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Unformatted text preview: ε C = | W | | Q h | = | Q h | − | Q l | | Q h | = T h − T l T h Solving for heat coming from the hot reservoir: Q l = | Q h | − | Q h | ε C = | Q h | (1 − ε C ) = ⇒ Q h = | Q l | 1 − ε C Substituting in values: Q h = 5000 J 1 − . 3 = 5000 J . 7 ±inally: Q h = 7142 . 9 J (b) How much work can be done with this engine? Again, using the equation for eFciency of a Carnot engine, one can solve for the work accomplished: W = ε C | Q h | = 0 . 30 (7142 . 9 J ) ±inally: W = 2142 . 9 J (c) What is the change in entropy oF the process? A Carnot is a closed cycle process, i.e. it begins and ends at the same point. As such, the change in entropy must be zero for the entire system. Stated di²erently, the change in entropy in the hot reservoir must equal the change in entropy in the low temperature reservoir: | Q h | T h = | Q l | T l ±inally: Δ S = 0 J/K 2...
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This note was uploaded on 05/01/2010 for the course PHYS 224 taught by Professor Hanlet during the Spring '10 term at Illinois Tech.
- Spring '10