Question
Download Solution PDF500 kJ of heat is removed from a cooled space, in a reversed Carnot cycle refrigerator, by an evaporator by refrigerant at – 23oC. The change in entropy of the refrigerant is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In the reversed Carnot cycle, the refrigerant absorbs heat, QL, isothermally from a low-temperature source at TL (process 4 – 1), is compressed isentropically to state 2 (temperature rises to TH), rejects heat, QH, isothermally to a high – temperature sink at TH (process 2 – 3), and expands isentropically to state 4 (temperature drops to TL).
Change in entropy (dS) = dQ/T
dQ = Heat transferred
T = Temperature in kelvin
Calculation:
Given:
dQ = 500 kJ
T = -23 + 273 = 250 K
dS = (500/250) = 2 kJ/K
Here entropy change of refrigerant is asked and refrigerant receives heat from the surrounding. Therefore, its entropy increases.
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