Show that the change in entroy is

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The discussion focuses on deriving the change in entropy for a heat engine cycle, expressed as ΔS = Q_c/T_c - Q_h/T_h. Participants explore the relationship between work done (W) and heat transfer (Q), emphasizing the isothermal nature of the heat exchanges. The calculations for changes in entropy for both the hot and cold registers are presented, highlighting the negative change for the hot register and the positive change for the cold register. The total change in entropy is calculated by combining these individual changes. The conversation illustrates the fundamental principles of thermodynamics applied to heat engines.
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Show that the change in entropy for a cycle of a heat engine is
\Delta S = \frac{Q_{cold}}{T_{cold}} - \frac{Q_{hot}}{T_{hot}}
 
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Please show some working or thoughts...
 
Well, I was thinking about
W = Q_{in} - Q_{out} = Q_{hot} - Q_{cold}
or that
\Delta S = S_{f} - S_{i}
But I'm not sure where to go from here...
 
endeavor said:
Well, I was thinking about
W = Q_{in} - Q_{out} = Q_{hot} - Q_{cold}
or that
\Delta S = S_{f} - S_{i}
But I'm not sure where to go from here...
You are to assume an isothermal heat transfer from the hot register to the gas and an isothermal flow from the gas to the cold register.

The change in entropy for the hot register in the transfer from the hot register to the gas is:

dS_h = -dQ_h/T_h

Similarly, the change in entropy of the cold register in extracting the heat from the gas to the cold register results in a change of entropy to the cold register of:

dS_c = +dQ_c/T_c

The total change in entropy of the system (hot register + cold register) is:

dS_{total} = dS_h + dS_c = dQ_c/T_c-dQ_h/T_h[/tex]

AM
 
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