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.© BrainMass Inc. brainmass.com June 23, 2018, 12:29 am ad1c9bdddf
The explanations are in the attached file.
Referenced web ...
The solution determines the change in specific entropy of a copper block.