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这是挪威科技大学 热力学课程的代写成功案例。
AMME2262课程简介
This unit aims to teach the fundamentals of thermal engineering with a particular emphasis on thermodynamics with some elements of heat transfer and thermal system instrumentation and measurement. At the end of this unit students will have an understanding of the basic laws of thermodynamics, heat transfer and measurement of thermal systems and how to apply them in an engineering context. The course content will include topics covering thermal and phase properties of substances, heat and work interactions, the first law of thermodynamics, the second law of thermodynamics, entropy and its application to engineering systems, gas power and refrigeration cycles, heat transfer by conduction, convection and radiation, thermal sensors and data acquisition.
Prerequisites
At the completion of this unit, you should be able to:
- LO1. write a formal laboratory report
- LO2. work in groups to complete laboratory experiments and analyse the results
- LO3. evaluate the relevant parameters for heat flow in internal engineering systems such as refrigeration and heating systems
- LO4. demonstrate an understanding of some of the basic equations governing thermodynamics
- LO5. analyze the thermodynamics of a simple open or closed engineering system
- LO6. analyze and determine the heat fluxes governing a control volume.
AMME2262 Thermodynamics HELP（EXAM HELP， ONLINE TUTOR）
It has been said that “six monkeys set to strum unintelligently on typewriters for millions of years would be bound in time to produce all the books in the British Museum.” This statement is nonsense, for it gives a misleading conclusion about very, very large numbers. Could all the monkeys in the world have typed out even a single specified book in the age of the universe?
Suppose that $10^{10}$ monkeys have been seated at typewriters for the age of the universe, $10^{18} \mathrm{~s}$. Suppose that a monkey can hit 10 keys per second on the typewriter per second. Assume the typewriter has 44 keys (we won’t distinguish between upper and lower case). Assume that Shakespeare’s Hamlet has $10^5$ characters.
Will the monkeys hit upon Hamlet?
(a) Show that the probability that any given sequence of $10^5$ characters typed at random will come out in the correct sequence (the sequence of Hamlet) is
$$
\left(\frac{1}{44}\right)^{100,000} \simeq 10^{-164,345},
$$
where we used that $\log _{10} 44=1.64345$.
(b) Hence show that the probability that one of the monkeys will type Hamlet in the age of the universe is about $10^{-164321}$.
Note that multiplying by the number of sequences that are typed during the age of the universe makes only a small difference to the power of 10 in the probablity. The probability is zero in any operational sense, so the statement made at the beginning of this question is nonsense: one book, much less an entire library, will never occur in the total literary production of the monkeys.
(c) Would the monkeys even produce “to be or not to be” (18 characters including spaces)?
(a) There are $44^{10^5}$ possible sequences of $10^5$ characters. Hence the probability that the correct sequence will emerge is the inverse of this:
$$
\left(\frac{1}{44}\right)^{100000} \simeq 10^{-164345} .
$$
(b) We are given $10^{10}$ monkeys and $10^{18}$ seconds with $10^4$ seconds per trial. Hence the number of trials is $10^{10} \times 10^{14}=10^{24}$. Hence the probability of the correct sequence is
$$
10^{-164345} \times 10^{24}=10^{-164321} .
$$
This probability is zero in any operational sense. Note that multiplying by the number of trials makes almost no difference to the power of 10 in Eq. (3).
(c) “To be or not to be” has 18 characters, including spaces. The probablity that this would be typed at random is $1 / 44^{18} \simeq 2.6 \times 10^{-30}$. We are given $10^{10}$ monkeys and $10^{18}$ seconds and lets say that it takes a monkey 1 second to type the 18 characters. The probability that one of the monkeys would type the phrase in the age of the universe is therefore $2.6 \times 10^{-30} \times 10^{28}=$ 0.026 . Hence the probability of the monkeys producing “To be or not to be” in the age of the universe is small, but not astronomically small.
Free energy of a 2-state system
Consider system with just two states, which is the simplest model one can study in statistical physics. Take the two states to have energies 0 and $\epsilon$.
(a) Find the free energy as a function of temperature.
(b) From the free energy determine the energy and entropy.
(c) Determine the heat capacity (also known as the specific heat) at constant volume and sketch it as a function of temperature.
(a) If the two states have energy 0 and $\epsilon$, the partition function is given by
$$
Z=1+e^{-\beta \epsilon},
$$
where $\beta=1 / k_B T$, and so the free energy is given by
$$
F=-k_B T \ln Z=-k_B T \ln \left(1+e^{-\beta \epsilon}\right) .
$$
(b) The energy is given by
$$
U=\frac{\partial}{\partial \beta}(\beta F)=\epsilon \frac{1}{e^{\beta \epsilon}+1} .
$$
The entropy is given by
$$
S=-\frac{\partial F}{\partial T}=\frac{\epsilon}{T} \frac{1}{e^{\beta \epsilon}+1}+k_B \ln \left(1+e^{-\beta \epsilon}\right) .
$$
Note that $S=(U-F) / T$ as required.
(c) The heat capacity at constant volume is given by
$$
C=\left(\frac{\partial U}{\partial T}\right)_V .
$$
Differentiating the above expression for $U$ gives
$$
C=k_B\left(\frac{\epsilon}{k_B T}\right)^2 \frac{e^{\beta \epsilon}}{\left[e^{\beta \epsilon}+1\right]^2} .
$$
The specific heat tends to zero both as $T \rightarrow 0$ and $T \rightarrow \infty$, and has a peak when $k_B T$ is about $0.42 \epsilon$. A plot is shown below.
Consider the model system of $N$ non-interacting magnetic moments discussed in class and in the book. The energy of magnetic moment $i$ is $-\mu_i B= \pm \mu B$ where $\mu_i= \pm \mu$ and $\mu$ is the size of the magnetic moment.
(a) Use the partition function to determine the magnetization $M$ exactly as a function of temperature and magnetic field.
(b) For $\mu B \ll k_B T$ show that
$$
M=N \mu^2 \frac{B}{k_B T} .
$$
Note: You derived this by a more complicated method, using results appropriate for a closed system rather than a system in thermal contact with a reservoir as here, in Qu. 2, HW 1.
(a) The energy levels are $\pm \mu B$, and so the partition function for one spin, $z$, is given by
$$
z=e^{\beta \mu B}+e^{-\beta \mu B}=2 \cosh (\beta \mu B) .
$$
The probabilities of the +1 and -1 states are given by
$$
P_1=\frac{e^{\beta \mu B}}{z}, \quad P_{-1}=\frac{e^{-\beta \mu B}}{z},
$$
and hence the magnetization per spin is given by
$$
m=\mu \tanh (\beta \mu B),
$$
and the total magnetization is
$$
M=N \mu \tanh (\beta \mu B) .
$$
(b) For $\mu B \ll k_B T$ we have, expanding the tanh in Eq. (1),
$$
m=\mu^2 \frac{B}{k_B T},
$$
which is known as Curie’s law, as discussed in class.
MY-ASSIGNMENTEXPERT™可以为您提供 sydney AMME2262 Thermodynamics热力学课程的代写代考和辅导服务！
