assignment Homework

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.

assignment Homework

Energy, Entropy, and Probabilities

The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

assignment Homework

Energy, Entropy, and Probabilities

The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align} We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

assignment Homework

Mass of a Slab

Determine the total mass of each of the slabs below.

1. A square slab of side length \(L\) with thickness \(h\), resting on a table top at \(z=0\), whose mass density is given by \begin{equation} \rho=A\pi\sin(\pi z/h). \end{equation}
2. A square slab of side length \(L\) with thickness \(h\), resting on a table top at \(z=0\), whose mass density is given by \begin{equation} \rho = 2A \Big( \Theta(z)-\Theta(z-h) \Big) \end{equation}
3. An infinitesimally thin square sheet of side length \(L\), resting on a table top at \(z=0\), whose surface density is given by \(\sigma=2Ah\).
4. An infinitesimally thin square sheet of side length \(L\), resting on a table top at \(z=0\), whose mass density is given by \(\rho=2Ah\,\delta(z)\).
5. What are the dimensions of \(A\)?
6. Write several sentences comparing your answers to the different cases above.