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Change in Temperature

by tan90
Tags: temperature change
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tan90
#1
Apr4-09, 07:47 PM
P: 22
1. The problem statement, all variables and given/known data
This is an experiment done on solid at high pressure. If the pressure is increased by an amount [tex]\Delta[/tex]p, this being done under condition where the sample is thermally insulated and at a sufficiently slow rate that the process can be regarded as quasi-static, what is the resulting change of temperature [tex]\Delta[/tex]T of the sample ? If[tex]\Delta[/tex]p is fairly small, derive an expression for [tex]\Delta[/tex]T in terms of [tex]\Delta[/tex]p, the absolute temperature T of the sample, its specific heat at constant pressure cp (in ergs g-1 deg -1), its density rho (in g/cm3), and its volume coefficient of thermal expansion [tex]\alpha[/tex](in deg -1)


2. Relevant equations

dv = [tex]\alpha[/tex]V dt

Maxwell's equations



3. The attempt at a solution

I started from an expression for coefficient of thermal expansion and tried to relate it with maxwell's equation. The constant terms in maxwell's equations are very confusing.
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Mapes
#2
Apr5-09, 06:58 AM
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Well, first of all, what variable are you going to hold constant?

(P.S. They're usually referred to as Maxwell relations to distinguish them from all that B-field and E-field stuff.)
tan90
#3
Apr5-09, 10:58 AM
P: 22
Since the questions says "If [tex]\Delta[/tex] p is fairly small, " is it ok to consider p as a constant? Other variables [tex]\Delta[/tex]T and [tex]\Delta[/tex]V do change. I am rather confused.

Thank you

Mapes
#4
Apr5-09, 11:10 AM
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Change in Temperature

Unfortunately, we can't assume [itex]\Delta P[/itex] is zero because we have to divide [itex]\Delta T[/itex] by [itex]\Delta P[/itex] (that is, we want to find [itex](\partial T/\partial P)_X[/itex], where X is the variable we hold constant). Since we've eliminated pressure, temperature, and volume, what's left? We're blocking heat transfer; besides energy, what flows during heat transfer?
tan90
#5
Apr5-09, 11:16 AM
P: 22
P,V,T all change. It means that the only entity left in Maxwell's relations is S. So are you suggesting[tex]\Delta[/tex]S is a constant?
Mapes
#6
Apr5-09, 11:18 AM
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Yes, S is constant, or [itex]\Delta S[/itex] is zero.
tan90
#7
Apr5-09, 11:26 AM
P: 22
Thank you ! Can you please explain a bit further about what is actually going on in the problem? Its rather hard to picture the problem for me.
Mapes
#8
Apr5-09, 11:35 AM
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The experiment involves pressurizing a solid. Will the solid heat up or cool down, and by how much for a given increase in pressure? The effect is somewhat related to thermal expansion, in which we observe how the volume of a solid changes for a given temperature increase at constant pressure. You should be able to write the differential term (e.g., [itex](\partial Y/\partial Z)_X[/itex]) for each effect, and manipulate them via Maxwell relations.

It turns out that even the sign of the answer can vary between materials--some materials heat up under pressure, some cool down--so I wouldn't try to intuit the answer through visualization. You have to go through the (admittedly abstract) calculations.
Count Iblis
#9
Apr5-09, 12:00 PM
P: 2,157
To find the partial derivative at constant S, you can write:

[tex]dS =\left(\frac{\partial S}{\partial P}\right)_{T}dP + \left(\frac{\partial S}{\partial T}\right)_{P}dT[/tex]

If you equate dS to zero and solve for the ratio dT/dP you got your partial derivative at constant S. Now, you can express the partial derivate of S w.r.t. T at constant P in terms of the heat capacity at constant pressure. But what to make of the partial derivative of S w.r.t. P at constant T? Well, we have the findamental thermodynamic relation:

dE = T dS - P dV

We can write:

T dS = d(TS) - S dT

and

P dV = d(PV) - V dP

This means that:

dE = d[TS - PV] -SdT + V dP ---------->

d[E + PV - TS] = -S dT + V dP

Define G = E + PV - TS. The above relatiomn implies that minus S is the partial derivative of G w.r.t. T and constant P and that V is the partial derivative of G w.r.t. P at constant T. Then the fact that te second derivative of G w.r.t. P and T does not depend on the order of differentiation implies that:

[tex]\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P} [/tex]

So, the partial derivative of S can be expressed in terms of the thermal expansion coefficient.
tan90
#10
Apr5-09, 04:25 PM
P: 22
Brilliant !!!


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