Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.
Due to difference in conductivity, metals having high conductivity compared to wood. On touch with a finger, heat from the surrounding flows faster to the finger from metals and so one feels the heat. Similarly, when one touches a cold metal the heat from the finger flows away to the surroundings faster.
Calculate the temperature which has numeral value on Celsius and Fahrenheit scale.
Let $Q$ be the value of temperature having same value an Celsius and Fahrenheit scale.
Now, we can write
$$\begin{aligned} \frac{{ }^{\circ} F-32}{180} & =\frac{{ }^{\circ} \mathrm{C}}{100} \\ \Rightarrow \text{Let}\quad F & =C=Q \\ \Rightarrow \quad\frac{Q-32}{180} & =\frac{Q}{100}=Q=-40^{\circ} \mathrm{C} \text { or }-40^{\circ} \mathrm{F} \end{aligned}$$
These days people use steel utensiles with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.
As copper is a good conductor of heat as compared to steel. The steel utensils with copper bottom absorbs heat more quickly than steel and give it to the food in utensil. As a result, of it, the food in utensil is heated uniformly and quickly.
Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$ ) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.
Let the mass and length of a uniform rod be $M$ and $l$ respectively.
Moment of inertia of the rod about its perpendicular bisector. $(I)=\frac{M l^2}{12}$
$$\begin{aligned} &\text { Increase in length of the rod when temperature is increased by } \Delta T \text {, is given by }\\ &\begin{aligned} \Delta l & =l \cdot \alpha \Delta T \quad \text{... (i)}\\ \therefore \text { New moment of inertia of the } \operatorname{rod}(I) & =\frac{M}{12}(l+\Delta l)^2=\frac{M}{12}\left(l^2+\Delta l^2+2 I \Delta l\right) \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { As change in length } \Delta l \text { is very small, therefore, neglecting }(\Delta l)^2 \text {, we get }\\ &\begin{aligned} I^{\prime} & =\frac{M}{12}\left(l^2+2 l \Delta l\right) \\ & =\frac{M l^2}{12}+\frac{M I \Delta l}{6}=l+\frac{M I \Delta l}{6} \end{aligned} \end{aligned}$$
$\therefore$ Increase in moment of inertia $$ \begin{aligned} & \Delta I=l-I=\frac{M l \Delta l}{6}=2 \times\left(\frac{M l^2}{12}\right) \frac{\Delta l}{l} \\ & \Delta I=2 \cdot I \alpha \Delta T \quad \text{[Using Eq. (i)]} \end{aligned}$$
During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of $p-T$ diagram of water.
Refer to the $\mathrm{p}-T$ diagram of water and double headed arrow. Increasing pressure at $0^{\circ} \mathrm{C}$ and 1 atm takes ice into liquid state and decreasing pressure in liquid state at $0^{\circ} \mathrm{C}$ and 1 atm takes water to ice state.
When crushed ice is squeezed, some of it melts, filling up gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.