# Lect10 - Lecture 10 Heat Pumps Refrigerators Available Work...

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Physics 213: Lecture 10, Pg 1 Lecture 10 Lecture 10 Heat Pumps, Heat Pumps, Refrigerators, Refrigerators, Available Work and Free Energy Available Work and Free Energy Agenda for today Agenda for today Pumping Heat – Refrigerators Free Energy and Available Work Work from Hot and Cold Bricks Re fe re nc e  fo r this   Le c ture : Ele m e nts   C h 10 Re fe re nc e  fo r  Le c ture  11: Ele m e nts   C h 11

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Physics 213: Lecture 10, Pg 2 Ideal Refrigerators and Heat Pumps Ideal Refrigerators and Heat Pumps The Carnot cycle is reversible (each step is reversible) : T c T h p V Q h Q c Hot reservoir at T h Cold reservoir at T c Q h Q c W on Fridge: Heat is flowing from cold to hot by action of work on the gas. Q c / Q h = T c / T h is still true. Coefficient of performance (efficiency): Refrigerator: K = ------------------------------ = --------- = -------- Heat Pump: K = ------------------------------ = --------- = -------- Heat transferred from T c Q c T c Work input Q h - Q c T h - T c Work input Q h - Q c T h - T c Heat delivered to T h Q h T h
Physics 213: Lecture 10, Pg 3 Helpful Hints in Dealing with Engines and Fridges Helpful Hints in Dealing with Engines and Fridges Quickly sketch the process. Define Q h and Q c and W by (or W on ) as positive and show directions of flow. Determine which Q is given. Write the First Law of Thermodynamics (FLT). There are only 3 configurations of Carnot ‘engines’: Engine (Q h given) Refrigerator (Q C given) Heat Pump (Q h or Q C given) FLT: W by = Q h - Q C W on = Q h - Q C W on = Q h - Q C W by = Q h (1- T C /T h ) W on = Q C (T h /T C - 1) W on = Q h (1- T C /T h ) or Fridge eqn. Q C Q h W by T h T C Q C Q h W on T h T C Q C Q h W on T h T C Q leak =Q C Q leak =Q h

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Physics 213: Lecture 10, Pg 4 Example: Refrigerator There is a 70 W heat leak from a room at temperature 22 °C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10 °C? Hint: For the fridge, Q c must exactly compensate the heat leak. So Q c = 70 J for each second of operation. (Watt = J/s) Assume Q c /Q h = T c /T h (Carnot) Heat Leak Hot reservoir at T h Cold reservoir at T C Q h Q C W on Fridge: Food, 263K Kitchen, 295K
Physics 213: Lecture 10, Pg 5 Supplement: Peltier cooler Supplement: Peltier cooler Driving a current (~amps) through generates a temperature difference. 20- 50˚C typical Not so common – they’re more costly, take a lot of power, and you still have to get rid of the heat! How’s it work…

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Lect10 - Lecture 10 Heat Pumps Refrigerators Available Work...

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