face Lecture

10 min.

Basics of heat engines
This brief lecture covers the basics of heat engines.
  • Found in: Contemporary Challenges course(s)
A short lecture introducing the idea that most of the energy loss when driving is going into the kinetic energy of the air.

group Small Group Activity

30 min.

Grey space capsule
In this small group activity, students work out the steady state temperature of an object absorbing and emitting blackbody radiation.

face Lecture

120 min.

Work, Heat, and cycles
These lecture notes covering week 8 of Thermal and Statistical Physics include a small group activity in which students derive the Carnot efficiency.

None

Heat pump
  1. Show that for a reversible heat pump the energy required per unit of heat delivered inside the building is given by the Carnot efficiency: \begin{align} \frac{W}{Q_H} &= \eta_C = \frac{T_H-T_C}{T_H} \end{align} What happens if the heat pump is not reversible?

  2. Assume that the electricity consumed by a reversible heat pump must itself be generated by a Carnot engine operating between the even hotter temperature \(T_{HH}\) and the cold (outdoors) temperature \(T_C\). What is the ratio \(\frac{Q_{HH}}{Q_H}\) of the heat consumed at \(T_{HH}\) (i.e. fuel burned) to the heat delivered at \(T_H\) (in the house we want to heat)? Give numerical values for \(T_{HH}=600\text{K}\); \(T_{H}=300\text{K}\); \(T_{C}=270\text{K}\).

  3. Draw an energy-entropy flow diagram for the combination heat engine-heat pump, similar to Figures 8.1, 8.2 and 8.4 in the text (or the equivalent but sloppier) figures in the course notes. However, in this case we will involve no external work at all, only energy and entropy flows at three temperatures, since the work done is all generated from heat.

  • Found in: Thermal and Statistical Physics course(s)
We have the following equations of state for the total magnetization \(M\), and the entropy \(S\) of a paramagnetic system: \begin{align} M&=N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\ S&=Nk_B\left\{\ln 2 + \ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right) +\frac{\mu B}{k_B T} \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}} \right\} \end{align}
  1. List variables in their proper positions in the middle columns of the charts below.

  2. Solve for the magnetic susceptibility, which is defined as: \[\chi_B=\left(\frac{\partial M}{\partial B}\right)_T \]

  3. Using both the differentials (zapping with d) and chain rule diagram methods, find a chain rule for:

    \[\left(\frac{\partial M}{\partial B}\right)_S \]

  4. Evaluate your chain rule. Sense-making: Why does this come out to zero?

  • Found in: Energy and Entropy course(s)