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Change in specific entropy of a copper block

The temperature of a block of copper is increased from T0 to T without any appreciable change in its volume. Show that the change in its specific entropy is

Δs = cp ln (T/T0) - v0β2/κ (T-T0) (equation is attached in a better form)

Δs = change in specific entropy

cp = specific heat at constant pressure = d(bar)q/dT at constant P or partial h/partial T at constant P

(T/T0) = finial Temperature / initial temperature

v0 = initial volume

β2 = expansivity squared (β = 1/v(partial v/partial T)constant P)

κ = isothermal compressibility = -1/v (partial v/partial P)constant T

(T-T0) = final temperature - initial temperature

I know I have to use the second TdS equation (Tds = cp dT - Tvβ dP) and change variable from P to T by using the fact that the volume is constant and expressing dv in terms of T and P.

Just don't know where to go from here.


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The explanations are in the attached file.

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Solution Summary

The solution determines the change in specific entropy of a copper block.