November 2020 – Physicspupil

November 11, 2020July 25, 2022

202011110822 Short Review II (Heat and Internal Energy)

Internal Energy

Each matter has three states:

$\begin{tabular}{p{1cm}|p{4cm}|p{5cm}|p{4cm}} & Property & Molecular arrangement & Molecular movement \\ \hline \hline Solid & fixed volume and shape & closely packed and fixed in positions & vibrate about a fixed position \\ \hline Liquid & fixed volume, unfixed shape & closely packed and not fixed in positions & move freely \\ \hline Gas & unfixed volume and shape & far apart and not fixed in positions & move freely at high speed\\ \hline \end{tabular}$

Each object contains energy, measured in joules ( $\textrm{J}$ ).

Kinetic energy () of a body is the energy due to its motion.
- The average kinetic energy is the kinetic energy of each molecule.
- The total kinetic energy is the kinetic energy of all the molecules.
- $\mathrm{Total\,KE}=\mathrm{no.\,of\,molecules}\times \mathrm{average\,KE}$
Potential energy ( $\textrm{PE}$ ) of a body is the energy due to its position or configuration.
Internal energy () of a body is the sum of the (total molecular) kinetic energy (KE) and (total molecular) potential energy (PE) of all its molecules. It is the total energy stored in a body.
- $\mathrm{Internal\,energy}=\mathrm{Total\,KE}+\mathrm{Total\,PE}$

Temperature,

in common sense, is a measure of hotness;
in scientific sense, is a measure of the average kinetic energy of a body.

Heat $Q$ is the energy transferred from a hot object to a cold object as a result of temperature difference. Its unit is also joule ( $\textrm{J}$ ).

REMARK. It would be to speak of heat as a form of energy because by definition, heat is the energy transferred , or, the change in internal energy.

Power $P$ is the rate of transferring energy, and it is measured in watts ( $\textrm{W}$ ).

$\mathrm{Power}=\displaystyle{\frac{\mathrm{energy\,transferred}}{\mathrm{time}}}\qquad\qquad \bigg( P=\frac{Q}{t}\bigg)$

REMARK. If energy is transferred not only by heating (e.g. but also by doing work ), we may not assume and we may not write .

Specific Heat Capacity

Specific heat capacity $c$ of a substance is the energy $Q$ transferred by heating in order to raise the temperature of the substance of mass $m=1\,\mathrm{kg}$ by $\Delta T=1\,\mathrm{C^\circ }$ .

By making an experiment, we can find out the specific heat capacity $c$ of a substance of mass $m$ by measuring how much energy $Q$ is needed when we increase its temperature by an arbitrary degree $\Delta T$ :

$\displaystyle{c=\frac{Q}{m\Delta T}}$

$\textrm{Unit:}\quad \mathrm{J\,kg^{-1}\,C^{\circ -1}}$

Or, given the specific heat capacity $c$ and the mass $m$ of an unknown substance, and having decided to what degree $\Delta T$ we wish to raise its temperature, we can figure out how much energy $Q$ is needed:

$Q=mc\Delta T$

REMARK.

Heat capacity is denoted by a capital , defined . Heat capacity is related to specific heat capacity by or .

When an object is measured to be of a certain temperature, the unit we say degree Celsius or degree Fahrenheit. When an object has increased or decreased by a certain degree, the temperature change/difference we say is of certain Celsius degree(s) or Fahrenheit degree(s).

When two objects of different temperatures are in contact, energy is transferred from the hotter one to the colder one, until they reach the same temperature, in which there is no $\textrm{\scriptsize NET}$ flow of heat, i.e., thermal equilibrium.

Before the objects reach the thermal equilibrium, the energy lost by the hotter object will be gained by the colder object if there is no loss of energy to the surroundings

because the law of conservation of energy states that:

The total amount of energy in a closed system is conserved, i.e., it is always kept constant. Energy cannot be created or destroyed, although it can change from one form into another.

CONCEPT TEST (Internal Energy)

Which of the following statement(s) is/are correct?
1. Temperature is a measure of the internal energy of an object.
2. Heat is the average kinetic energy of an object.
3. Heat is the energy gained by an object as a result of heat flow.
  1. (I) only
  2. (III) only
  3. (I) and (II) only
  4. (II) and (III) only
Which of the following deductions must be correct?
1. Average KE $\uparrow$ $\Rightarrow$ Total KE $\uparrow$ .
2. Internal energy $\downarrow$ $\Rightarrow$ Total KE $\downarrow$ .
3. Total PE internal energy .
  1. (I) only
  2. (II) only
  3. (I) and (III) only
  4. (I), (II) and (III)
Which of the following statements is/are wrong?
1. Heat is a description of the stored energy in a body.
2. Temperature is a measure of the total KE of a body.
3. Total KE Average KE .
  1. (I) only
  2. (I) and (II) only
  3. (II) and (III) only
  4. (I), (II) and (III)
Which of the following is correct?
1. Molecules in random motion possess molecular kinetic energy.
2. The molecules of a gas are very far apart, so that they have less intermolecular potential energy than when they are in liquid or solid state.
3. The molecules of a liquid move at a faster speed than that of a gas.
4. The molecules of a solid vibrate about fixed positions, so that they have no molecular kinetic energy.

Answers:

Explanation:

(I) is wrong: temperature is a measure of the average kinetic energy. (II) is wrong: heat is the energy transferred from a hotter object to a colder object as a result of temperature difference. (III) is correct.
(I) is correct: total KE $=$ no. of molecules $\times$ average KE. (II) is wrong: a decrease in internal energy may be due to $\textrm{{\scriptsize EITHER}}$ a decrease in total KE, $\textrm{{\scriptsize OR}}$ a decrease in total PE. (III) is correct.
(I) and (II) are wrong by definition. (III) is wrong: total KE $\uparrow$ $\Rightarrow$ no. of molecules $\uparrow$ $\textrm{{\scriptsize OR}}$ average KE $\uparrow$ .
A is correct: molecular KE is related to the random motion of molecules. B is wrong: the molecules of a gas have the farthest separation between them, hence the greatest PE to overcome the attractive intermolecular forces. C is wrong: it should be the converse. D is wrong: when the molecules vibrate, there is molecular KE.

CONCEPT TEST (Specific Heat Capacity)

Which of the following statements is/are correct?
1. Heat capacity $C$ is a property of a substance.
2. Heat capacity $C$ depends on the mass of substance.
3. Specific heat capacity is a property of a substance.
  1. (I) and (II)
  2. (I) and (III)
  3. (II) and (III)
  4. (I), (II) and (III)
Which of the following statements is/are incorrect?
1. Specific heat capacity of water will be changed if it is mixed with salt.
2. Specific heat capacity $c$ can be negative.
3. Specific heat capacity of a substance is the energy that is released to lower the temperature of of the substance by .
  1. (II) only
  2. (III) only
  3. (I) and (II) only
  4. (II) and (III) only
Which of the following statements is incorrect?
1. When object $A$ is in thermal equilibrium with object $B$ , and $B$ is in thermal equilibrium with object $C$ , then $A$ and $C$ are in thermal equilibrium.
2. When two bodies are in thermal equilibrium, there is no heat flow between them.
3. Heat flows spontaneously from a hotter region to a colder region.
4. All liquid-in-glass thermometers have systematic errors because in reaching thermal equilibrium with the object to measure, the thermometer must have transferred some heat to/from the object.

Answers:

Explanation:

(I) is wrong; (II) correct: since heat capacity of a substance varies with its mass, it is not a (specific) property of the substance. (III) is correct.
(I) is correct: specific heat capacity depends on the material; when water is mixed with salt, the material is changed. (II) Specific heat capacity must be positive: as $\textrm{{\scriptsize EITHER}}$ $Q,\Delta T<0$ $\textrm{{\scriptsize OR}}$ $Q,\Delta T>0$ , so that $c=\displaystyle{\frac{Q}{m\Delta T}}>0$ . (III) is correct by definition.
B is incorrect because heat can be flowing from one body to another even though the net flow of heat is zero.

November 11, 2020July 25, 2022

202011110657 Sidenote of Separation of Variables

Two-particle system: $\displaystyle{i\hbar\frac{\partial \Psi}{\partial t}=H\Psi}$

in which $\displaystyle{H=-\frac{\hbar^2}{2m_1}\nabla_1^2-\frac{\hbar^2}{2m_2}\nabla_2^2+V(\mathbf{r_1},\mathbf{r_2},t)}$ .

Assuming that the potential is time-independent.

Let $\Psi (\mathbf{r}_1,\mathbf{r}_2,t)=\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)$ ,

$\begin{aligned} & i\hbar \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)}{\partial t} \\ = & -\frac{\hbar^2}{2m_1}\nabla_1^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)\bigg) -\frac{\hbar^2}{2m_2}\nabla_2^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \bigg) \\ &\qquad\enspace +V(\mathbf{r_1},\mathbf{r_2})\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \\ & \\ & \textrm{\scriptsize OR} \\ & \\ & i\hbar\Big[ \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)}{\partial t}\varphi (t)+\psi (\mathbf{r}_1,\mathbf{r}_2)\frac{\partial \varphi (t)}{\partial t}\Big]\\ = & -\frac{\hbar^2}{2m_1}\varphi (t)\nabla_1^2\psi (\mathbf{r}_1,\mathbf{r}_2)-\frac{\hbar^2}{2m_2}\varphi (t)\nabla_2^2\psi (\mathbf{r}_1,\mathbf{r}_2) +V\psi \varphi \\ \end{aligned}$

November 11, 2020July 25, 2022

202011110613 Sidenote of Dummy Variables

Q: What is a dummy variable?

A: $f(x)=x^2$ and $g(y)=y^2$ are dummy variables because they describe the same pattern.

The indefinite integrals $\int x\,\mathrm{d}x=x^2/2+C$ and $\int y\,\mathrm{d}y=y^2/2+C$ are $\textrm{\scriptsize \textbf{NOT}}$ dummy variables because they are functionals admitting of different functions.

But if we put an upper and a lower limit to make it a definite integral,

i.e.,

$\begin{aligned} \int_0^1x\,\mathrm{d}x=\bigg[ \frac{x^2}{2} \bigg]_0^1=\frac{1}{2} \\ \int_0^1 y\,\mathrm{d}y=\bigg[ \frac{y^2}{2} \bigg]_0^1=\frac{1}{2} \\ \end{aligned}$ ,

they are dummy variables as the structure preserves the value.

November 9, 2020October 24, 2022

202011091423 Exercise 1 (Q2)

If $f(x)=(x-1)(x+5)$ , find the values of $f(2)$ , $f(1)$ , $f(0)$ , $f(a+1)$ , $f\bigg(\displaystyle{\frac{1}{a}}\bigg)$ , $f(-5)$ .

Solution.

$\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}$

On reflection.

Suppose you are given an unknown quadratic function $f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ such that

$\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}$

and you are asked to make polynomial interpolation.

$\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}$
$\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}$

And then is the interpolated polynomial $f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1)$ .

November 5, 2020October 24, 2022

202011051527 Exercise 1 (Q1)

If $f(x)=2x^2-4x+1$ , find the values of $f(1)$ , $f(0)$ , $f(2)$ , $f(-2)$ , $f(a)$ , $f(x+\delta x)$ .

Solution.

Given $f(x)=2x^2-4x+1$ .

$\begin{aligned} f(1) & =2(1)^2-4(1)+1=-1 \\ f(0) & = 2(0)^2 - 4(0) +1 = 1 \\ f(2) & = 2(2)^2-4(2)+1 =1 \\ f(-2) & = 2(-2)^2-4(-2)+1=17\\ f(a) & = 2a^2 - 4a +1 \\ f(x+\delta x) & = 2(x+\delta x)^2 - 4 (x+\delta x) +1 \end{aligned}$

This exercise is done.

On reflection.

Suppose you are given the following conditions:

$\begin{aligned} x_0 = 0 & \qquad f(x_0) = 1 \\ x_1 = 1 &\qquad f(x_1) = -1 \\ x_2 = 2 &\qquad f(x_2) =1 \end{aligned}$

and you are asked to interpolate by Lagrange polynomials over the range $[0,2]$ .

$\begin{aligned} \mathcal{L}(x) & = (1)\bigg( \displaystyle{\frac{x-1}{0-1}} \bigg)\bigg( \displaystyle{\frac{x-2}{0-2}} \bigg) + (-1)\bigg( \displaystyle{\frac{x-0}{1-0}} \bigg) \bigg( \displaystyle{\frac{x-2}{1-2}} \bigg) + (1)\bigg( \displaystyle{\frac{x-0}{2-0}} \bigg) \bigg( \displaystyle{\frac{x-1}{2-1}} \bigg) \\ & = \displaystyle{\frac{(x-1)(x-2)}{2}} + x(x-2) + \displaystyle{\frac{x(x-1)}{2}} \\ & = \displaystyle{\frac{(x-1)(x-2)+2x(x-2)+x(x-1)}{2}} \\ & = \displaystyle{\frac{x^2-3x+2+2x^2-4x+x^2-x}{2}} \\ & = \displaystyle{\frac{4x^2-8x+2}{2}} \\ & = 2x^2-4x+1\\ \end{aligned}$

The interpolating polynomial $\mathcal{L}(x)$ checks with the original function $f(x)$ .