NCERT Exemplar Solutions Class 11 Physics Thermal Properties of Matter — Very Short, Short and Long Answer Questions Q13–27

NCERT Exemplar Problems, Class 11 Physics — Thermal Properties of Matter, Very Short Answer Questions 13 to 17, Short Answer Questions 18 to 21 and Long Answer Questions 22 to 27. All 15 questions solved. The book numbers them 11.13 to 11.27; the matching textbook chapter is Chapter 10.

Three ideas carry this page. Expansion is proportional: $\Delta l = l\alpha\Delta T$, so two rods can be chosen to grow by exactly the same amount, and a clamped body that is stopped from growing feels a stress equal to the modulus times the strain it was denied. Heat lost equals heat gained: every calorimetry problem is one balance sheet of $ms\Delta T$ and $mL$ terms. Heat flows faster across larger differences: conduction depends on conductivity, cooling follows Newton’s law, and radiation grows as $T^4$.

Key insight. In Questions 22 and 23 the trick is to make the changes equal, $\alpha_1 l_1 = \alpha_2 l_2$, so the difference of two growing lengths never changes. In Questions 20 and 21 the trick is to write one heat balance before reaching for a formula.

Very Short Answer Questions

Question 13

Is the bulb of a thermometer made of diathermic or adiabatic wall?

Solution. Diathermic. A thermometer works by coming to the same temperature as the body it touches, which needs heat to pass freely through the bulb. An adiabatic wall blocks heat flow, so a thermometer behind one would never register the temperature outside.

Question 14

A student records the initial length $l$, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

S.No.	$l$ (m)	$\Delta T$ ($^\circ$C)	$\Delta l$ (m)
1.	2	10	$4\times10^{-4}$
2.	1	10	$4\times10^{-4}$
3.	2	20	$2\times10^{-4}$
4.	3	10	$6\times10^{-4}$

If the first observation is correct, what can you say about observations 2, 3 and 4?

Solution. For one rod, $\alpha = \dfrac{\Delta l}{l\,\Delta T}$ must be the same in every row. The first row gives $\alpha = \dfrac{4\times10^{-4}}{2\times10} = 2\times10^{-5}\ ^\circ\text{C}^{-1}$. Using it to predict $\Delta l$:

• Row 2: $1\times2\times10^{-5}\times10 = 2\times10^{-4}$ m, but $4\times10^{-4}$ m was recorded. Wrong.
• Row 3: $2\times2\times10^{-5}\times20 = 8\times10^{-4}$ m, but $2\times10^{-4}$ m was recorded. Wrong.
• Row 4: $3\times2\times10^{-5}\times10 = 6\times10^{-4}$ m, as recorded. Correct.

Question 15

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.

Solution. The hand does not sense temperature; it senses how fast heat flows into or out of the skin. Metal is a far better conductor than wood. When both bars are hotter than the hand, the metal delivers heat to the finger much faster, both from the point of contact and from the rest of the bar behind it, so it feels hotter. When both are colder, the metal draws heat away from the finger faster, so it feels colder. Wood, a poor conductor, lets the small region under the finger come quickly to near skin temperature, so it feels closer to neutral either way.

Question 16

Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

Solution. The scales are related by $\dfrac{t_C}{100} = \dfrac{t_F – 32}{180}$, that is $t_F = \frac{9}{5}t_C + 32$. Put $t_F = t_C = x$:

$$x = \frac{9}{5}x + 32 \;\Rightarrow\; -\frac{4}{5}x = 32 \;\Rightarrow\; x = -40.$$

Question 17

These days people use steel utensils 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.

Solution. A flame heats only a small patch of the base. Copper conducts so well that the heat it receives spreads quickly sideways across the whole bottom before it passes up into the steel, so the base is at nearly the same temperature everywhere. Steel conducts more slowly, so it then lets this evenly spread heat into the food gradually rather than in hot spots over the flame. The result is uniform heating without local burning.

Short Answer Questions

Question 18

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 $\Delta T$.

Solution. About the perpendicular bisector, $I = \frac{1}{12}Ml^2$. The mass does not change; only the length does, to $l + \Delta l$ with $\Delta l = l\alpha\Delta T$. Then

$$I’ = \frac{1}{12}M(l+\Delta l)^2 = \frac{1}{12}M\left(l^2 + 2l\,\Delta l + (\Delta l)^2\right).$$

For a small rise the $(\Delta l)^2$ term is negligible, so

$$I’ \approx \frac{1}{12}Ml^2\left(1 + 2\alpha\Delta T\right) = I + 2I\alpha\Delta T.$$

The factor $2$ appears because $I$ depends on the square of a length.

Question 19

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.

Solution. Water is unusual: its solid–liquid line on the $P$–$T$ diagram slopes slightly backwards, so a higher pressure lowers the melting point. Ice at about $0^\circ$C and $1$ atm sits just on the solid side of that line.

When the crushed ice or snow is squeezed in the palm, the pressure at the points where flakes touch rises, the melting point there drops below the ice’s temperature, and a little ice melts. The water flows into the gaps between flakes. When the hand releases the pressure, the melting point returns to $0^\circ$C and this water refreezes, welding the flakes together into a firm ball. This melting under pressure and refreezing on release is called regelation.

Question 20

$100$ g of water is supercooled to $-10^\circ$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? $[S_w = 1\ \text{cal/g}/^\circ\text{C}$ and $L^w_{\text{fusion}} = 80\ \text{cal/g}]$

Solution. Supercooled water is liquid below its freezing point, which is not a stable state. When freezing starts, the latent heat released by the water that freezes warms the whole mass. It stops at $0^\circ$C, where ice and water can coexist; it cannot go higher, because then the ice would melt and absorb the heat back.

Heat needed to bring $100$ g of water from $-10^\circ$C to $0^\circ$C:

$$Q = m S_w \Delta T = 100 \times 1 \times 10 = 1000\ \text{cal}.$$

This must come from freezing a mass $m’$:

$$m’ L = 1000 \;\Rightarrow\; m’ = \frac{1000}{80} = 12.5\ \text{g}.$$

(Treating the specific heat of the frozen part as that of water, as the data do, is accurate enough here.)

Question 21

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 minutes 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.

Solution. Option (i). By Newton’s law of cooling, the rate at which the water loses heat is proportional to how far it is above room temperature. In option (i) the water is mixed straight away to a comfortable, lukewarm temperature, only a little above the room, so it loses heat slowly while Ramesh is away. In option (ii) the hot water sits alone well above room temperature for those minutes and loses heat much faster. The cold water would add no heat in either case, so the bucket that lost less heat while waiting, the one filled at once, ends up warmer.

Long Answer Questions

Question 22

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 lengths remain constant. If $\alpha_{iron} = 1.2\times10^{-5}\ /\text{K}$ and $\alpha_{brass} = 1.8\times10^{-5}\ /\text{K}$, what should we take as length of each strip?

Solution. The difference $l_{iron} – l_{brass} = 10$ cm must hold at every temperature. After a rise $\Delta T$,

$$l_{iron}(1 + \alpha_{iron}\Delta T) – l_{brass}(1 + \alpha_{brass}\Delta T) = 10\ \text{cm},$$

which stays $10$ cm only if the two strips grow by the same amount:

$$l_{iron}\,\alpha_{iron} = l_{brass}\,\alpha_{brass} \;\Rightarrow\; \frac{l_{iron}}{l_{brass}} = \frac{1.8}{1.2} = \frac{3}{2}.$$

The iron strip must be the longer one, because it expands less per unit length. Then $\frac{3}{2}l_{brass} – l_{brass} = 10$, so $l_{brass} = 20$ cm and $l_{iron} = 30$ cm.

Question 23

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 ($\beta_{v\,brass} = 6\times10^{-5}\ /\text{K}$ and $\beta_{v\,iron} = 3.55\times10^{-5}\ /\text{K}$) to create a volume of $100$ cc. How do you think you can achieve this.

Solution. Use the same idea in three dimensions. Make an iron vessel of internal volume $V_{iron}$ and fix inside it a solid brass block (or rod) of volume $V_{brass}$. The usable volume is $V_{iron} – V_{brass} = 100$ cc. It stays $100$ cc if both volumes grow by the same amount:

$$V_{iron}\,\beta_{iron} = V_{brass}\,\beta_{brass} \;\Rightarrow\; \frac{V_{iron}}{V_{brass}} = \frac{6}{3.55}.$$

Then

$$V_{brass}\left(\frac{6}{3.55} – 1\right) = 100 \;\Rightarrow\; V_{brass} = \frac{355}{2.45} \approx 144.9\ \text{cc},\qquad V_{iron} \approx 244.9\ \text{cc}.$$

As the temperature rises the inside of the iron vessel grows, but the brass block grows by exactly the same volume, so the free space between them is unchanged.

Question 24

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^\circ$C is drunk. You can take body (tooth) temperature to be $37^\circ$C and $\alpha = 1.7\times10^{-5}/^\circ$C, bulk modulus for copper $= 140\times10^9\ \text{N/m}^2$.

Solution. The copper filling is warmed by $\Delta T = 57 – 37 = 20^\circ$C and tries to expand in volume by the fraction $3\alpha\Delta T$. The cavity, taken as rigid, stops it, so the filling is compressed back by that volume strain. The stress is the bulk modulus times that strain:

$$\text{stress} = K\cdot 3\alpha\Delta T = 140\times10^9 \times 3\times1.7\times10^{-5}\times 20 = 1.428\times10^8\ \text{N/m}^2.$$

That is about $1.4\times10^3$ times atmospheric pressure, which is why fillings and teeth can crack under repeated hot and cold drinks.

Question 25

A rail track made of steel having length $10$ m is clamped on a railway line at its two ends (Fig. 11.3). On a summer day due to rise in temperature by $20^\circ$C, it is deformed as shown in figure. Find $x$ (displacement of the centre) if $\alpha_{steel} = 1.2\times10^{-5}/^\circ$C.

Solution. The clamps hold the ends $L = 10$ m apart, but the rail now has length $L + \Delta L$ with $\Delta L = L\alpha\Delta T$. As in the figure, it bends into two straight halves, each of length $\frac{1}{2}(L + \Delta L)$, meeting at the centre a distance $x$ off the line. Each half is the hypotenuse of a right triangle with legs $L/2$ and $x$:

$$x = \sqrt{\left(\frac{L+\Delta L}{2}\right)^2 – \left(\frac{L}{2}\right)^2} \approx \frac{1}{2}\sqrt{2L\,\Delta L} = \frac{L}{2}\sqrt{2\alpha\Delta T},$$

dropping the tiny $(\Delta L)^2$ term. With $\alpha\Delta T = 1.2\times10^{-5}\times20 = 2.4\times10^{-4}$,

$$x = 5\sqrt{4.8\times10^{-4}} \approx 5\times0.0219 \approx 0.11\ \text{m}.$$

A rail that grows by only $2.4$ mm bows out by about $11$ cm: the square root turns a small strain into a large sideways shift, which is why rails are laid with expansion gaps.

Question 26

A thin rod having length $L_0$ at $0^\circ$C and coefficient of linear expansion $\alpha$ has its two ends maintained at temperatures $\theta_1$ and $\theta_2$, respectively. Find its new length.

Solution. In the steady state the temperature falls linearly along a uniform rod. At distance $x$ from the $\theta_1$ end,

$$\theta(x) = \theta_1 + \frac{x}{L_0}(\theta_2 – \theta_1).$$

A small element of original length $dx_0$ at that point becomes $dx = dx_0\left(1 + \alpha\theta\right)$. Adding up all the elements,

$$L = \int_0^{L_0}\left[1 + \alpha\theta_1 + \frac{\alpha(\theta_2 – \theta_1)}{L_0}x\right]dx_0 = L_0 + L_0\alpha\theta_1 + \frac{\alpha(\theta_2-\theta_1)}{L_0}\cdot\frac{L_0^2}{2}.$$

So

$$L = L_0\left[1 + \frac{\alpha(\theta_1 + \theta_2)}{2}\right].$$

This is what one gets by taking the whole rod at its average temperature $\frac{1}{2}(\theta_1 + \theta_2)$, which is valid because expansion is linear in temperature and the temperature is linear along the rod.

Question 27

According to Stefan’s law of radiation, a black body radiates energy $\sigma T^4$ from its unit surface area every second where $T$ is the surface temperature of the black body and $\sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$ is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius $0.5$ m. When detonated, it reaches temperature of $10^6$ K and can be treated as a black body.

(a) Estimate the power it radiates.

(b) If surrounding has water at $30^\circ$C, how much water can $10\%$ of the energy produced evaporate in $1$ s? $[S_w = 4186.0\ \text{J/kg K}$ and $L_v = 22.6\times10^5\ \text{J/kg}]$

(c) If all this energy $U$ is in the form of radiation, corresponding momentum is $p = U/c$. How much momentum per unit time does it impart on unit area at a distance of $1$ km?

Solution.

(a) Power $=$ (energy per unit area per second) $\times$ (surface area):

$$P = \sigma T^4\cdot 4\pi r^2 = 5.67\times10^{-8}\times(10^6)^4\times4\pi(0.5)^2 = 5.67\times10^{16}\,\pi \approx 1.78\times10^{17}\ \text{W}.$$

(b) In $1$ s, $10\%$ of this is $1.78\times10^{16}$ J. Each kilogram of water must be heated from $30^\circ$C to $100^\circ$C and then boiled:

$$4186\times70 + 22.6\times10^5 = 2.93\times10^5 + 22.6\times10^5 = 2.553\times10^6\ \text{J/kg}.$$

$$m = \frac{1.78\times10^{16}}{2.553\times10^6} \approx 7.0\times10^9\ \text{kg}.$$

(c) At $R = 1$ km the power is spread over a sphere of area $4\pi R^2$. Momentum per unit time is force, so the force per unit area is

$$\frac{P}{c\cdot 4\pi R^2} = \frac{1.78\times10^{17}}{3\times10^8\times4\pi\times10^6} \approx 47\ \text{N/m}^2.$$

(The printed key gives $47.7\ \text{N/m}^2$ because it rounds the power to $1.8\times10^{17}$ W first; carrying more figures gives $47.25$. Both round to about $47$–$48\ \text{N/m}^2$.)

Common mistakes

• Checking only $\Delta l$ against row 1 in Question 14. Compare $\Delta l/(l\Delta T)$, since both $l$ and $\Delta T$ change between rows.
• Saying metal “is” hotter in Question 15. Both bars are at the same temperature; the metal conducts heat to or from the skin faster.
• Letting the mixture end below $0^\circ$C in Question 20. Freezing releases heat until the whole mass reaches $0^\circ$C and stops there.
• Choosing option (ii) in Question 21. Hot water left alone loses heat faster because it is further above room temperature.
• Making the brass strip longer in Question 22. The metal that expands less (iron) must be longer so that both grow equally.
• Using $\alpha$ for a volume strain in Question 24. Bulk stress uses the volume strain $3\alpha\Delta T$.
• Expecting $x$ to be of the order of $\Delta L$ in Question 25. $x$ goes as $\sqrt{L\,\Delta L}$, which is much larger.
• Forgetting to spread the power over $4\pi R^2$ in Question 27(c). The force per unit area falls as $1/R^2$.

Practise next

• Problem. $50$ g of ice at $-20^\circ$C is dropped into $200$ g of water at $40^\circ$C. Taking the specific heat of ice as $0.5$ cal/g$^\circ$C, of water $1$ cal/g$^\circ$C and the latent heat of fusion as $80$ cal/g, find the final temperature.

Show answer

$14^\circ$C. Warming and melting the ice takes $50\times0.5\times20 + 50\times80 = 4500$ cal, less than the $8000$ cal the water gives up in cooling to $0^\circ$C, so all the ice melts. Then $200(40 – T) = 4500 + 50T$ gives $T = 14^\circ$C, the same heat balance as Question 20.

• Problem. A copper strip ($\alpha = 1.7\times10^{-5}\ \text{K}^{-1}$) and a steel strip ($\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$) are to differ in length by $5$ cm at every temperature. Find their lengths.

Show answer

Steel $17$ cm, copper $12$ cm. Equal growth needs $1.2\,l_s = 1.7\,l_c$, so $l_s = \frac{17}{12}l_c$ and $\frac{5}{12}l_c = 5$ cm, giving $l_c = 12$ cm and $l_s = 17$ cm, as in Question 22.

• Problem. A black sphere of radius $10$ cm is kept at $1000$ K. Find the power it radiates, using $\sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$.

Show answer

About $7.1$ kW. $P = \sigma T^4\cdot4\pi r^2 = 5.67\times10^{-8}\times10^{12}\times4\pi(0.1)^2 \approx 7.13\times10^3$ W, the same calculation as Question 27(a).

• Exemplar Thermal Properties of Matter: Multiple Choice Questions — the first twelve questions of this unit, on expansion, temperature scales and cooling.
• NCERT textbook Chapter 10 — Thermal Properties of Matter for the theory of expansion, calorimetry, change of state and heat transfer.
• Exemplar Thermodynamics — the next unit, which turns these heat balances into the first law.