Quiz04a-s - C = | W | | Q h | = | Q h | | Q l | | Q h | = T...

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Quiz 4a 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 does S = 0? (a) at STP (b) in an irreversible process (c) when Δ V =0 and Δ T n =0 (d) when Δ T =0 and Δ V n =0 (e) in a reversible process 2. Consider an ideal gas in a volume of 0 . 05 m 3 . In an isochoric process with 20 J of heat transferred and Δ p =2 atm , what is the change in internal energy of the gas? (a) 0 J (b) 20 J (c) 0 . 1 J (d) 10100 J (e) 10120 J Isochoric means Δ V =0 which means that W =0 . Hence, by the frst law oF thermodynamics Δ E = Q 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 efciency oF ε =25% and absorbs 2500 J oF energy From a “hot” thermal reservoir oF temperature 350 K . (a) How much heat is dumped into the cold reservoir? The equation for eFciency of a Carnot engine is:
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Unformatted text preview: C = | W | | Q h | = | Q h | | Q l | | Q h | = T h T l T h Solving for heat going to the cold reservoir: Q l = | Q h | | Q h | C = | Q h | (1 C ) Substituting in values: Q l = (2500 J )(1 . 25) = 0 . 75 (2500 J ) inally: Q l = 1875 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 = C | Q h | = 0 . 25 (2500 J ) inally: W = 625 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 dierently, 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: S = 0 J/K 2...
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Quiz04a-s - C = | W | | Q h | = | Q h | | Q l | | Q h | = T...

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