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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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 Spring '10
 Hanlet
 Thermodynamics, Entropy, Heat, 0 j, hot reservoir

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