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Prvi zakon termodinamike

Prvi zakon termodinamike. Termodinamika. Termodinamički sustav. Q =  U + W. Prvi zakon termodinamike :. , t 2  t 1  Q  0. Q = mc(t 2 – t 1 ). , t 2  t 1  Q  0. W = p(V 2 – V 1 ). , V 2  V 1  W  0. , V 2  V 1  W  0.

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Prvi zakon termodinamike

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  1. Prvi zakon termodinamike Termodinamika Termodinamički sustav Q = U + W Prvi zakon termodinamike : , t2t1 Q  0 Q = mc(t2 – t1) , t2t1 Q  0 W = p(V2 – V1) , V2V1  W  0 , V2V1  W  0

  2. Primjer: Pri 20 oC dva mola vodika nalaze se pod tlakom 300 kPa. Nakon širenja pri stalnom tlaku obujam plina je 17 litara. a) Koliki je rad obavio plin pri širenju? b) Kolika je promjena unutarnje energije plina ako je on primio 700 J topline? Rješenje: n = 2 pV1= nRT1 t1 = 20 oC , T = 293K p = 300 kPa = 3105 Pa V2 = 17 l = 0,017 m3 V1= 0,016 m3 a) W = ? W = p(V2 – V1) = 3105 Pa(0,017 m3 – 0,016 m3) W = 300 J

  3. Q = 700J b) U = ? Q = U + W U = Q – W = 700J – 300J U = 400J

  4. p V Adijabatski procesi Termodinamički proces Adijabatski procesi Q = 0 Q = U + W izoterma 0 = U + W U = - W adijabata

  5. Zadatak 1: Za koliko se promijeni unutarnja energija sustava kojemu dovedemo 300 kJ toplinei istodobno na njemu obavimo rad od 150 kJ? Rješenje: Q = 300 kJ -150 kJ W = U = ? Q = U + W = 300 kJ – (- 150 kJ) U = Q - W U = 450 kJ

  6. Zadatak 2: Za koliko će se promijeniti unutarnja energija sustava koji pri adijabatskoj ekspanzijiobavi rad od 500 kJ? Rješenje: W = 500 kJ U = ? U = - W U = - 500 kJ

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