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Specific Heat Thermodynamics Professor Lee Carkner Lecture 8

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PAL # 7 Work Net work of 3 process cycle of 0.15 kg of air in a piston Isothermal expansion at 350 C from 2 MPa to 500 kPa: isothermal work = PVln(V 2 /V 1 ) Get V from PV = mRT V 1 = (0.15)(0.287)(623) / (2000) = 0.01341 m 3 V 2 = (0.15)(0.287)(623) / (500) = 0.05364 m 3 W = (2000)(0.01341)ln(0.95364/0.01341) =

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PAL # 7 Work Polytropic compression with n =1.2 Need the final volume P 2 V 2 n = P 3 V 3 n V 3 = ((500)(0.05364) 1.2 / 2000) (1/1.2) = W = (P 3 V 3 -P 2 V 2 )/1-n = (2000)(0.01690)- (500)(0.05364) /(1-1.2) = Isobaric compression: W = P V = (2000)(0.01341-0.01690) = Net work = 37.18-34.86-6.97 =

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Internal Energy of Ideal Gases We have defined the enthalpy as: but Pv = RT, so: So if u is just a function of T then h is too

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Temperature Dependence of c P

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Ideal Gas Specific Heats We define the specific heat as: So then we can solve for the change in internal energy du = c v dT If the change in temperature is small: Where c v is the average over the temperature range

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Linear Approximation of c

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Using Specific Heats We can write a similar equation for h h = c p T Either specific heat: Is tabulated Are generally referenced to 0 at 0K

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c v is Universal

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Specific Heat Relations We can relate c p and c v dh = du + RdT c p = c v + R For molar specific heats The specific heat ratio: k = c p /c v

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Solids and Liquids Volume is constant This means: c still is temperature dependent

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Incompressible Solid

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Incompressible Enthalpy We can write out the enthalpy change expression for constant v h = du + vdP + Pdv = du + vdP For solids the pressure does not change much and so:

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Enthalpy of Liquids Heaters (constant pressure) P = 0 Pumps (constant temperature) T = 0

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Next Time Test 1 For Monday: Read: 5.1-5.3 Homework: Ch 5, P: 12, 15, 20

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