ch20-p009 - 9. This problem is similar to Sample Problem...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
() ( ) 300.0 K = ln = 0.101 kg 386 J/kg K ln = 0.710 J/K. 305.5 K f L iL T Sm c T §· Δ⋅ ¨¸ ©¹ (b) Since the temperature of the reservoir is virtually the same as that of the block, which gives up the same amount of heat as the reservoir absorbs, the change in entropy L S Δ of the reservoir connected to the left block is the opposite of that of the left block: L S Δ = −Δ S L = +0.710 J/K. (c) The entropy change for block R is ( ) 300.0 K = ln = 0.101 kg 386 J/kg K ln = +0.723 J/K. 294.5 K f R iR T c T (d) Similar to the case in part (b) above, the change in entropy R S Δ of the reservoir connected to the right block is given by R S Δ = Δ S R = 0.723 J/K. (e) The change in entropy for the two-block system is Δ S L + Δ S R = 0.710 J/K + 0.723 J/K = +0.013 J/K. (f) The entropy change for the entire system is given by
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online