30 Sharciga ugu horreeya ee thermodynamics - dhibaatooyinka iyo xalalka
1. 3000 J oo kuleyl ah ayaa lagu daraa nidaam, 2500 J oo shaqo ahna waxaa qabta nidaamku. Waa maxay isbeddelka tamarta gudaha ee nidaamka?
La yaqaan:
Kulayl (Q) = +3000 Joule
Shaqada (W) = +2500 Joule
La Doonayo: isbeddelka tamarta gudaha ee nidaamka
Xalka:
Isle'egta sharciga koowaad ee thermodynamics
ΔU = QW
Xeerarka calaamadda:
Q waa togan haddii kulaylka lagu daro nidaamka
W waa togan haddii shaqada uu qabto nidaamku
Q waa taban haddii kulaylku ka baxo nidaamka
W waa taban haddii shaqada laga qabto nidaamka
Isbeddelka tamarta gudaha ee nidaamka:
ΔU = 3000-2500
ΔU = 500 Joule
Tamarta gudaha waxay ku kortaa 500 Joules.
2. 2000 J oo kuleyl ah ayaa lagu daraa nidaamka, 2500 J oo shaqo ahna waa laga qabtaa nidaamka. Waa maxay isbeddelka tamarta gudaha ee nidaamka?
La yaqaan:
Kulayl (Q) = +2000 Joule
Shaqada (W) = -2500 Joule
La Doonayo: Isbeddelka tamarta gudaha ee nidaamka
Xalka:
ΔU = QW
ΔU = 2000-(-2500)
ΔU = 2000+2500
ΔU = 4500 Joule
Tamarta gudaha waxay ku kortaa 4500 Joules.
3. 2000 J oo kulayl ah ayaa ka baxaya nidaamka, 2500 J oo shaqo ahna waa la qabtaa nidaamka. Waa maxay isbeddelka tamarta gudaha ee nidaamka?
La yaqaan:
Kulayl (Q) = -2000 Joule
Shaqada (W) = -3000 Joule
La Doonayo: Isbeddelka tamarta gudaha ee nidaamka
Xalka:
ΔU = QW
ΔU = -2000-(-3000)
ΔU = -2000+3000
ΔU = 1000 Joule
Tamarta gudaha waxay ku kortaa 4500 Joules.
Gabagabo:
– Haddii kulaylka lagu daro nidaamka, markaa tamarta gudaha ee nidaamku way kordheysaa
– Haddii kulaylku ka baxo nidaamka, markaa tamarta gudaha ee nidaamku way yaraanaysaa
– Haddii shaqada uu qabto nidaamku, markaa tamarta gudaha ee nidaamku way yaraanaysaa
– Haddii shaqada lagu qabto nidaamka, markaas tamarta gudaha ee nidaamku way kordheysaa
4. Xisaabi isbeddelka tamarta gudaha ee 2 moles oo gaas ah oo ku habboon marka lagu daro 400 J oo kuleyl ah, gaaskuna uu ballaarto, isagoo sameynaya 300 J oo shaqo ah oo ku saabsan hareeraheeda.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics: \( \Delta U = Q - W \):
\[ \Delta U = 400 – 300 = 100\, \text{J} \]
5. Go'aami wareejinta kulaylka ee nidaam qabta 200 J oo shaqo ah oo ku xeeran oo leh isbeddel ku yimaada tamarta gudaha ee 50 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[ Q = \Delta U + W = 50 + 200 = 250\, \text{J} \]
6. Xisaabi shaqada uu nidaamku qabto marka uu nuugo 600 J oo kuleyl ah tamartiisa gudahana ay kordho 150 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = 600 – 150 = 450\, \text{J}
\]
7. Go'aami wareejinta kulaylka ee nidaam qabta 500 J oo shaqo ah oo tamartiisa gudaha ay hoos u dhacdo 100 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = (-100) + 500 = 400\, \text{J}
\]
8. Xisaabi isbeddelka tamarta gudaha marka 300 J oo kuleyl ah la waayo oo 200 J oo shaqo ah lagu qabto nidaamka.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = -300 + 200 = -100\, \qoraal{J}
\]
9. Go'aami shaqada laga qabtay nidaamka marka uu lumiyo 400 J oo kuleyl ah oo tamartiisa gudaha ay hoos u dhacdo 200 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = \Delta U – Q = (-200) – (-400) = 200\, \text{J}
\]
10. Xisaabi wareejinta kulaylka marka tamarta gudaha ee nidaamku ay kordho 100 J iyo 50 J oo shaqo ah oo lagu sameeyo nidaamka.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = 100 + 50 = 150\, \text{J}
\]
11. Go'aami isbeddelka tamarta gudaha marka nidaamku nuugo 250 J oo kuleyl ah isla markaana uu sameeyo 150 J oo shaqo ah oo ku xeeran.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = Q – W = 250 – 150 = 100\, \text{J}
\]
12. Xisaabi shaqada uu nidaamku qabtay marka uu waayo 300 J oo kuleyl ah, tamartiisa gudahana ay hoos u dhacdo 100 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = -300 – (-100) = -200\, \qoraal{J}
\]
13. Go'aami wareejinta kulaylka ee nidaam sameeya 400 J oo shaqo ah oo ku saabsan hareerihiisa, tamartiisa gudahana ay kordho 150 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = 150 + 400 = 550\, \text{J}
\]
14. Xisaabi isbeddelka tamarta gudaha marka lagu daro 500 J oo kuleyl ah, nidaamkuna wuxuu sameeyaa 300 J oo shaqo ah oo ku xeeran.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = Q – W = 500 – 300 = 200\, \text{J}
\]
15. Go'aami shaqada uu nidaamku qabto marka uu nuugo 600 J oo kuleyl ah, tamartiisa gudahana ay kordho 200 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = 600 – 200 = 400\, \text{J}
\]
16. Xisaabi wareejinta kulaylka marka nidaamku sameeyo 700 J oo shaqo ah oo tamartiisa gudaha ay hoos u dhacdo 300 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = (-300) + 700 = 400\, \text{J}
\]
17. Go'aami isbeddelka tamarta gudaha marka 800 J oo kuleyl ah la waayo oo 400 J oo shaqo ah lagu qabto nidaamka.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = -800 + 400 = -400\, \qoraal{J}
\]
18. Xisaabi shaqada laga qabtay nidaamka marka uu waayo 900 J oo kuleyl ah, tamartiisa gudahana ay hoos u dhacdo 500 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = \Delta U – Q = (-500) – (-900) = 400\, \text{J}
\]
19. Go'aami wareejinta kulaylka marka tamarta gudaha ee nidaamku ay kordho 600 J, iyo shaqada 300 J ee nidaamka lagu sameeyo.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = 600 + 300 = 900\, \text{J}
\]
20. Xisaabi isbeddelka tamarta gudaha marka nidaamku nuugo 700 J oo kuleyl ah isla markaana uu sameeyo 350 J oo shaqo ah oo ku xeeran.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = Q – W = 700 – 350 = 350\, \text{J}
\]
21. Go'aami shaqada uu nidaamku qabto marka uu waayo 800 J oo kuleyl ah, tamartiisa gudahana ay hoos u dhacdo 400 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = -800 – (-400) = -400\, \qoraal{J}
\]
22. Xisaabi wareejinta kulaylka marka nidaamku sameeyo 900 J oo shaqo ah oo ku saabsan hareerihiisa, tamartiisa gudahana ay kordho 450 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = 450 + 900 = 1350\, \text{J}
\]
23. Go'aami isbeddelka tamarta gudaha marka 1000 J oo kuleyl ah lagu daro, nidaamkuna uu sameeyo 500 J oo shaqo ah oo ku xeeran.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = Q – W = 1000 – 500 = 500\, \text{J}
\]
24. Xisaabi shaqada uu nidaamku qabtay marka uu nuugo 1100 J oo kuleyl ah, tamartiisa gudahana ay kordho 550 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = 1100 – 550 = 550\, \text{J}
\]
25. Go'aami wareejinta kulaylka marka nidaamku sameeyo 1200 J oo shaqo ah, tamartiisa gudahana ay hoos u dhacdo 600 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = (-600) + 1200 = 600\, \text{J}
\]
26. Xisaabi isbeddelka tamarta gudaha marka 1300 J oo kuleyl ah la waayo, oo 650 J oo shaqo ah lagu qabto nidaamka.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = -1300 + 650 = -650\, \qoraal{J}
\]
27. Go'aami shaqada laga qabtay nidaamka marka uu lumiyo 1400 J oo kuleyl ah, tamartiisa gudahana ay hoos u dhacdo 700 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = \Delta U – Q = (-700) – (-1400) = 700\, \text{J}
\]
28. Xisaabi wareejinta kulaylka marka tamarta gudaha ee nidaamku ay kordho 800 J, iyo 400 J oo shaqo ah oo lagu sameeyo nidaamka.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
Q = \Delta U + W = 800 + 400 = 1200\, \text{J}
\]
29. Go'aami isbeddelka tamarta gudaha marka nidaamku nuugo 1500 J oo kuleyl ah isla markaana uu sameeyo 750 J oo shaqo ah oo ku xeeran.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
\Delta U = Q – W = 1500 – 750 = 750\, \text{J}
\]
30. Xisaabi shaqada uu nidaamku qabtay marka uu waayo 1600 J oo kuleyl ah, tamartiisa gudahana ay hoos u dhacdo 800 J.
Solution:
Iyadoo la adeegsanayo sharciga koowaad ee thermodynamics:
\[
W = Q – \Delta U = -1600 – (-800) = -800\, \qoraal{J}
\]
Dhibaatooyinkan iyo xalalkan waxaa loogu talagalay inay bixiyaan faham dhammaystiran oo ku saabsan sharciga koowaad ee thermodynamics, kaas oo sheegaya in isbeddelka tamarta gudaha ee nidaamka xiran uu la mid yahay kulaylka lagu daray nidaamka marka laga reebo shaqada uu nidaamku qabtay.
