PHYS 2101 Chapter : H15 S13
Document Summary
These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. Temperatures ta and tb are equal so tia= tib and therefore, because both are negative. S and magnitude of qid is larger because the process is isobaric and temperature change for both processes is the same. Both entropy changes are negative and therefore smaller than those for processes ia and ib, and also. , or b > a > c > d. because of the fact that cp > cv, and qia=ncv(ta-ti) and. S ib ia ia ic ib ia ic id ic id. 2. three carnot engines operate between the following temperatures limits. If all three engines extract the same amount of energy during one cycle, qh, the net work for each cycle is: eff.
