# Specific Heat Thermodynamics Professor Lee Carkner Lecture 8.

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Specific Heat Thermodynamics Professor Lee Carkner Lecture 8

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) =

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 =

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

Temperature Dependence of c P

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

Linear Approximation of c

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

c v is Universal

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 

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

Incompressible Solid

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: 

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

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