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PHYS427fa08_HW11_Solution

# PHYS427fa08_HW11_Solution - UIUC Physics 427 Homework 11...

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UIUC Physics 427 Homework 11 Solutions Philip Powell November 10, 2008 1. Problem 1 (K&K 8.1) (a) Here we consider a reversible heat pump which takes in a heat Q c from the (cold) outside at T l and puts a heat Q h into a (warm) building at T h . Since the pump is reversible, there is no net gain in entropy during the process, which means that σ h = σ l or: Q h T h = Q l T l (1) The work required to pump the heat, by conservation of energy is simply W = Q h - Q l , so using the expression above to substitute for Q c we obtain W = 1 - T l T h Q h = T h - T l T h Q h (2) Thus, we find W Q h = η c = T h - T l T h (3) (b) Next, we assume that the electricity consumed by this reversible heat pump (i.e. the source of the work) is itself generated by a Carnot engine operating between a higher temperature T hh and T l . The work generated by a Carnot heat engine is W = T hh - T l T hh Q hh (4) Using this expression with our result from part a, we obtain Q hh Q h = T h - T l T h T hh T hh - T l = 1 - T l T h 1 - T l T hh (5) Using the values T hh = 600 K, T h = 300 K, and T l = 270 K we obtain Q hh Q h = 0 . 182 (6) 2. Problem 2 (K&K 8.5) Consider a river at temperature T l = 20 C, which is used as the low temperature reservoir of a large 1

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power plant, whose steam temperature is T h = 500 C. If the plant operates at maximal (Carnot) efficiency, the electrical output is given by W = T h T l - 1 Q l (7)
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PHYS427fa08_HW11_Solution - UIUC Physics 427 Homework 11...

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