Solution to Thermodynamics Cengel/Boles Chapter 7 Question 61

Quick Answers

a)
∆Schip = −6.87 × 10−4 kJ/K
b)
∆SR−134a = 7.72 × 10−4 kJ/K
c)
∆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:
∆Uchip = Q − Wb
∆U + Wb = ∆H
[mcp (T − Ti )]chip = Q
Given mchip = 0.010 kg, cpchip = 0.3 kJ/kg.K, T = −40◦ C, Tichip = 20◦ C:

Q = −0.18 kJ
With this heat being gained by the refrigerant, from its energy balance:

[m∆hv ]R−134a = Q
@−40◦ C
From table A-11 ∆hvR−134a = 225.86 kJ/kg, so the mass vaporized is:

mR−134a,vap = 7.97 × 10−4 kg = 0.797 g
Since this mass is lower than the original 5 g, the assumption that the
refrigerant is only partially vaporized is correct.



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