Homework Assignment 13 (due 12/11)

PHYS111Q : Homework Assignments

Homework Assignment 13 (due 12/11)

Heat engines, refrigerators, entropy, and the second law

Goals :

Gain a qualitative understanding of entropy and the second law of thermodynamics. Here are several statements of the second law.

$\begin{displaymath} \Delta S \geq 0 \end{displaymath}$

Energy tends to disperse.

There is no such thing as a perfect (100% efficient) engine.
Apply the first law of thermodynamics

$\begin{displaymath} \Delta E^\mathrm{int} = Q - W \end{displaymath}$

to simple systems.
Apply the first law to ideal gases undergoing various thermodynamic processes,
- For any process,
  
  $\begin{displaymath} \Delta E^\mathrm{int} = n C_V \Delta T \end{displaymath}$
- In a constant-volume process
  
  $\begin{displaymath} W = 0 \hspace{2 cm} Q = n C_V \Delta T \end{displaymath}$
- In an isobaric process (constant pressure)
  
  $\begin{displaymath} Q = n C_P \Delta T \hspace{2 cm} W = P \Delta V \end{displaymath}$
- In an isothermal process (constant temperature),
  
  $\begin{displaymath} \Delta E^\mathrm{int} = 0 \hspace{2 cm} Q = W = nRT \ln\left( \frac{V_f}{V_i} \right) \end{displaymath}$
- In an adiabatic process,
  
  $\begin{displaymath} Q = 0 \hspace{2 cm} W = -\Delta E^\mathrm{int}~( = - n C_V \Delta T ) \end{displaymath}$
  
  $\begin{displaymath} PV^\gamma = \mathrm{constant} \hspace{2cm} TV^{\gamma - 1... ...tant} \hspace{2cm} \left( \gamma = \frac{C_P}{C_V} \right) \end{displaymath}$
Calculate the efficiency

$\begin{displaymath} \epsilon = \frac{W^\mathrm{net}}{Q^\mathrm{in}} = \frac{\... ...^\mathrm{in}\vert - \vert Q^\mathrm{out}\vert}{Q^\mathrm{in}} \end{displaymath}$

of a heat engine and the coefficient of performance (COP)

$\begin{displaymath} K = \frac{\vert Q^\mathrm{in}\vert}{\vert W^\mathrm{net}\ve... ...}\vert}{\vert Q^\mathrm{out}\vert - \vert Q^\mathrm{in}\vert} \end{displaymath}$

of a refrigerator, and compare the results with those of the Carnot engine/refrigerator

$\begin{displaymath} \epsilon = 1 - \frac{T_C}{T_H} \hspace{2 cm} K = \frac{T_C}{T_H - T_C} \end{displaymath}$

the maximum values allowed by the second law.

Your assignment awaits you at the UT Homework Service (Class no. 99941)

Hints