100 g of water is supercooled to $-10^{\circ} \mathrm{C}$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $\left[S_w=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\right.$ and $\left.L_{\text {Fusion }}^w=80 \mathrm{cal} / \mathrm{g}\right]$
Given, mass of water $(m)=100$
Change in temperature $\Delta T=0-(-10)=10^{\circ} \mathrm{C}$
Specific heat of water $\left(S_w\right)=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Latent heat of fusion of water $L_{\text {fusion }}^w=80 \mathrm{cal} / \mathrm{g}$
Heat required to bring water in super cooling from $-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$,
$$\begin{aligned} Q & =m s_w \Delta T \\ & =100 \times 1 \times 10=1000 \mathrm{cal} \end{aligned}$$
$$\begin{aligned} &\text { Let } m \text { gram of ice be melted. }\\ &\begin{array}{ll} \therefore & Q=m L \\ \text { or } & m=\frac{Q}{L}=\frac{1000}{80}=12.5 \mathrm{~g} \end{array} \end{aligned}$$
As small mass of ice is melted, therefore the temperature of the mixture will remain 0$$^\circ$$C.
One day in the morning. Ramesh filled up $1 / 3$ bucket of hot water from geyser, to take bath. Remaining $2 / 3$ was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 min before he could take bath. Now, he had two options (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain
The first option would have kept water warmer because according to Newton's law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of the body and the surrounding and in the first case the temperature difference is less, so, rate of loss of heat will be less.
We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm . We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their length $B$ remain constant. If $\alpha_{\text {iron }}=1.2 \times 10^{-5} / \mathrm{K}$ and $\alpha_{\text {brass }}=1.8 \times 10^{-5} / \mathrm{K}$, what should we take as length of each strip?
According to question $l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm}=$ constant at all temperatures
Let $l_0$ be length at temperature $0^{\circ} \mathrm{C}$ and $l$ be the length after change in temperature of $\Delta t$.
Now, we can write $\quad l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm}$ at all temperatures
$$\begin{aligned} l_{\text {iron }}\left(1+\alpha_{\text {iron }} \Delta t\right)-l_{\text {brass }}\left(1+\alpha_{\text {brass }} \Delta t\right) & =10 \mathrm{~cm} \\ l_{\text {iron }} \alpha_{\text {iron }} & =l_{\text {brass }} \alpha_{\text {brass }} \\ \therefore\quad \frac{l_{\text {iron }}}{l_{\text {brass }}} & =\frac{1.8}{1.2}=\frac{3}{2} \\ \therefore \quad \frac{1}{2} l_{\text {brass }} & =10 \mathrm{~cm} \\ \Rightarrow \quad l_{\text {brass }} & =20 \mathrm{~cm} \text { and } l_{\text {iron }} 30 \mathrm{~cm} \end{aligned}$$
We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron $\left(\beta_{\text {vbrass }}=6 \times 10^{-5} / K\right.$ and $\left.\beta_{\text {viron }}=3.55 \times 10^{-5} / \mathrm{K}\right)$ to create a volume of 100 cc . How do you think you can achieve this?
In the previous problem the difference in the length was constant.
In this problem the difference in volume is constant.
The situation is shown in the diagram.
Let $V_{i o}, V_{b o}$ be the volume of iron and brass vessel at $0^{\circ} \mathrm{C}$
$V_{i,}, V_b$ be the volume of iron and brass vessel at $\Delta \theta^{\circ} \mathrm{C}$,
$\gamma_i, \gamma_b$ be the coefficient, of volume expansion of iron and brass.
$$\text { As per question, } \quad V_{i o}-V_{b o}=100 \mathrm{cc}=V_i-V_b\quad \text{... (i)}$$
$$\text { Now, } \quad V_i=V_{i 0}\left(1+\gamma_i \Delta \theta\right)$$
$$\begin{aligned} V_b & =V_{b o}\left(1+\gamma_b \Delta \theta\right) \\ V_i-V_b & =\left(V_{i o}-V_{b o}\right)+\Delta \theta\left(V_{i o} \gamma_i-V_{b o} \gamma_b\right) \end{aligned}$$
Since, $V_i-V_b=$ constant.
So, $$V_{i o} \gamma_i=V_{b o} \gamma_b$$
$$\begin{aligned} & \Rightarrow \quad \frac{V_{i o}}{V_{b o}}=\frac{\gamma_b}{\gamma_i}=\frac{\frac{3}{2} \beta_b}{\frac{3}{2} \beta_i}=\frac{\beta_b}{\beta_i}=\frac{6 \times 10^{-5}}{3.55 \times 10^{-5}}=\frac{6}{3.55} \\ & \frac{V_{i o}}{V_{b o}}=\frac{6}{3.55}\quad \text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { Solving Eqs. (i) and (ii), we get }\\ &\begin{aligned} V_{i o} & =244.9 \mathrm{cc} \\ V_{b o} & =144.9 \mathrm{cc} \end{aligned} \end{aligned}$$
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^{\circ} \mathrm{C}$ is drunk. You can take body (tooth) temperature to be $37^{\circ} \mathrm{C}$ and $\alpha=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ bulk modulus for copper $=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Given, decrease in temperature $(\Delta t)=57-37=20^{\circ} \mathrm{C}$
Coefficient of linear expansion $(\alpha)=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Bulk modulus for copper $(B)=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Coefficient of cubical expansion $(\gamma)=3 \alpha=5.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Let initial volume of the cavity be $V$ and its volume increases by $\Delta V$ due to increase in temperature.
$$\begin{array}{ll} \therefore & \Delta V=\gamma V \Delta t \\ \Rightarrow & \frac{\Delta V}{V}=\gamma \Delta t\quad \text{... (i)} \end{array}$$
Thermal stress produced $=B \times$ Volumetric strain
$$\begin{aligned} & =B \times \frac{\Delta V}{V}=B \times \gamma \Delta t \\ & =140 \times 10^9 \times\left(5.1 \times 10^{-5} \times 20\right) \\ & =1,428 \times 10^8 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$$
This is about $10^3$ times of atmospheric pressure.