Share
Explore BrainMass

# 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

&#916;s = cp ln (T/T0) - v0&#946;2/&#954; (T-T0) (equation is attached in a better form)

&#916;s = 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

&#946;2 = expansivity squared (&#946; = 1/v(partial v/partial T)constant P)

&#954; = 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 - Tv&#946; 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.

#### Solution Preview

The explanations are in the attached file.

Referenced web ...

#### Solution Summary

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

\$2.19