Cengel_Thermodynamics_7-61.pdf - Quick Answers a Schip =...

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Quick Answers a) Δ S chip = - 6 . 87 × 10 - 4 kJ/K b) Δ S R - 134 a = 7 . 72 × 10 - 4 kJ/K c) Δ S = +8 . 5 × 10 - 5 kJ/kg Since the entropy change for the system is positive, entropy increases and therefore the process is possible. Step-by-step solution From the problem statement, the chip will be cooled by the refrigerant and, since they are in contact, the final temperature for both must be the same, from the zeroth law. Since the refrigerant is saturated liquid, we must assume part of it will vaporize and also that all the heat lost from the chip goes to the R-134a. The final temperature for the chip must be the initial temperature for the refrigerant (saturated state), so the heat lost from the chip, from its energy balance: Δ U chip = Q - W b Δ U + W b = Δ H [ mc p ( T - T i )] chip = Q Given m chip = 0 . 010 kg , c p chip = 0 . 3 kJ/kg.K , T = - 40 C , T i chip = 20 C : Q = - 0 . 18 kJ With this heat being gained by the refrigerant, from its energy balance: [ m Δ h v ] R - 134 a = Q From table A-11 Δ h @ - 40 C v R - 134 a = 225 . 86 kJ/kg
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