30 Umthetho wokuqala we-thermodynamics - izinkinga nezixazululo
1. 3000 J ka ukushisa ingeziwe ohlelweni kanye no-2500 J we umsebenzi kwenziwa uhlelo. Luyini ushintsho emandleni angaphakathi ohlelweni?
Kwaziwa:
Ukushisa (Q) = +3000 Joule
Umsebenzi (W) = +2500 amaJoule
Okufunwayo: ushintsho lwamandla angaphakathi ohlelweni
Isixazululo:
Isibalo se umthetho wokuqala we-thermodynamics
ΔU = QW
Imithetho yesibonakaliso:
U-Q ulungile uma ukushisa kwengezwa ohlelweni
U-W ulungile uma umsebenzi wenziwa uhlelo
U-Q ungemuhle uma ukushisa kuphuma ohlelweni
U-W ubonisa ukuthi awulungile uma umsebenzi wenziwa ohlelweni
Ushintsho emandleni angaphakathi ohlelweni:
ΔU = 3000-2500
ΔU = 500 amaJoule
Amandla angaphakathi anda ngama-Joules angu-500.
2. Kufakwa ama-J angu-2000 okushisa ohlelweni bese kwenziwa umsebenzi ongu-J angu-2500 ohlelweni. Luyini ushintsho lwamandla angaphakathi ohlelweni?
Kwaziwa:
Ukushisa (Q) = +2000 Joule
Umsebenzi (W) = -2500 amaJoule
Okufunwayo: Ushintsho lwamandla angaphakathi ohlelweni
Isixazululo:
ΔU = QW
ΔU = 2000-(-2500)
ΔU = 2000+2500
ΔU = 4500 amaJoule
Amandla angaphakathi anda ngama-Joules angu-4500.
3. Ukushisa okungu-2000 J kuyaphuma ohlelweni bese kwenziwa umsebenzi ongu-2500 J ohlelweni. Luyini ushintsho lwamandla angaphakathi ohlelweni?
Kwaziwa:
Ukushisa (Q) = -2000 Joule
Umsebenzi (W) = -3000 amaJoule
Okufunwayo: Ushintsho lwamandla angaphakathi ohlelweni
Isixazululo:
ΔU = QW
ΔU = -2000-(-3000)
ΔU = -2000+3000
ΔU = 1000 amaJoule
Amandla angaphakathi anda ngama-Joules angu-4500.
Isiphetho:
– Uma kufakwa ukushisa ohlelweni, khona-ke amandla angaphakathi ohlelweni ayanda
– Uma ukushisa kuphuma ohlelweni, khona-ke amandla angaphakathi ohlelweni ayancipha
– Uma umsebenzi wenziwa uhlelo, khona-ke amandla angaphakathi ohlelo ayancipha
– Uma umsebenzi wenziwa ohlelweni, khona-ke amandla angaphakathi ohlelweni ayakhula
4. Bala ushintsho emandleni angaphakathi ama-moles amabili egesi elifanele uma kufakwa ama-J angu-400 okushisa, futhi igesi iyanda, yenza umsebenzi wama-J angu-300 endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics: \( \Delta U = Q – W \):
\[ \Delta U = 400 – 300 = 100\, \umbhalo{J} \]
5. Nquma ukudluliselwa kokushisa kwesistimu eyenza umsebenzi ongu-200 J endaweni ezungezile futhi eshintsha amandla angaphakathi angu-50 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[ Q = \Delta U + W = 50 + 200 = 250\, \umbhalo{J} \]
6. Bala umsebenzi owenziwe uhlelo lapho lumunca ama-J angu-600 okushisa futhi amandla alo angaphakathi anda ngo-150 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = 600 – 150 = 450\, \text{J}
\]
7. Nquma ukudluliselwa kokushisa kwesistimu eyenza umsebenzi ongu-500 J bese amandla ayo angaphakathi ehla ngo-100 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = (-100) + 500 = 400\, \umbhalo{J}
\]
8. Bala ushintsho lwamandla angaphakathi lapho kulahleka u-300 J wokushisa kanye nomsebenzi ongu-200 J ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = -300 + 200 = -100\, \umbhalo{J}
\]
9. Thola umsebenzi owenziwe ohlelweni uma lulahlekelwa ukushisa okungu-400 J futhi amandla alo angaphakathi ehla ngo-200 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = \Delta U – Q = (-200) – (-400) = 200\, \text{J}
\]
10. Bala ukudluliselwa kokushisa lapho amandla angaphakathi esistimu enyuka ngo-100 J kanye no-50 J womsebenzi owenziwa ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = 100 + 50 = 150\, \umbhalo{J}
\]
11. Thola ushintsho emandleni angaphakathi lapho uhlelo lumunca ukushisa okungu-250 J futhi lwenza umsebenzi ongu-150 J endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = Q – W = 250 – 150 = 100\, \umbhalo{J}
\]
12. Bala umsebenzi owenziwe uhlelo uma lulahlekelwa ukushisa okungu-300 J, futhi amandla alo angaphakathi ehla ngo-100 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = -300 – (-100) = -200\, \text{J}
\]
13. Nquma ukudluliselwa kokushisa kwesistimu eyenza umsebenzi ongu-400 J endaweni ezungezile futhi amandla ayo angaphakathi akhuphuke ngo-150 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = 150 + 400 = 550\, \umbhalo{J}
\]
14. Bala ushintsho lwamandla angaphakathi uma kufakwa u-500 J wokushisa, bese uhlelo lwenza u-300 J womsebenzi endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = Q – W = 500 – 300 = 200\, \umbhalo{J}
\]
15. Nquma umsebenzi owenziwe uhlelo uma lumunca ama-J angu-600 okushisa, futhi amandla alo angaphakathi anda ngo-J angu-200.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = 600 – 200 = 400\, \text{J}
\]
16. Bala ukudluliselwa kokushisa lapho uhlelo lwenza umsebenzi ongu-700 J futhi amandla alo angaphakathi ehla ngo-300 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = (-300) + 700 = 400\, \umbhalo{J}
\]
17. Nquma ushintsho lwamandla angaphakathi lapho kulahleka u-800 J wokushisa kanye nomsebenzi ongu-400 J ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = -800 + 400 = -400\, \umbhalo{J}
\]
18. Bala umsebenzi owenziwe ohlelweni uma lulahlekelwa ukushisa okungu-900 J, futhi amandla alo angaphakathi ehla ngo-500 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = \Delta U – Q = (-500) – (-900) = 400\, \text{J}
\]
19. Nquma ukudluliselwa kokushisa lapho amandla angaphakathi esistimu enyuka ngo-600 J, futhi umsebenzi ongu-300 J wenziwa ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = 600 + 300 = 900\, \umbhalo{J}
\]
20. Bala ushintsho lwamandla angaphakathi lapho uhlelo lumunca ukushisa okungu-700 J futhi lwenza umsebenzi ongu-350 J endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = Q – W = 700 – 350 = 350\, \umbhalo{J}
\]
21. Nquma umsebenzi owenziwe uhlelo uma lulahlekelwa ngu-800 J wokushisa, futhi amandla alo angaphakathi ehla ngo-400 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = -800 – (-400) = -400\, \text{J}
\]
22. Bala ukudluliselwa kokushisa lapho uhlelo lwenza umsebenzi ongu-900 J endaweni oluzungezile, futhi amandla alo angaphakathi anda ngo-450 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = 450 + 900 = 1350\, \umbhalo{J}
\]
23. Nquma ushintsho lwamandla angaphakathi lapho kufakwa u-1000 J wokushisa, futhi uhlelo lwenza u-500 J womsebenzi endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = Q – W = 1000 – 500 = 500\, \umbhalo{J}
\]
24. Bala umsebenzi owenziwe uhlelo uma lumunca ukushisa okungu-1100 J, futhi amandla alo angaphakathi anda ngo-550 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = 1100 – 550 = 550\, \text{J}
\]
25. Nquma ukudluliselwa kokushisa lapho uhlelo lwenza umsebenzi ongu-1200 J, futhi amandla alo angaphakathi ehla ngo-600 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = (-600) + 1200 = 600\, \umbhalo{J}
\]
26. Bala ushintsho lwamandla angaphakathi lapho kulahleka u-1300 J wokushisa, futhi umsebenzi ongu-650 J wenziwa ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = -1300 + 650 = -650\, \umbhalo{J}
\]
27. Nquma umsebenzi owenziwe ohlelweni uma lulahlekelwa ukushisa okungu-1400 J, futhi amandla alo angaphakathi ehla ngo-700 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = \Delta U – Q = (-700) – (-1400) = 700\, \text{J}
\]
28. Bala ukudluliselwa kokushisa lapho amandla angaphakathi esistimu enyuka ngo-800 J, futhi umsebenzi ongu-400 J wenziwa ohlelweni.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
Q = \Delta U + W = 800 + 400 = 1200\, \umbhalo{J}
\]
29. Thola ushintsho emandleni angaphakathi lapho uhlelo lumunca ukushisa okungu-1500 J futhi lwenza umsebenzi ongu-750 J endaweni ezungezile.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
\Delta U = Q – W = 1500 – 750 = 750\, \umbhalo{J}
\]
30. Bala umsebenzi owenziwe uhlelo uma lulahlekelwa ukushisa okungu-1600 J, futhi amandla alo angaphakathi ehla ngo-800 J.
Isixazululo:
Ukusebenzisa umthetho wokuqala we-thermodynamics:
\[
W = Q – \Delta U = -1600 – (-800) = -800\, \text{J}
\]
Lezi zinkinga nezixazululo zenzelwe ukunikeza ukuqonda okuphelele komthetho wokuqala we-thermodynamics, othi ushintsho lwamandla angaphakathi ohlelweni oluvaliwe lulingana nokushisa okwengezwe ohlelweni ngaphandle komsebenzi owenziwe uhlelo.
