Quiz04b-s

# Quiz04b-s - ε C = | W | | Q h | = | Q h | − | Q l | | Q...

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Quiz 4b Name: Date: Phys224 Spring 2010 Dr. P. Hanlet Concept (3 points each) 1. The second law of thermodynamics, can be stated succintly as: S 0 Under which condition is S > 0? (a) always at STP (b) in a reversible process (c) when Δ T =0 and Δ V negationslash =0 (d) all of the above (e) none of the above 2. Consider compression of an ideal gas. In an adiabatic process with 20 J of work performed on it, and Δ S =10 J/K , what is the change in internal energy of the gas? (a) 0 J (b) 20 J (c) 10 J (d) 20 J (e) none of the above Adiabatic means Q =0 which means that W =0 . Hence, by the first law of thermodynamics Δ E = W . Since work is done “on” the gas (it is being compressed), then the internal energy increases. 1

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Problem (5 pts) Show your work!!! If I can read it, I will give you partial credit!!! Correct answers without work will NOT get full credit. 3. A Carnot heat engine has an efficiency of ε = 30% and dumps 5000 J of energy into the “cold” thermal reservoir of temperature 150 K . (a) How much heat is taken from the hot reservoir? The equation for efficiency of a Carnot engine is:
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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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