1. Using the definition of entropy change, dQ = T dS, what is the entropy change

Question

1. Using the definition of entropy change, dQ = T dS, what is the entropy change of (1) the hot reservoir, and (2) the cold reservoir, when a total amount of heat Q_H is extracted from the hot reservoir and used to run a Carnot Engine?

2. If an amount of heat Q_H is extracted from a hot reservoir (at temperature T_H) but is simply wasted by dumping into a cold reservoir (at temperature T_C) instead of using it to run a Carnot cycle to extract useful mechanical work, by how much has the entropy of the Universe increased? Compare this to the amount of work which could have been generated by this same amount of heat running a Carnot engine.

Please give detailed explanation(s) for both qustions.

Thanks...

Explanation / Answer

1)

entropy cahnge of hot reservoir = ds = dQ/T = -Qh/Th (- sign as heat is lost)

where Th = hot reservoir temperature

let Q_c be heat rejected to the low temperature resevoit at Tctemp

then entropy change of cold resrvoir = dS = Q_c/Tc

for carnot engine

Qh/Qc = Th/Tc

so Qc = Qh.Tc/Th

so entropy cahnge of cold for carnot engine = Qh.Tc/Th.Tc = Qh/Th

2) )entropy cahnge of hot reservoir = ds = - dQ/T = -Qh/Th

where Th = hot reservoir temperature

2) let Q_c be heat rejected to the low temperature resevoit at Tc temp

then entropy change of cold resrvoir = dS = Q_c/Tc

so entrop change of universe = -Qh/Th + Qc/Tc

Work done = Qh - Qc

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