Specific Heat
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Transcript of Specific Heat
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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(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) =
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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 =
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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 cP
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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 = cv dT
If the change in temperature is small:
Where cv 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 = cp T
Either specific heat: Is tabulated
Are generally referenced to 0 at 0K
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cv is Universal
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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
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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