Answer any two questions. a. State Newton's law of cooling. Use it to find the specific heat capacity of a unknown liquid. b. Derive an expression for the pressure exerted by an ideal gas on the…

a. Newton’s Law of Cooling and Determination of Specific Heat Capacity of an Unknown Liquid

Newton’s Law of Cooling

Newton’s law of cooling states that:

The rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small.

Mathematically, it is expressed as: $$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$$ where:

• $\theta$ = temperature of the body at time $t$,
• ${\theta }_0$ = temperature of the surroundings,
• $k$ = cooling constant (depends on the nature of the body and surroundings),
• $t$ = time.

Experimental Setup to Find Specific Heat Capacity

To determine the specific heat capacity ($c$) of an unknown liquid using Newton’s law of cooling, we follow these steps:

1. Heat the Liquid: A known mass ($m$) of the liquid is heated to a temperature (${\theta }_1$) above the surrounding temperature (${\theta }_0$).
2. Insulate and Cool: The liquid is then allowed to cool in an insulated container (to minimize heat loss to surroundings other than air).
3. Measure Temperature vs. Time: Record the temperature ($\theta$) of the liquid at regular time intervals ($t$).
4. Plot a Cooling Curve: Plot $\theta$ vs. $t$ and determine the cooling constant ($k$) from the slope of the curve.
5. Relate to Specific Heat Capacity: The heat lost by the liquid in a small time interval $dt$ is given by: $$dQ=mcd\theta$$ According to Newton’s law, the rate of heat loss is also: $$\frac{dQ}{dt}=-k(\theta -{\theta }_0)$$ Equating the two expressions: $$mc\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$$ Rearranging: $$c=-\frac{k(\theta -{\theta }_0)}{m\frac{d\theta }{dt}}$$ Since $k$ can be determined from the cooling curve, $c$ can be calculated.

Example Calculation

Suppose:

• Mass of liquid, $m=0.5\text{kg}$,
• Initial temperature, ${\theta }_1=80^{\circ }\text{C}$,
• Surrounding temperature, ${\theta }_0=25^{\circ }\text{C}$,
• Cooling constant, $k=0.05{\text{s}}^{-1}$ (from the slope of the cooling curve),
• Rate of temperature change at $\theta =50^{\circ }\text{C}$, $\frac{d\theta }{dt}=-2^{\circ }\text{C/s}$.

Then: $$c=-\frac{0.05\times (50-25)}{0.5\times (-2)}=\frac{0.05\times 25}{1}=1.25{\text{J/g}}^{\circ }\text{C}$$

---

b. Derivation of Pressure Exerted by an Ideal Gas Using Kinetic Theory

Assumptions of Kinetic Theory

1. A gas consists of a large number of tiny, identical molecules in random motion.
2. The volume of the molecules is negligible compared to the volume of the gas.
3. The molecules obey Newton’s laws of motion.
4. Collisions between molecules and with the walls are perfectly elastic.
5. The time of collision is negligible compared to the time between collisions.

Derivation

Consider a gas of $N$ molecules, each of mass $m$, moving randomly in a cubical container of side length $L$. Let the molecules have velocities $v_x,v_y,v_z$ along the $x,y,z$ axes.

Collision with the Wall

When a molecule collides with a wall perpendicular to the $x$-axis, it exerts a force due to the change in momentum. The momentum change per collision is: $$\Delta p_x=2m{v}_x$$ The number of collisions per second with one wall is: $$\frac{v_x}{2L}$$ Thus, the force exerted by one molecule on the wall is: $$F_x=\Delta p_x\times \text{collisions per second}=2m{v}_x\times \frac{v_x}{2L}=\frac{m{v}_x^2}{L}$$ For $N$ molecules, the total force (pressure $\times$ area) is: $$F_{\text{total}}=\frac{Nm\overline{v_x^2}}{L}$$ where $\overline{v_x^2}$ is the mean square velocity in the $x$-direction.

Since the motion is random, $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\frac{1}{3}\overline{v^2}$, where $\overline{v^2}$ is the mean square speed.

Thus: $$F_{\text{total}}=\frac{Nm\overline{v^2}}{3L}$$ The pressure $P$ is force per unit area ($L^2$): $$P=\frac{F_{\text{total}}}{L^2}=\frac{Nm\overline{v^2}}{3L^3}=\frac{Nm\overline{v^2}}{3V}$$ where $V=L^3$ is the volume of the container.

Using the Ideal Gas Law

From kinetic theory, the average kinetic energy of a molecule is: $$\frac{1}{2}m\overline{v^2}=\frac{3}{2}{k}_{B}T$$ where $k_B$ is Boltzmann’s constant and $T$ is the absolute temperature.

Substituting: $$P=\frac{2}{3}\frac{N}{V}\left(\frac{1}{2}m\overline{v^2}\right)=\frac{2}{3}\frac{N}{V}\left(\frac{3}{2}{k}_{B}T\right)=\frac{N{k}_{B}T}{V}$$ Since $N{k}_B=nR$ (where $n$ is the number of moles and $R$ is the gas constant), we get: $$P=\frac{nRT}{V}$$ This is the ideal gas equation.

---

c. Working Principle of a Diesel Engine with $PV$ Diagram

Working Principle

A diesel engine operates on the diesel cycle, which consists of four processes:

1. Suction Stroke (Intake): Air is drawn into the cylinder.
2. Compression Stroke (Adiabatic Compression): Air is compressed adiabatically (no heat exchange), increasing its temperature.
3. Power Stroke (Combustion and Expansion): Fuel is injected and ignites due to high compression temperature, causing rapid expansion and doing work.
4. Exhaust Stroke (Rejection): Burnt gases are expelled.

$PV$ Diagram of Diesel Engine

Processes Explained

1. 1 → 2 (Adiabatic Compression):

   • Air is compressed from $V_1$ to $V_2$ (volume ratio $r=V_1/V_2$).
   • Pressure and temperature rise without heat exchange ($Q=0$).
   • Work done on the gas: $W_{12}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$, where $\gamma =C_p/C_v$.

2. 2 → 3 (Constant Pressure Heat Addition):

   • Fuel is injected at constant pressure, increasing volume to $V_3$.
   • Heat added: $Q_{23}=P_2(V_3-V_2)$.

3. 3 → 4 (Adiabatic Expansion):

   • Gas expands, doing work on the piston.
   • Work done by the gas: $W_{34}=\frac{P_3{V}_3-P_4{V}_4}{\gamma -1}$.

4. 4 → 1 (Exhaust):

   • Burnt gases are expelled at constant volume (or pressure in real engines).

Efficiency of Diesel Engine

The thermal efficiency ($\eta$) is given by: $$\eta =1-\frac{1}{r^{\gamma -1}}\left(\frac{r_c^{\gamma }-1}{\gamma (r_c-1)}\right)$$ where $r_c=V_3/V_2$ is the cutoff ratio.

---

Note: The $PV$ diagram above is a simplified representation. In reality, the processes are not perfectly adiabatic or isobaric, but the ideal cycle provides a theoretical basis for efficiency calculations.