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Entropy Irreversible Heat Flow Quality of Heat

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Entropy Irreversible Heat Flow Quality of Heat Reversible /Irreversible processes Ball drop Second Law of Thermodynamics • • • 10.2 Irreversible process • • • • • • Entropy Mechanical Heat energy Energy Quality of Heat Irreversible process -proceeds spontaneously Order / Disorder -associated with increasing disorder The reversal of an irreversible process is improbable ( even though it may be possible according to the First Law of Thermodynamics) Entropy Irreversible Heat Flow Heat Entropy is a state function that governs the availability of heat to do work. Disorder/Order Q The entropy of the system is a measure of the disorder of the system. The entropy increases for irreversible (spontaneous) processes. The entropy of the universe is increasing leading to greater disorder. Heat flow from high T to low T is irreversible The initial (a) and final (b) states have the same energy but have different ability to do work Quality of Heat Irreversible free expansion of a gas W=0 Th vacuum Q=0 Heat flows from high temperature to low Tc temperature. Q TT High temperature sources of heat can be converted to work with higher efficiency. High temperature is a measure of the quality of The gas is allowed to expand into a vacuum, in an insulated container the heat. from a to b without doing work or taking up heat. eg by poking a hole in the barrier. For this case ∆U =0. The two states a and b have the same internal energy but have different abilities to do work 1 Work could be done by the gas Entropy is a state function To study the properties of a system we use a reversible Carnot cycle. Define the change in Entropy Th dQ T ∆=S c A 1 ∫ Q T P h B 4 The change in a cyclic process is 2 D Q dQ c C = 0 3 ∫ T V For the Carnot cycle Here Q is negative since heat goes dQ Q Q c ∫ = h +=c 0 out of the gas. TTThc Any cycle can be constructed from a sum of Entropy change in irreversible heat Carnot cycles. flow Since S is a state function we can calculate ∆S for irreversible processes from the properties of the initial and final states. For irreversible heat flow from state (a) to state (b) the change in entropy can be calculated by changing the temperature reversibly. T T T h f f dQ∆ Q average T ∆=S = heat bath dQ h ∫ T TT2 T2 >Tf T varies h Therefore, for any arbitrary cycle Tf dQ∆ Q average T Tc Tf ∆=S = dQ c ∫ TT1 T <T dS ==0 Tc 1 f ∫∫T dQ ∆S>0 for Entropy S is a state function for any system. Qh QQQccc− ⎛⎞11 ∆SS=∆hc +∆ S = + = + = Q c⎜⎟ − + > 0irreversible heat flow TTTT21 2 1⎝⎠ TT 21 Entropy change in irreversible free The change in entropy reflects the loss of expansion of a gas ability to do work. Calculate ∆S for transition from (a) to (b). Free expansion, with no heat transfer. How does T change? T remains unchanged Work that could have been done by (a) -> (b) is How can we go reversibly from a to b? By an isothermal expansion. V WnRTln= 2 =∆ TS Calculate ∆S. since Q=W for isothermal expansion V1 V Eunavailable = Tmin ∆S QW and WnRTln= 2 ∆=S = V TT 1 During an irreversible process in which the entropy of the system V2 increases by ∆S, the energy E=T ∆S becomes unavailable to ∆=SnRln since V > V min 2 1 ∆S >0 do work. T is the coolest temperature available to the system. V1 min Entropy Increases for the irreversible process 2 Loss of available energy Question in irreversible heat transfer. o o Suppose a quantity of heat Q flows from A 250 g pot of water is heated from 10 C to 95 C. T T to T The quantity of heat is much smaller h How much does the entropy of the water h c. Q than the heat capacity of the reservoirs so that change? (c =4184 J/kg K) the temperatures of the reservoirs don’t change. water T Find the change in the ability to do work. c ∆S QQ⎛⎞ 1 1 =− + =Q − + ∆=∆+∆SShc S ⎜⎟> 0 TThc⎝⎠ TT hc Eunavailable ⎛⎞11 ⎛⎞Tc ETSTQunavailable=∆=−+= min c ⎜ ⎟ Q1⎜ − ⎟ ⎝⎠Th Tc ⎝ Th ⎠ Lowering the temperature lowers the Carnot efficiency of the heat Question Entropy and Disorder In an adiabatic free expansion 8.7 mol of an ideal free expansion of a gas gas at 450 K expands 10-fold in volume. How much energy becomes unavailable to do work? The gas molecules have a larger accessible volume in b than in a. The state (a) is more improbable than (b) - How probable is it to go from b to a by random motion of the gas molecules? Increased entropy in a -> b is due to an increased disorder. ∆S0> Decreased in entropy for one system must Entropy and disorder be coupled to an entropy increase in another system ice melting Q heat bath disordered T ice ordered water Tm Q Q ∆S0water =< Q ∆S0heatbath => T Tm ∆=>S0ice Tm but T < T m The change in entropy for the total system must The increase in entropy upon ice melting is due to the disordering ∆S= ∆Swater + ∆Sheat bath > 0 increase. in the liquid state. 3 Entropy statement of the second law of Living things are open systems thermodynamics The entropy of a closed system can never decrease. ∆S > 0 A closed system does not exchange heat or energy with its environment. This is equivalent to the statement that heat does not flow from low The decrease in entropy in growth is driven by coupling to to high temperatures and that heat cannot be converted to work with energy inputs. 100% efficiency since these would result in a decrease in entropy. The earth is an open system The decrease in entropy is driven by energy from the sun. 4.
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