MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Thermodynamics of Materials

3.00 Fall 2001

W. Craig Carter

Department of Materials Science and Engineering

Massachusetts Institute of Technology

77 Massachusetts Ave.

Cambridge, MA 02139

Problem Set 5: Due Fri. Oct. 26, Before 5PM in 4-047

Exercise 5.1

Consider heating a body $A$, of constant heat capacity $1$J/${}^\circ$C and initially at temperature 100K, to a final temperature of 200K. The heating takes place by sequentially placing it in thermal contact with $N$ different large thermal reservoirs (so large that there temperature does not change during thermal contact).

For example,

$N$=1

The body $A$ is placed in contact with a single reservoir at 200K.

$N$=2

The body $A$ first attains thermal equilibrium with a one reservoir at $T$=150K; subsequently, it attains thermal equilibrium with the second at $T$=200K.

$N=M$

The body $A$ first attains thermal equilibrium with a one reservoir at $T= (100 + 100/M)$K. Next, it attains equilibrium with a reservoir at $T= (100 + 2{\times}100/M)$K. $\ldots$. Next, it attains equilibrium with a reservoir at $T= (100 + j{\times}100/M)$K. $\ldots$. Finally, $A$ attains equilibrium with a reservoir at $T= (100 + M{\times}100/M)$K.

On the same graph, plot the change in (1) entropy of the body $A$ and (2) the change in entropy of the universe as a function of $1/N$.

Exercise 5.2

Consider the phase transformation of pure carbon graphite to pure carbon diamond at atmospheric pressure and at temperatures between 298 and 1200K.

Using the data:

graphite	Standard molar enthalpy	$\ensuremath{\overline{H}}(T=298,P=1$   atm$) = 0$
graphite	Molar entropy	$\ensuremath{\overline{S}}(T=298,P=1$   atm$) = 5.694 {\ensuremath{\frac{\mbox{J}}{\mbox{mole} {}^\circ\mbox{K}}}}$
graphite	Molar heat capacity	$\ensuremath{\overline{C_P}}(T=298,P=1$   atm$) = 17.2 + 4.27 T - \frac{8.79 \times 10^{-5}}{T^2} {\ensuremath{\frac{\mbox{J}}{\mbox{mole} {}^\circ\mbox{K}}}}$
diamond	Standard molar enthalpy	$\ensuremath{\overline{H}}(T=298,P=1$   atm$) = 1900 {\ensuremath{\frac{\mbox{J}}{\mbox{mole}}}}$
diamond	Molar entropy	$\ensuremath{\overline{S}}(T=298,P=1$   atm$) = 2.44 {\ensuremath{\frac{\mbox{J}}{\mbox{mole} {}^\circ\mbox{K}}}}$
diamond	Molar heat capacity	$\ensuremath{\overline{C_P}}(T=298,P=1$   atm$) = 9.12 + 13.2 T - \frac{6.19 \times 10^{-5}}{T^2} {\ensuremath{\frac{\mbox{J}}{\mbox{mole} {}^\circ\mbox{K}}}}$

1. Draw an accurate plot of the molar change in molar enthalpy for graphite transforming to diamond at 1 atm pressure for temperatures between 298 and 1200K.
2. Draw an accurate plot of the molar change in Gibb's free energy for graphite transforming to diamond at 1 atm pressure for temperatures between 298 and 1200K.
3. Draw an accurate plot of each of the molar free energies for diamond and for graphite as a function of the entropy of the system at 1 atm pressure for temperatures between 298 and 1200K.

Exercise 5.3

A quantity of super-cooled liquid Mn at 800K is adiabatically solidified into its equilibrium solid phase at constant pressure. From the data below, calculate what fraction of the Mn will have solidified and the final temperature of the system.

Transition (phase 1 $\rightarrow$ phase 2)	$\Delta \ensuremath{\overline{H}}$ ( $\frac{\mbox{J}}{\mbox{mole}}$)	$T^{trans}_{eq}$ (K)
$\alpha \rightarrow \beta$	2010	993
$\beta \rightarrow \gamma$	2300	1373
$\gamma \rightarrow \delta$	1800	1409
$\delta \rightarrow$   liquid$$	13400	1517

phase of Mn	$\ensuremath{\overline{C_p}}$ ( $\frac{\mbox{J}}{\mbox{mole} {}^\circ\mbox{K}}$)	temperature range (K)
$\alpha$	$21.6 + 15.9T$	298-993
$\beta$	$34.9 + 2.8T$	993-1373
$\gamma$	$44.8$	1373-1409
$\delta$	$47.3$	1409-1517
liquid	$46.0$	1517-$T_{boil}$