15 Tusaalooyin oo ku saabsan Su'aalaha Heerkulka iyo Kulaylka
Kala-hagaajinta Heerkulbeeg
1. Heerkulbeegga la hagaajiyay ee X wuxuu muujinayaa -30 °C barta barafowga biyaha iyo 90 °C barta karkaraya biyaha. Heerkulka 60 °C ee X wuxuu la mid yahay...
A. 20 o C
B. 45 o C
C. 50 o C
D. 75 o C
E. 80 o C
Dood
Waa la ogyahay in:
Barta qaboojinta biyaha ee heerkulbeegga X = -30 o
Barta karkaraya ee biyaha heerkulbeegga X = 90 o
Su'aal: 60 o X = ….. o C
Jawaab:
Miisaanka Fahrenheit, barta qaboojinta biyaha waa 32 o F halka barta karkaraya biyuhuna ay tahay 212 o F. Inta u dhaxaysa barta qaboojinta iyo barta karkaraya biyaha waxaa jira farqi u dhexeeya 212 o – 32 o = 180 o.
Miisaanka Celsius, barta qaboojinta biyaha waa 0 o C halka barta karkaraya biyuhuna ay tahay 100 o C. Inta u dhaxaysa barta qaboojinta iyo barta karkaraya biyaha waxaa jira farqi u dhexeeya 100 o – 0 o = 100 o.
Miisaanka X, barta qaboojinta biyaha waa -30 o X barta karkarana biyuhu waa 90 o X. Inta u dhaxaysa barta qaboojinta iyo barta karkaraya biyaha waxaa jira farqi u dhexeeya 90 o – (-30 o ) = 90 o + 30 o = 120 o.
U beddel miisaanka X una beddel cabbirka Celsius:
Jawaabta saxda ah waa D.
Ballaarinta
2. Bir bir ah ayaa lagu kululeeyaa heerkul ah 80 o C dhererkeeduna wuxuu noqdaa 115 cm. Haddii isku-dhafka ballaarinta toosan ee birta uu yahay 3.10 -3 o C -1 heerkulka bilowga ah ee birtuna uu yahay 30 o C, markaa dhererka bilowga ah ee birtu waa….
A. 100 cm
B. 101,5 cm
C. 102 cm
D. 102,5 cm
E. 103 cm
Dood
Waa la ogyahay in:
Heerkulka bilowga ah (T 1 ) = 30 o C
Heerkulka kama dambaysta ah (T2 ) = 80 o C
Isbeddelka heerkulka (ΔT) = 80 o C – 30 o C = 50 o C
Isugeynta ballaarinta toosan ee birta (α) = 3.10 -3 o C -1
Dhererka kama dambaysta ah ee birta (L) = 115 cm
La Rabay: Dhererka bilowga ah ee birta (L o )
Jawaab:
Qaacidada ballaarinta dhererka:
Jawaabta saxda ah waa A.
3. Usha naxaasta ah waxay marka hore dhererkeedu yahay 40 cm. Marka lagu kululeeyo heerkul ah 80 o C dhererkeedu wuxuu noqonayaa 40,04 cm. Haddii isku-dhafka ballaarinta toosan ee naxaasta uu yahay 2,0 x 10 -5 o C -1 markaas heerkulka bilowga ah ee usha naxaasta ah waa….
A. 20 o C
B. 22 o C
C. 25 o C
D. 30 o C
E. 50 o C
Dood
Waa la ogyahay in:
Heerkulka kama dambaysta ah (T2 ) = 80 o C
Dhererka bilowga ah (L o ) = 40 cm
Dhererka kama dambaysta ah (L) = 40,04 cm
Kordhinta dhererka (ΔL) = 40,04 cm – 40 cm = 0,04 cm
Isugeynta ballaarinta toosan ee naxaasta (α) = 2,0 x 10 -5 o C -1
Su'aal: Heerkulka bilowga ah (T 1 )
Jawaab:
Qaacidada ballaarinta birta:
0,04 = (2,0 x 10 -5 )(40)(80 – T 1 )
0,04 = (80 x 10 -5 )(80 – T 1 )
0,04 = 0,0008 (80 – T 1 )
0,04 = 0,064 – 0,0008 T 1
0,0008 T 1 = 0,064 – 0,040
0,0008 T 1 = 0,024
T 1 = 30 o C
Jawaabta saxda ah waa D.
Gudbinta Kulaylka iyadoo la adeegsanayo Gudbinta
4. Ulaha birta ah ee isku cabbir ah, laakiin laga sameeyay biraha kala duwan, ayaa la isku xiraa sida ka muuqata sawirka hoose. Haddii kulaylka kulaylka ee biraha I uu yahay 4 jibaar ku celcelinta birta II, markaa heerkulka isgoyska labada biraha waa…
A. 450 C
B. 40 0 C
C. 35 0 C
D. 30 0 C
E. 25 0 C
Dood
Waa la ogyahay in:
Cabbir isku mid ah oo jirridda ah
Qaboojinta kulaylka ee birta I = 4k
Qaboojinta kulaylka ee birta II = k
Heerkulka cidhifka birta ah I = 50 0 C
Heerkulka cidhifka birta II = 0 0 C
La weydiiyay: Heerkulka isgoyska labada biraha
Jawaab:
Qaacidada heerka wareejinta kulaylka iyada oo loo marayo gudbinta:
Sharaxaad: Q/t = heerka gudbinta kulaylka, k = conductivity kulaylka, A = bedka dusha sare, T 1 -T 2 = isbeddelka heerkulka, l = dhererka usha
Heerkulka diyaaradda xadka P iyo Q:
Laba ulo bir ah A iyo B waa isku cabbir sidaa darteed bedka dusha sare (A) iyo dhererka (l) ee ulaha ayaa laga saarayaa isle'egta.
Jawaabta saxda ah waa B.
5. Fiiro gaar ah u yeelo weedha soo socota!
(1) Kontoroolka birta
(2) Farqiga heerkulka ee u dhexeeya dhammaadka birta
(3) Dhererka birta
(4) Cufnaanta birta
Arrimaha go'aamiya heerka faafitaanka kulaylka ee birta waa...
A. (1), (2) iyo (3)
B. (1) iyo (4)
C. (2) iyo (4)
D. (3) iyo (4)
E. (4) oo keliya
Dood
Iyada oo ku saleysan qaacidada heerka wareejinta kulaylka iyada oo loo marayo gudbinta, arrimaha go'aamiya heerka wareejinta kulaylka waa gudbinta birta (k), farqiga heerkulka ee u dhexeeya dhammaadka birta (T) iyo dhererka birta (l). Cufka birta saameyn kuma laha.
Jawaabta saxda ah waa A.
6. Laba ulood oo PQ ah oo isku cabbir ah, laakiin noocyo kala duwan oo bir ah ayaa ku lifaaqan sida ku cad sawirka hoose. Haddii isku-dhafka gudbinta kulaylka P uu labanlaabmo isku-dhafka gudbinta kulaylka Q, markaa heerkulka xadka u dhexeeya P iyo Q waa...
A. 84°C
B. 78°C
Heerkulbeeg 72°C
D. 70°C
E. 90°C
Dood
Waa la ogyahay in:
Ulaha PQ waxay leeyihiin cabbir isku mid ah.
Isugeynta gudbinta kulaylka ee birta P (k P ) = 2k
Isku-darka gudbinta kulaylka ee birta Q (k Q ) = k
Su'aal: Heerkulka diyaaradda xadka P iyo Q
Jawaab:
Qaacidada heerka wareejinta kulaylka iyada oo loo marayo gudbinta:
Sharaxaad: Q/t = heerka gudbinta kulaylka, k = conductivity kulaylka, A = bedka dusha sare, T 1 -T 2 = isbeddelka heerkulka, l = dhererka usha
Heerkulka diyaaradda xadka P iyo Q:
Ulaha PQ waa isku cabbir sidaa darteed A iyo l waa laga saarayaa isla'egta.
Jawaab sax ah ma jirto.
Isbeddelka Foomka
7. 1 kg oo baraf ah heerkulkiisu yahay 0 o C ayaa lagu qasaa 0,5 kg oo biyo ah heerkulkiisuna yahay 0 o C, ka dibna...
A. Qaar ka mid ah biyuhu way qaboojiyaan
B. qaar ka mid ah barafku wuu dhalaalaa
C. dhammaan barafku wuu dhalaalaa
D. dhammaan biyuhu way qaboojiyaan
E. cufka barafka ee biyaha ku jira ayaa weli ah mid aan joogto ahayn
Dood
Ereyga barafku wuxuu tilmaamayaa biyo qaab adag ah, halka ereyga biyuhu uu tilmaamayo biyo qaab dareere ah.
Marka heerkulku yahay 0 ° C, barafku wuxuu isu beddelaa mid adag una beddelaa dareere. Isbeddelkan xaaladda, barafku waa inuu nuugaa kulaylka. Barafku wuxuu ku dhex milmaa biyaha, sidaa darteed barafku waa inuu nuugaa kulaylka biyaha. Si kastaba ha ahaatee, biyuhu sidoo kale waa 0 ° C, sidaa darteed kuleylku ma nuugi karo barafku. Sidaa darteed, cufka barafka ee biyaha ku jira ayaa weli ah mid joogto ah.
Jawaabta saxda ah waa E.
Mabda'a Madow
8. Weel bir ah oo miisaankiisu yahay 200 garaam, waxaa ku jira 100 garaam oo saliid ah heerkulkiisu yahay 20 o C. Weelka dhexdiisa, 50 garaam oo bir ah oo heerkulkiisu yahay 75 o C ayaa la gelinayaa. Haddii heerkulka weelku uu kordho 5 o C kulaylka gaarka ah ee saliiddana uu yahay 0,43 kal/g o C, markaa kulaylka gaarka ah ee birtu waa...
A. 0,143 kal/g o C
B. 0,098 kal/g o C
C. 0,084 kal/g o C
D. 0,075 kal/g o C
E. 0,064 kal/g o C
Dood
Waa la ogyahay in:
Cufka weelka birta (m) = 200 g
Heerkulka bilowga ah ee weelka birta ah (T 1 ) = 20 o C
Saliiddu waxay ku jirtaa weelka birta ah sidaa darteed heerkulka saliidda = heerkulka weelka birta ah.
Heerkulka kama dambaysta ah ee weelka birta ah (T2 ) = 20 o C + 5 o C = 25 o C
Cufka saliidda (m) = 100 garaam
Kulayl gaar ah oo saliid ah (c saliid) = 0,43 kalori/g o C
Heerkulka saliidda bilowga ah (T 1 ) = 20 o C
Saliiddu waxay ku jirtaa weel si saliiddu ugu dheelitirnaato kulaylka weelka birta. Markaa haddii heerkulka ugu dambeeya ee weelka birta ahi yahay 25 o C, markaa heerkulka ugu dambeeya ee saliiddu waa 25 o C.
Heerkulka saliidda kama dambaysta ah (T 2 ) = 20 o C + 5 o C = 25 o C
Cufka birta (m) = 50 garaam
Heerkulka bilowga ah ee birta (T1 ) = 75 o C
Birta waxaa lagu dhex geliyaa saliid weel, sidaas darteed birtu waxay ku jirtaa dheelitirka kulaylka ee saliidda iyo weelka. Markaa haddii heerkulka ugu dambeeya ee weelku yahay 25 o C, markaa heerkulka ugu dambeeya ee birtu waa 25 o C.
Su'aal: Kulayl gaar ah oo bir ah (c iron)
Jawaab:
Kulayl ay sii dayso birtu:
Q = mc ΔT = (50)(c)(75-25) = (50)(c)(50) = 2500c kalooriyo
Kulaylka uu nuugo weelka birta ah:
Q = mc ΔT = (200)(c)(25-20) = (200)(c)(5) = 1000c kalooriyo
Kulaylka uu nuugo saliiddu:
Q = mc ΔT = (100)(0,43)(25-20) = (43)(5) = 215 kaloori
Mabda'a Black wuxuu sheegayaa in nidaam xiran oo go'doonsan, kulaylka uu sii daayo shay heerkul sare leh uu nuugo shay heerkul hooseeya.
Q la sii daayay = Q la nuugo
2500c = 1000c + 215
2500c – 1000c = 215
1500c = 215
c = 215/1500
c = 0,143 kalori/g o C
Jawaabta saxda ah waa A.
9. Koob ay ku jiraan 200 garaam oo biyo ah heerkul ah 20°C waxaa lagu buuxiyaa 50 garaam oo baraf ah heerkul ah -2°C. Haddii is-weydaarsiga kulaylka oo keliya uu ka dhaco biyaha iyo barafka, ka dib marka dheelitirku dhaco, waxaa la heli doonaa kuwa soo socda: (c biyo = 1 cal/gr°C; c baraf = 0,5 cal/gr°C; L = 80 cal/gr)
A. barafku wuu dhalaalaa heerkulkuna wuxuu ka sarreeyaa 0°C
B. barafku wuu dhalaalaa heerkulkuna waa 0°C
C. Barafku ma dhalaalo dhammaan heerkulkuna waa 0°C
D. heerkulka nidaamka oo dhan wuxuu ka hooseeyaa 0°C
E. Qaar ka mid ah biyuhu way qaboojiyaan heerkulka nidaamkuna waa 0°C
Dood
Waa la ogyahay in:
Cufka biyaha (m ) = 200 garaam
Heerkulka biyaha (T biyo ) = 20 o C
Kulayl gaar ah oo biyo ah (c biyo ) = 1 kalori/gr°C
Cufka barafka (m es ) = 50 garaam
Heerkulka barafka (Ts ) = -2 o C
Kulayl gaar ah oo baraf ah (c es ) = 0,5 kalori/gr°C
Kulaylka biyaha dhalaalaya (L) = 80 kalori/gr
Jawaab:
Kuleyli si aad heerkulka barafka uga kordiso -2 o C ilaa 0 o C:
Q = mc ΔT
Q = (50 garaam)(0,5 kalori/gr°C)(0 o C – (-2 o C))
Q = (50)(0,5 kaloori)(2)
Q = 50 kaloori
Kuleyli si aad barafka oo dhan ugu dhalaaliso biyo:
Q = m L = (50 garaam)(80 kalori/garaam) = 4000 kaloori
Kuleyli si aad hoos ugu dhigto heerkulka biyaha oo dhan laga bilaabo 20 o C ilaa 0 o C:
Q = mc ΔT
Q = (200 garaam)(1 kalori/gr°C)(0 o C – (20 o C))
Q = (200)(1 kal)(-20)
Q = -4000 kaloori
Calaamad togan macnaheedu waa kulayl ayaa lagu daray, calaamad taban macnaheedu waa kulayl ayaa la sii daayay.
Kulaylka loo baahan yahay si heerkulka barafka loo gaarsiiyo 0 ° C waa 50 kalori, kulaylka loo baahan yahay si loo dhalaaliyo dhammaan barafka waa 4000 kalori. Sidaa darteed, kulaylka guud ee loo baahan yahay si loo dhalaaliyo dhammaan barafka waa 4050 kalori. Kulaylka la heli karo waa kulaylka ay sii daayaan biyuhu, kaas oo ah 4000 kalori.
Waxaa la soo gabagabeyn karaa in kulaylka la heli karo uusan ku filnayn inuu barafka oo dhan ku dhalaaliyo biyo. Inta badan barafku wuxuu u dhalaalay biyo, laakiin qayb yar ma dhalaalin. Biyahan iyo barafka soo haray waxay joogaan 0 ° C.
Jawaabta saxda ah waa C.
10. Qayb aluminium ah oo miisaankeedu yahay 200 garaam oo heerkulkeedu yahay 20 o C ayaa lagu shubayaa weel biyo ah oo miisaankiisu yahay 100 garaam iyo heerkulkiisu yahay 80 o C. Haddii kulaylka gaarka ah ee aluminiumku yahay 0,22 cal/g o C iyo kulaylka gaarka ah ee biyuhu yahay 1 cal/g o C, markaa heerkulka ugu dambeeya ee aluminiumku wuxuu u dhow yahay...
A. 20 o C
B. 42 o C
C. 62 o C
D. 80 o C
E. 100 o C
Dood
Waa la ogyahay in :
Culeyska aluminiumka = 200 garaam
Heerkulka aluminiumka = 20 o C
Cufka biyaha = 100 garaam
Heerkulka biyaha = 80 o C
Kulayl gaar ah oo aluminium ah = 0,22 kalori/g o C
Kulayl gaar ah oo biyo ah = 1 kalori/g o C
La weydiiyay : heerkulka ugu dambeeya ee aluminiumka
Jawaab :
Aluminiumku wuxuu ku jiraa biyaha sidaa darteed heerkulka ugu dambeeya ee aluminiumku = heerkulka ugu dambeeya ee biyaha.
Kulaylka ay sii daayaan biyaha heerkulka sare (Q oo la sii daayo) = kulaylka ay nuugaan aluminiumka heerkulka hoose (Q oo la nuugo)
biyo m c (ΔT) = m aluminium c (ΔT)
(100)(1)(80 – T) = (200)(0,22)(T – 20)
(100)(80 – T) = (44)(T – 20)
8000 – 100T = 44T – 880
8000 + 880 = 44T + 100T
8880 = 144T
T = 62 o C
Jawaabta saxda ah waa C.
11. 50g oo qadaadiic ah heerkulkiisu yahay 85°C ayaa lagu shubaa 50g oo biyo ah heerkulkiisuna yahay 29,8°C (kuleyl gaar ah oo biyo ah = 1 kal.g —1 .°C —1 ). Haddii heerkulka kama dambaysta ahi yahay 37°C oo weelkuna aanu nuugin kulaylka, markaa kulaylka gaarka ah ee birta waa…
A. 0,15 kal.g -1 .°C -1
B. 0,30 kal.g -1 .°C -1
C. 1,50 kalori.g -1 .°C -1
D. 4,8 kal.g -1 .°C -1
E. 7,2 kal.g -1 .°C -1
Dood
Waa la ogyahay in :
Cufka birta (m bir ) = 50 garaam
Heerkulka birta = 85 o C
Cufka biyaha (m ) = 50 garaam
Heerkulka biyaha = 29,8 o C
Kulayl gaar ah oo biyo ah (c biyo ) = 1 kalori.g -1 .°C -1
Heerkulka kama dambaysta ah ee isku darka = 37 o C
La weydiiyay : kulayl gaar ah oo bir ah (birta c)
Jawaab :
Kulaylka ay sii dayso birta heerkulka sare leh (Q la sii daayo) = kulaylka ay nuugto biyaha heerkulka hoose (Q la nuugo)
m bir c (ΔT) = m biyo c (ΔT)
(50)(c)(85 – 37) = (50)(1)(37 – 29,8)
(c)(85 – 37) = (1)(37 – 29,8)
48 c = 7,2
c = 0,15 kalori.g -1 .°C -1
Jawaabta saxda ah waa A.
12. 50 garaam oo baraf ah oo heerkulkiisu yahay 0°C ayaa lagu shubaa 200 garaam oo biyo ah heerkulkiisuna yahay 30°C oo lagu ridayo weel gaar ah. U qaado in weelku aanu nuugin kulaylka. Haddii kulaylka gaarka ah ee biyuhu uu yahay 1 kalori – 1 °C –1 kulaylka isku-darka barafkuna uu yahay 80 kalori –1 , markaa heerkulka ugu dambeeya ee isku-darka waa….
A. 5°C
B. 8°C
Heerkulbeeg 11°C
D. 14°C
E. 17°C
Dood
Waa la ogyahay in :
Cufka barafka (m es ) = 50 garaam
Heerkulka barafka = 0°C
Cufka biyaha (m ) = 200 garaam
Heerkulka biyaha = 30 o C
Kulayl gaar ah oo biyo ah (c biyo ) = 1 kalorig – 1 °C –1
Kulaylka barafka dhalaalaya (L ) = 80 kalori.g –1
La weydiiyay : heerkulka ugu dambeeya ee isku darka
Jawaab :
Marka hore qiyaas xaaladda kama dambaysta ah:
Kulaylka ay sii daayaan biyuhu si loo yareeyo heerkulkiisa laga bilaabo 30 o C ilaa 0 o C:
Q dabacsan = m biyo c biyo (ΔT) = (200)(1)(30-0) = (200)(30) = 6000
Kulaylka loo baahan yahay si uu barafka oo dhan u dhalaaliyo:
Q dhalaali = m es L es = (50) (80) = 4000
Kulaylka loo isticmaalo in lagu dhalaaliyo barafka oo dhan waa 4000 oo keliya halka kulaylka la heli karo uu yahay 6000. Waxaa la soo gabagabeyn karaa in heerkulka ugu dambeeya ee isku darka uu ka sarreeyo 0 o C.
Mabda'a Madow :
Kuleylka ay sii daayaan biyaha = kulayl si uu u dhalaaliyo barafka oo dhan + kulayl si uu u kordhiyo heerkulka biyaha barafka
(m biyo )(c biyo )(ΔT) = (m es )(L es ) + (m es )(c biyo )(ΔT)
(200)(1)(30-T) = (50)(80) + (50)(1)(T-0)
(200)(30-T) = (50)(80) + (50)(T-0)
6000 – 200T = 4000 + 50T – 0
6000 – 4000 = 50T + 200T
2000 = 250T
T = 2000/250
T = 8 o C
Jawaabta saxda ah waa B.
Isbeddellada xaaladda
13. Jaantuska hoose wuxuu muujinayaa xiriirka ka dhexeeya heerkulka (T) iyo kulaylka (Q) ee lagu dabaqay 1 garaam oo adag. Xaddiga kulaylka uumiga ee adag waa…
A. 60 kaloori/gram
B. 70 kaloori/gram
C. 80 kaloori/gram
D. 90 kaloori/gram
E. 100 kaloori/gram
Dood:
Kulaylka uumiga waa xaddiga kulaylka ee nuuga (ama la sii daayo) 1 garaam oo shay ah si uu xaaladdiisa uga beddelo dareere una beddelo gaas (ama laga bilaabo gaas ilaa dareere).
Waa la garanayaa :
Kuleylka la nuugo ama la sii daayo: Q = 140 kaloori - 60 kaloori = 80 kaloori
Cufka adag: m = 1 garaam
La weydiiyay :
Kulaylka uumiga (L)v) adkaha?
Jawab :
Qaaciddada lagu go'aaminayo kulaylka uumiga :
Q = m Lv
Sharaxaad: Q = kulayl la nuugo ama la sii daayo, m = cufka walaxda, Lv = kulaylka uumiga
Lv = Q / m
Lv = 80 kaloori / 1 garaam
Lv = 80 kaloori/gram
Jawaabta saxda ah waa C.
14. Hoos waxaa ku yaal jaantuska kulaylka iyo heerkulka 1 kg oo uumi ah oo cadaadis caadi ah. Barta karkaraya ee biyaha waa 2256 x 103 J/kg kulaylka gaarka ah ee biyuhu waa 4,2 x 103 J/kg K, markaa kulaylka lagu sii daayo isbeddelka uumiga ilaa biyaha waa…
A. 4,50 × 103 Joule
B. 5,20 × 103 Joule
C. 2,00 × 106 Joule
D. 2,26 × 106 Joule
E. 4,40 × 106 Joule
Dood:
Waa la garanayaa :
Kulaylka uumiga ama kulaylka karkaraya (L)v) = 2.256 x 103 J/kg
Kulayl gaar ah oo biyo ah (c) = 4200 J/kg K
Cufka uumiga (m) = 1 kg
La weydiiyay :
Kulayl la sii daayay (Q)?
Jawab :
Q = m Lv
Q = (1 kg)(2.256 x 103 J/kg)
Q = 2256 x 103 Joule
Q = 2,256 x 106 Joule
Jawaabta saxda ah waa D.
15. Xaddiga kulaylka la nuugo si kor loogu qaado heerkulka 2 kg oo biyo ah laga bilaabo -2 oC ilaa 10 oC waa… Kulayl gaar ah oo biyo ah = 4.200 J/kg Co, kulayl gaar ah oo baraf ah = 2.100 J/kg Co, kulaylka isku-darka biyaha (LF) = 334.000 J/kg
A. 760.400 J
B. 750.000 J
C. 668.000 J
D. 600.000 J
E. 540.000 J
Dood:
Waa la garanayaa :
Cufka (m) ee biyaha = 2 kg
Heerkulka bilowga ah (T) = -2 oC
Heerkulka kama dambaysta ah (T) = 10 oC
Kulayl gaar ah oo baraf ah (c es) = 2100 J/kg Co
Kulayl gaar ah oo biyo ah (c biyo) = 4200 J/kg Co
Kulaylka isku-darka biyaha (L)F) = 334.000 J/kg
La weydiiyay :
Kulaylka ma nuugo (Q)?
Jawab :
Isbeddelka heerkulka laga bilaabo -2 oC ilaa 10 oC waxay martaa dhowr marxaladood.
Marxaladda 1aad, heerkulka barafku wuxuu ka kordhaa -2 oC ilaa 0 oC (kordhinta heerkulka barafku waxay joogsataa barta barafowga biyaha, taas oo ah 0 oC)
Marxaladda 2, dhammaan barafku wuu dhalaalaa (xaaladda adag waxay isu beddeshaa xaalad dareere ah heerkulka barta barafka ee biyaha, kaasoo ah 0 oC)
Marxaladda 3aad, heerkulka biyuhu wuu kordhaa mar kale laga bilaabo 0 oC ilaa 10 oC)
Marka laga bilaabo heerkulka -2 oC ilaa 0 oC, biyuhu wali way adag yihiin. Heerkulka 0 oC, waxaa jira isbeddel ka yimid adke una beddelay dareere. Ka dib marka xaaladda adag isu beddesho xaalad dareere ah, heerkulka biyuhu mar kale wuu kordhaa laga bilaabo 0 oC ilaa 10 oC.
Q1 = (m)(c es)(delta T) = (2 kg)(2100 J/kg Co(0) oC – (-2) oC)) = (2) (2100 J) (2) = 8400 J
Q2 = (m)(L)F) = (2 kg) (334.000 J/kg) = 668.000 J.
Q3 = (m)(c biyo)(delta T) = (2 kg)(4200 J/kg Co(10) oC - 0 oC)) = (2) (4200 J) (10) = 84000 J
Kulaylka la nuugo:
Q = Q1 +Q2 +Q3
Q = 8400 J + 668.000 J + 84000 J
Q = 760.400 Joule
Jawaabta saxda ah waa A.
Isha su'aasha:
Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda
