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(V2/V1)

Get V from PV = mRT V1 = (0.15)(0.287)(623) / (2000) = 0.01341 m3

V2 = (0.15)(0.287)(623) / (500) = 0.05364 m3

W = (2000)(0.01341)ln(0.95364/0.01341) =

PAL # 7 Work Polytropic compression with n =1.2

Need the final volume P2V2

n = P3V3n

V3 = ((500)(0.05364)1.2 / 2000)(1/1.2) =

W = (P3V3-P2V2)/1-n = (2000)(0.01690)-(500)(0.05364) /(1-1.2) =

Isobaric compression: W = PV = (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 cP

Ideal Gas Specific Heats

We define the specific heat as:

So then we can solve for the change in internal energy

du = cv dT

If the change in temperature is small:

Where cv is the average over the temperature range

Linear Approximation of c

Using Specific Heats

We can write a similar equation for h

h = cp T

Either specific heat: Is tabulated

Are generally referenced to 0 at 0K

cv is Universal

Specific Heat Relations We can relate cp and cv

dh = du + RdT

cp = cv + R

For molar specific heats

The specific heat ratio:

k = cp/cv

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