1 o C -1 , markaa kulaylka gaarka ah ee birta waa...
A. 10,5 J.kg -1o C -1
B. 105 J.kg -1o C -1
C. 150 J.kg -1o C -1
D. 1050 J.kg -1o C -1
E. 1500 J.kg -1o C -1
Dood
Beddelka cutubyada cufka laga bilaabo garaam ilaa kiilogaraam (Nidaamka Caalamiga ah ee Cutubyada)
Waa la ogyahay in:
Cufka birta (m 1 ) = 200 garaam = 0,2 kg
Heerkulka birta (T 1 ) = 120 o C
Cufka biyaha (m2 ) = 100 garaam = 0,1 kg
Heerkulka biyaha (T 2 ) = 30 o C
Heerkulka kama dambaysta ah ee isku darka (T) = 60 o C
Kulayl gaar ah oo biyo ah (c 2 ) = 4200 J.kg -1 o C -1
La weydiiyay: Kulayl gaar ah oo bir ah (c 1 )
Jawaab:
Q la sii daayay = Q la nuugo
m1 c1 (ΔT1) = m2 c2 (ΔT2)
(0,2)(c 1 )(120 – 60) = (0,1)(4200)(60 – 30)
(0,2)(c 1 )(60) = (0,1)(4200)(30)
12 c 1 = 12600
c 1 = 12600 / 12
c 1 = 1050
Jawaabta saxda ah waa D.
4. Su'aalaha Imtixaanka Qaranka ee SMA/MA 2018
250 ml oo biyo karkaraya (100 o C) ayaa lagu shubaa digsi ay ku jiraan 400 ml oo biyo ah heerkul ah 35 o C. Ka dib marka dheelitirka kulaylka dhaco, heerkulka isku darka waa… (kuleylka gaarka ah ee biyaha 1,0 kalori -1 . o C -1 )
Dood
Waa la ogyahay:
Heerkulka biyaha kulul = 100 o C
Mugga biyaha kulul = 250 ml = 0,25 litir = 0,25 dm 3 = (0,25)(1000 cm 3 ) = 250 cm 3
Heerkulka biyaha qabow = 35 o C
Mugga biyaha qabow = 400 ml = 0,4 litir = 0,4 dm 3 = (0,4)(1000 cm 3 ) = 400 cm 3
Kulayl gaar ah oo biyo ah = 1,0 kalori.gr -1 . o C -1
Cufnaanta biyaha = 1 gr/cm 3
La weydiiyay: Heerkulka isku darka
Jawaab:
Xisaabi cufka biyaha kulul iyo kuwa qabow adigoo isticmaalaya qaacidada cufnaanta:
m = ρ V
m = cufnaan, ρ = cufnaan, V = mug
Cufka biyaha kulul: m = ρ V = ( 1 gr/cm 3 ) (250 cm 3 ) = 250 garaam
Cufnaanta biyaha qabow: m = ρ V = ( 1 gr/cm 3 ) (400 cm 3 ) = 400 garaam
Xisaabi heerkulka isku darka adoo isticmaalaya qaacidada mabda'a Black:
Kulaylka ay nuugaan biyaha qabow = kulayl ay sii daayaan biyaha kulul
mc Δ T = mc Δ T
Kulaylka gaarka ah waa isku mid sidaa darteed waxaa laga saarayaa isla'egta.
m Δ T = m Δ T
(400)(T-35) = (250)(100-T)
400T – 14000 = 25000 – 250T
400T + 250T = 25000 + 14000
650T = 39000
T = 39000 / 650
T = 60 o C
