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2

## Question 8801d

A08
Featured 2 months ago

#### Answer:

This is what I get

#### Explanation:

Voltage vs time relationship of a charging $R C$ circuit is shown in the figure above and is given as

$V \left(t\right) = {E}_{s} \left(1 - {e}^{- \frac{t}{\tau}}\right)$ ......(1)
where ${E}_{s}$ is the DC supply across $R C$ circuit of time constant $\tau$.

It is given that ${E}_{s} = {V}_{C} - {0}_{\text{ref}} = 8.262 - 2.385 = 5.877 \setminus V$

We also know that charge held in a capacitor can be written as

$Q = C V$.

Therefore (1) becomes

$Q \left(t\right) = C {E}_{s} \left(1 - {e}^{- \frac{t}{\tau}}\right)$ ......(2)

Total charge would be maximum at $t = \infty$. From (2) we get

${Q}_{\max} = C {E}_{s}$ ........(3)

Given condition is Charge on the capacitor must be $= \frac{15}{16} {Q}_{\max}$ $\implies$ at what time $t$ or how many times $R C$.
Let the desired time be $= x \tau$

$\frac{15}{16} {Q}_{\max} = C {E}_{s} \left(1 - {e}^{- \frac{x \tau}{\tau}}\right)$ .......(4)

From (3) and (4) we get

$\frac{15}{16} = \left(1 - {e}^{- x}\right)$

Inserting various values in (1) we get

$V \left(t\right) = 5.877 \times \frac{15}{16} = 5.510 \setminus V$

5

## A projectile launcher is set on a table so that the ball becomes a projectile at a height of 1.2 m above the floor. The mass of the ball is 0.01 kg. A plunger is used to push the ball into the barrel of the launcher compressing a spring a distance of...?

A08
Featured 2 months ago

Let the initial velocity of the projectile be $= u$. Let the origin of coordinates be located at the point of projection.

Horizontal motion.
Distance traveled $= 4 \setminus m$ during time of flight $t$, with initial horizontal velocity $= u \cos {30}^{\circ}$

$\therefore u \cos {30}^{\circ} t = 4$
$\implies \frac{\sqrt{3}}{2} u t = 4$ .......(1)

Vertical motion.
To calculate the time of flight we use the kinematic expression

$s = u t + \frac{1}{2} a {t}^{2}$

Taking $g = 9.8 \setminus m {s}^{-} 2$, noting that gravity acts in the downwards direction and inserting given values we get

$- 1.2 = u \sin {30}^{\circ} t + \frac{1}{2} \left(- 9.8\right) {t}^{2}$
$\implies 4.9 {t}^{2} - 0.5 u t - 1.2 = 0$ ........(2)

Eliminating $u t$ from (2) with the help of (1) we get
$\implies 4.9 {t}^{2} - 0.5 \times \frac{8}{\sqrt{3}} - 1.2 = 0$
$\implies t = \pm 0.846 \setminus s$ ........(3)

Ignoring the $- v e$ root as time can not be negative. From (1) we get the value of $u$ as

$u = \frac{8}{\sqrt{3} \times 0.846} = 5.46 \setminus m {s}^{-} 1$ ......(4)

Now the kinetic energy of the projectile at the time of projection is provided by the potential energy of the compressed spring. We know that

$K {E}_{\text{projectile}} = \frac{1}{2} m {u}^{2}$ and
$P {E}_{\text{spring}} = \frac{1}{2} k {x}^{2}$
where $m$ is mass of the projectile, $k$ is the spring constant and $x$ is the compression of the spring.

Equating both we get

$\frac{1}{2} k {x}^{2} = \frac{1}{2} m {u}^{2}$
$\implies k = m {u}^{2} / {x}^{2}$

Inserting various values we get

$\implies k = 0.01 {u}^{2} / {\left(0.1\right)}^{2}$
$\implies k = {u}^{2}$

Using (4)

$k = 29.8 \setminus N {m}^{-} 1$

2

## How does conservation of momentum explain how Newton's cradle works?

NJ
Featured 2 months ago

You lift any number of balls and let go. The moving ball(s) then collide(s) with the stationary ball(s) and causes an equal number of balls to move on the other side.

This demonstrates conservation of momentum, because the number of balls moving before the collision is equal to the number of balls moving after the collision.

Momentum is defined as: $\setminus \vec{p} = m \setminus \vec{v}$

Each ball has the same mass, so this equivalent to saying the mass of moving balls remains constant from collision to collision. That takes care of one term. What about velocity? Well, the velocity of the moving balls right before the collision and the velocity of the moving balls on the other side right after the collision are equal. This can be seen, because the moving balls after the collision will never surpass the starting height of the balls on pre-collision side.

So if mass and velocity are unchanging for the system right before and right after collision, then the momentum will necessarily be constant: hence, conserved.

2

## Find an expression for the displacement?

A08
Featured 2 months ago

Given expression for velocity of a particle by

v=(t−2)\ ms^-1

Comparing with kinematic expression

$v = u + a t$ ...............(1)

we get

$u = - 2 \setminus m {s}^{-} 1 \mathmr{and} a = 1 \setminus m {s}^{-} 2$

(1). Kinematic expression for displacement is

$s = {s}_{0} + u t + \frac{1}{2} a {t}^{2}$

Inserting given conditions we get the expression for displacement $r$ as

$r = {r}_{0} + \left(- 2\right) t + \frac{1}{2} \left(1\right) {t}^{2}$
$r = {r}_{0} - 2 t + \frac{1}{2} {t}^{2}$

To find ${r}_{0}$, use the given condition at $t = 6 \setminus s$

$r \left(6\right) = 11 = {r}_{0} - 2 \left(6\right) + \frac{1}{2} {\left(6\right)}^{2}$
$\implies 11 = {r}_{0} - 12 + 18$
$\implies {r}_{0} = 5 \setminus m$

$\therefore$ Expression for $r$ becomes

$r = 5 - 2 t + \frac{1}{2} {t}^{2}$ ........(2)

(2). From (2)

$r \left(8\right) = 5 - 2 \times 8 + \frac{1}{2} {\left(8\right)}^{2}$
$\implies r \left(8\right) = 5 - 16 + 32$
$\implies r \left(8\right) = 21 \setminus m$

(3). From (2)

$53 = 5 - 2 t + \frac{1}{2} {t}^{2}$
$\implies {t}^{2} - 4 t - 96 = 0$

Solving the quadratic by split the middle term we get

$t = - 8 \mathmr{and} 12$,

Ignoring $- v e$ root as time can not be negative and inserting in (1) we get

$v \left(12\right) = - 2 + 1 \times 12 = 10 \setminus m {s}^{-} 1$

3

## What volume of water at 0°C can a freezer make into ice cubes in 1.0 hour,if the coefficient of performance of the cooling unit is 7.0 and the power input is 1.0 kilowatt?

Jane
Featured 2 months ago

#### Answer:

$8.82 L$

#### Explanation:

Power input of the refrigerator is $1 k W$ means per second is supplies $1 \cdot {10}^{3} J$ of energy

So,in $1 h r$ it will supply ${10}^{3} \cdot 3600 = 36 \cdot {10}^{5} J = 8.56 \cdot {10}^{5}$ calories of heat energy.

Now,let the refrigerator will take the temperature of water to ${T}_{2}$ and the water is at ${T}_{1}$ temperature.

Given, $\beta = 7 , {T}_{1} = 0 + 273 = 273 K$

So,using the equation, $\beta = \frac{{T}_{2}}{{T}_{1} - {T}_{2}}$

We,get, ${T}_{2} = 238.875 K = - {34.125}^{\circ} C$

Now,heat energy required for conversion of $m$ $g r a m$ of water at ${0}^{\circ} C$ to same amount of ice at ${0}^{\circ} C$ i.e the latent heat required is $80 m$ (as,latent heat of water is $80$calories per gram)

So,to take this $m$ $g r a m$ of ice at ${0}^{\circ} C$ to $- {34.125}^{\circ} C$ heat energy required will be $m \cdot 0.5 \cdot \left(0 - \left(- 34.125\right)\right)$ calories (using the formula, heat required $H = m s d \theta$,where, $s$ is the specific heat,for ice it is $0.5$ C.G.S units and $d \theta$ is the change in temperature)

So,this entire amount of heat will be equals to $8.56 \cdot {10}^{5}$ calories

So, $80 m + m \cdot 0.5 \cdot 34.125 = 8.56 \cdot {10}^{5}$

we get,from the above equation, $m = 8819.06$ $g r a m$

Now,density of water is $1$ $\frac{g r a m}{c {m}^{3}}$

So,volume of this much water will be $\frac{8819.06}{1} = 8819.06 c {m}^{3}$ (as, $v = \frac{m}{d}$)

Now, $1 c {m}^{3} = 0.001 L$

so, $8819.06 c {m}^{3} = 8.81906 = 8.82 L$

2

## A hot coffee at 80 c is left in a room at 20 c.it is found that after 20 minutes in the room, the temperature of coffee has decreased by 20 c. Determine the temperature of coffee after 30 minutes?

Nam D.
Featured 1 month ago

#### Answer:

I get ${53}^{\circ} \text{C}$ after $30$ minutes have passed.

#### Explanation:

We use Newton's Law of Cooling, which states that

$\frac{\mathrm{dT}}{\mathrm{dt}} = K \left(T - {T}_{0}\right)$

Solving the differential equation, we get $T = {T}_{0} + A {e}^{K t}$.

where $t$ is the time in minutes, ${T}_{0}$ is the surrounding temperature, $T$ is the object temperature, and $K$ is the constant (varies).

Plugging in the values, we need to find $K$ and $A$ first.

We got:

${T}_{0} = 20$

$\therefore T = 20 + A {e}^{K t}$

When $t = 0 , T = 80$, and so

$80 = 20 + A {e}^{0}$

$80 = 20 + A$

We get:

$A = 60$

The equation becomes:

$\therefore T = 20 + 60 {e}^{K t}$

Now, we can find $K$.

Decreasing by ${20}^{\circ} \text{C}$, the new temperature is therefore $80 - 20 = {60}^{\circ} \text{C}$.

$60 = 20 + 60 {e}^{20 K}$

${e}^{20 K} = \frac{40}{60}$

$= \frac{2}{3}$

$20 K = \ln \left(\frac{2}{3}\right)$

$K = \frac{\ln \left(\frac{2}{3}\right)}{20}$

$\approx - 0.02$

So, the final equation is:

$T = 20 + 60 {e}^{- 0.02 t}$

After $30$ minutes, we plug in $t = 30$, and we get

$T = 20 + 60 {e}^{- 0.02 \cdot 30}$

$= 20 + 60 {e}^{-} 0.6$

$= 20 + 32.9$

$= 52.9$

$\approx {53}^{\circ} \text{C}$

So, the temperature after half an hour will be approximately $53$ degrees celsius.

2

## A tank open on top has a depth h and is filled with water. A hole is made three-quarter from the ground and another hole is made half the depth. What is the difference in the range of water ejected through these holes?

Mr. Mike
Featured 3 weeks ago

#### Answer:

The difference in range will be the height of the level of water in the tank times $\left(1 - \frac{\sqrt{3}}{2}\right)$.

#### Explanation:

You really want to use Torricelli's Law here.

$v = \sqrt{2 g \left(h - x\right)}$

where

$v$ = the horizontal velocity of the stream of water,

$g$ = the acceleration of gravity, and

$h$ = the vertical distance from the surface of the water to the ground, and

$x$ = is the vertical height of the hole from the ground (see diagram).

The range of the stream is the horizontal distance the stream lands away from the tank. For the halfway hole, I have labeled this ${R}_{1}$, and for the 3/4 hole, I have labeled this ${R}_{2}$.

How long will it take a water droplet to reach the ground once it has left the hole? Since it's vertical velocity component is controlled by gravity, the time it takes to land on the ground, $t$, will be

$t = \sqrt{\frac{2 x}{g}}$.

For the halfway hole, $x = \frac{h}{2}$, and for the 3/4 hole, $x = \frac{3 h}{4}$.

Now we need to figure out the horizontal distance that the drop will travel in that time. It will simply be its horizontal velocity times the time.

$R = v t$

We can find $v$ from Torricelli's Law and $t$ from above, so we will have

$R = \sqrt{2 g \left(h - x\right)} \sqrt{\frac{2 x}{g}} = 2 \sqrt{\left(h - x\right) x}$

The problem statement asks us to determine

R_1-R_2=2sqrt((h-h/2)h/2)-2sqrt((h-(3h)/4)(3h)/4

$= 2 h \left[\sqrt{\left(1 - \frac{1}{2}\right) \frac{1}{2}} - \sqrt{\left(1 - \frac{3}{4}\right) \frac{3}{4}}\right]$

$= 2 h \left(\frac{1}{2} - \frac{\sqrt{3}}{4}\right) = h \left(1 - \frac{\sqrt{3}}{2}\right)$

2

## A great conical mound of height h is built by workers. If the workers simply heap up uniform material found at ground level, and the total weight of the finished mound is M, show that the work they do is (1/4)hM?

dk_ch
Featured 1 month ago

Let the radius of the great conical mound of height $h$ built by workers be $r$.

Given that the weight of the finished mound is $M$. So the weight of finished mound per unit volume will be $m = \frac{M}{\frac{1}{3} \pi {r}^{2} h}$.

Now for the sake of our calculation let us consider the center of the circular base of the the mound $\left(O\right)$ as origin with diameter of the mound lying along X-axis and height along Y-axis.

It is obvious that the rate of decrease of radius of the conical mound with height will be given by $\frac{r}{h}$.

Hence at an arbitrary height $y$ its radius will be $\left(r - \frac{r y}{h}\right) = r \left(1 - \frac{y}{h}\right)$

So the volume of an imaginary circular disk of infinitesimal thickness $\mathrm{dy}$ at this height $y$ will be given by

$\mathrm{dv} = \pi {r}^{2} {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

So weight of this thin disk will be

$= m \cdot \mathrm{dv} = \frac{M}{\frac{1}{3} \pi {r}^{2} h} \cdot \pi {r}^{2} {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

$= \frac{3 M}{h} \cdot {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

So work done against gravitational pull for lifting this imaginary thin disk to a height of $y$ will be given by

$\mathrm{dw} = \text{weight"(mdv)xx"height} \left(y\right) = \frac{3 M}{h} \cdot {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy} \cdot y$

The total work done $W$ to finish the mound will be obtained by integrating this $\mathrm{dw}$ as follows where $y$ varies from $0 \to h$

$W = \frac{3 M}{h} {\int}_{0}^{h} {\left(1 - \frac{y}{h}\right)}^{2} y \mathrm{dy}$

$\implies W = \frac{3 M}{h} {\left[{y}^{2} / 2 - \frac{2 {y}^{3}}{3 h} + {y}^{4} / \left(4 {h}^{2}\right)\right]}_{0}^{h}$

$\implies W = \frac{3 M}{h} \left[{h}^{2} / 2 - \frac{2 {h}^{3}}{3 h} + {h}^{4} / \left(4 {h}^{2}\right)\right]$

$\implies W = \frac{3 M}{h} \cdot \frac{1}{12} \left[6 {h}^{2} - 8 {h}^{2} + 3 {h}^{2}\right]$

$\implies W = \frac{3 M}{h} \cdot \frac{1}{12} \cdot {h}^{2}$

$\implies W = \frac{1}{4} h M$

Proved

Alternative method

Considering the center of mass of the conical mound which is placed at $\frac{1}{4} h$ we can calculate the work done in simpler way.

So work done to heap up uniform material found at ground level which is equivalent to lift the COM to a height of $\frac{h}{4}$ from ground will be

$W = M \times \frac{h}{4} = \frac{1}{4} h M$

2

## A charge of 2 μC is located 48 cm above the origin. A second charge of -8 μC is located 48 cm below the origin. What is the magnitude and direction of the net electric field 20cm to the right of the origin?

Michael
Featured 1 month ago

#### Answer:

$\textsf{E = 316.7 \times {10}^{3} \textcolor{w h i t e}{x} \text{N/C}}$ at an angle of $\textsf{{75.9}^{\circ}}$ below the horizontal or bearing $\textsf{{284.1}^{\circ}}$

#### Explanation:

The electric field at a particular point is the force experienced by a test charge of +1 Coulomb.

I will assign a charge of +1C at the location specified and find the net force that it experiences. The electric field is found by dividing that force by unit charge.

I will show 2 ways that you can do this.

(1)

Coulombs Law gives us:

$\textsf{F = k . \frac{{q}_{1} {q}_{2}}{r} ^ 2}$

$\textsf{k = 9.00 \times {10}^{9} \textcolor{w h i t e}{x} N . {m}^{2} \text{/} {C}^{2}}$

Here are the forces acting:

From Pythagoras we get:

$\textsf{{r}^{2} = {48}^{2} + {20}^{2}}$

$\textsf{r = 52 \textcolor{w h i t e}{x} c m = 0.52 \textcolor{w h i t e}{x} m}$

$\therefore$$\textsf{{F}_{1} = \frac{9 \times {10}^{9} \times 2 \times {10}^{- 6} \times 1}{0.52} ^ 2 = 66.57 \times {10}^{3} \textcolor{w h i t e}{x} N}$

$\textsf{{F}_{1} = 66.57 \textcolor{w h i t e}{x} k N}$

By inspection you can see that $\textsf{{F}_{2}}$ must be 4 x this:

$\textsf{{F}_{2} = 66.57 \times 4 = 266.27 \textcolor{w h i t e}{x} k N}$

Now we find the resultant of these two forces.

We have a S ide A ngle S ide triangle ABC so we can apply The Cosine Rule:

$\textsf{{a}^{2} = {b}^{2} + {c}^{2} - 2 \text{bc} \cos A}$

From the diagram you can see that $\textsf{\tan \theta = \frac{48}{20} = 2.4}$

From which $\textsf{\theta = {67.38}^{\circ}}$

$\textsf{A = 2 \theta = 2 \times {67.38}^{\circ} = {134.76}^{\circ}}$

$\therefore$$\textsf{{F}_{r e s}^{2} = {66.57}^{2} + {266.27}^{2} - 2 \times 66.57 \times 266.27 \cos \left(134.76\right)}$

$\textsf{{F}_{r e s}^{2} = 4 , 431.5 + 70 , 899.7 - \left(- 24 , 962.5\right)}$

$\textsf{{F}_{r e s} = 316.69 \textcolor{w h i t e}{x} k N = 316.69 \times {10}^{3} \textcolor{w h i t e}{x} N}$

$\therefore$sf(color(red)(E=(316.69xx10^3)/(1)color(white)(x)"N/C")

To find angle C we can use The Sine Rule:

$\textsf{\frac{a}{\sin} A = \frac{c}{\sin} C}$

$\therefore$$\textsf{\frac{316.69}{\sin {134.76}^{\circ}} = \frac{266.7}{\sin} C}$

$\textsf{\sin C = \frac{266.27}{446.0} = 0.597}$

$\textsf{C = {36.65}^{\circ}}$

$\textsf{{36.65}^{\circ} + {134.76}^{\circ} + B = {180}^{\circ}}$

$\textsf{B = {8.58}^{\circ}}$

The angle the resultant makes with the horizontal is therefore sf(67.38^@+8.56^@=color(red)(75.94^@)

(2)

We can resolve $\textsf{{F}_{1}}$ and $\textsf{{F}_{2}}$ into their horizontal and vertical components which can be added and resolved using Pythagoras.

$\textsf{{F}_{x} = {F}_{1} \cos \theta - {F}_{2} \cos \theta}$

$\textsf{{F}_{1} = 66.57 \times 0.3846 - 266.27 \times 0.3846 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{x} = 25.603 - 102.407 = 76.80 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{y} = {F}_{1} \sin \theta + {F}_{2} \sin \theta}$

$\textsf{{F}_{y} = 66.57 \times 0.92307 + 266.27 \times 0.92307 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{y} = 61.449 + 245.785 = 307.23 \textcolor{w h i t e}{x} k N}$

Now we can apply Pythagoras:

$\textsf{{F}_{r e s}^{2} = {76.80}^{2} + {307.23}^{2}}$

$\textsf{{F}_{r e s} = \sqrt{100 , 288.5} = 316.68 \textcolor{w h i t e}{x} k N}$

$\therefore$sf(color(red)(E=(316.68xx10^3)/(1)color(white)(x)"N/C")

$\textsf{\tan \alpha = {307.23}^{\circ} / {76.8}^{\circ} = 4}$

sf(color(red)(alpha=75.96^@)#

As you can see there is close agreement between the 2 methods so thats all good.

2

## What is the internal energy for an isothermal process?

Truong-Son N.
Featured 1 month ago

$\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV} = {\int}_{{V}_{1}}^{{V}_{2}} - P + T {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$

Now decide what gas law to use, or what $\alpha$ and $\kappa$ corresponds to your substance.

Well, from the total differential at constant temperature,

$\mathrm{dU} = {\cancel{{\left(\frac{\partial U}{\partial T}\right)}_{V} \mathrm{dT}}}^{0} + {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV}$,

so by definition of integrals and derivatives,

$\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV}$ $\text{ } \boldsymbol{\left(1\right)}$

The natural variables are $T$ and $V$, which are given in the Helmholtz free energy Maxwell relation.

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$$\text{ } \boldsymbol{\left(2\right)}$

This is also related by the isothermal Helmholtz relation (similar to the isothermal Gibbs' relation):

$\mathrm{dA} = \mathrm{dU} - T \mathrm{dS}$$\text{ } \boldsymbol{\left(3\right)}$

Differentiating $\left(3\right)$ at constant temperature,

${\left(\frac{\partial A}{\partial V}\right)}_{T} = {\left(\frac{\partial U}{\partial V}\right)}_{T} - T {\left(\frac{\partial S}{\partial V}\right)}_{T}$

From $\left(2\right)$,

${\left(\frac{\partial A}{\partial V}\right)}_{T} = - P$

and also from $\left(2\right)$,

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from $\left(3\right)$ we get

$- P = {\left(\frac{\partial U}{\partial V}\right)}_{T} - T {\left(\frac{\partial P}{\partial T}\right)}_{V}$

or we thus go back to $\left(1\right)$ to get:

$\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2)-P + T((delP)/(delT))_VdV" }}{|}$

And what remains is to distinguish between the last term for gases, liquids and solids...

GASES

Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{n R}{V}$

and that just means

${\left(\frac{\partial U}{\partial V}\right)}_{T} = - P + \frac{n R T}{V}$

$= - P + P = 0$

which says that ideal gases have changes in internal energy as a function of only temperature. One would get

$\textcolor{b l u e}{\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} 0 \mathrm{dV} = 0}$.

Not very interesting.

Of course, if your gas is not ideal, this isn't necessarily true.

LIQUIDS AND SOLIDS

These data are tabulated as coefficients of volumetric thermal expansion $\alpha$ and isothermal compressibility $\kappa$,

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

$\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}$

$\frac{\alpha}{\kappa} = \left[. . .\right] = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

at VARIOUS temperatures for VARIOUS condensed phases. Some examples at ${20}^{\circ} \text{C}$:

• ${\alpha}_{{H}_{2} O} = 2.07 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{A u} = 4.2 \times {10}^{- 5} {\text{K}}^{- 1}$ (because that's REAL useful, right?)
• ${\alpha}_{E t O H} = 7.50 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{P b} = 8.7 \times {10}^{- 5} {\text{K}}^{- 1}$

• ${\kappa}_{{H}_{2} O} = 4.60 \times {10}^{- 5} {\text{bar}}^{- 1}$

• ${\kappa}_{A u} = 5.77 \times {10}^{- 7} {\text{bar}}^{- 1}$
• ${\kappa}_{E t O H} = 1.10 \times {10}^{- 4} {\text{bar}}^{- 1}$
• ${\kappa}_{P b} = 2.33 \times {10}^{- 6} {\text{bar}}^{- 1}$

In that case,

${\left(\frac{\partial U}{\partial V}\right)}_{T} = - P + \frac{T \alpha}{\kappa}$

Thus,

$\textcolor{b l u e}{\Delta U} = {\int}_{{V}_{1}}^{{V}_{2}} - P + \frac{T \alpha}{\kappa} \mathrm{dV}$

$= \textcolor{b l u e}{\frac{\alpha T \Delta V}{\kappa} - {\int}_{{V}_{1}}^{{V}_{2}} P \left(V\right) \mathrm{dV}}$

Next, the pressure will vary a lot if the volume varies a little for a solid or liquid...

Hence, to find $\Delta U$, we first need to measure the temperature of the system, and reference $\alpha$ and $\kappa$.

Then, we'll need to measure how the volume changes (to high precision!) with the change in pressure to empirically find the pressure as a function of the volume for the given solid or liquid.