A cyclic device works between three reservoirs

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · The laws of thermodynamics · The laws of thermodynamics

Thermal Physics, Ch The second law, entropy and irreversibility

Problem

A cyclic device works between three reservoirs, each large enough that its temperature does not change. Per cycle the working substance takes

Q_1 = 1500\ \mathrm{J}

from a reservoir at

T_1 = 600\ \mathrm{K}

, rejects

Q_2 = 300\ \mathrm{J}

to a reservoir at

T_2 = 400\ \mathrm{K}

, rejects an amount

Q_3

to a reservoir at

T_3 = 300\ \mathrm{K}

, and returns to its initial state. Use the book's convention in which every heat is counted as delivered to the working substance, so the rejected heats enter the sum with a minus sign. ((a)) Write the Clausius inequality

\oint \delta Q/T \le 0

as a sum over the three exchanges, each heat divided by the temperature of the reservoir it comes from or goes to, and derive from it the smallest possible value of

Q_3

. ((b)) Using energy conservation over one cycle, find the largest possible work output per cycle. ((c)) The actual device has

Q_3 = 800\ \mathrm{J}

. Evaluate the cycle sum

\oint \delta Q/T

, then state the total entropy created per cycle (the working substance returns to its initial state, so only the reservoirs change), the actual work output, and the lost work

T_0\,\Delta S_{\rm total}

with

T_0 = 300\ \mathrm{K}

the coldest available reservoir. Answer in the order: cycle sum, entropy created, actual work, lost work.

Browse the problem library