NCERT Solutions Class 11 Physics Chapter 10 — Thermal Properties of Matter

NCERT Class 11 Physics — Thermal Properties of Matter, Chapter 10 Exercises. All 20 questions solved.

The exercises at the end of Chapter 10 fall into four groups. Questions 10.1 to 10.5 are about temperature scales and thermometers. Questions 10.6 to 10.11 are about thermal expansion. Questions 10.12 to 10.16 are calorimetry — heat given, heat taken and latent heat. Questions 10.17 to 10.20 are about how heat travels: conduction, radiation and Newton’s law of cooling. The arithmetic runs on three sets of results:

• Thermal expansion — $\Delta l/l = \alpha_l\,\Delta T$ for a length and $\Delta V/V = \alpha_V\,\Delta T$ for a volume, with $\alpha_V \approx 3\alpha_l$ for a solid.
• Heat and change of state — $Q = ms\,\Delta T$ to change a temperature, $Q = mL$ to change a state, and in a mixture heat lost by the hot body equals heat gained by the cold one.
• Heat flow — conduction through a slab, $H = KA\,\dfrac{T_C – T_D}{L}$, and Newton’s law of cooling, $-\dfrac{dT}{dt} = k\,(T – T_s)$.

Temperature differences are the same in kelvin and in degrees Celsius, so $\Delta T$ can be taken in either.

Key insight. Almost every numerical question here is either a fraction or a balance. Expansion is a fractional change, $\Delta l/l$ or $\Delta V/V$, proportional to $\Delta T$, so the original size only scales the answer (10.6 to 10.11). Calorimetry is a balance: heat out of one body equals heat into another, or equals the work done (10.12 to 10.16). Conduction is a steady rate fixed by $H = KA\,\Delta T/L$ (10.17, 10.18). Decide which of the three a question is before touching a number, and it reduces to a single equation.

Question 10.1

The triple points of neon and carbon dioxide are $24.57\ \text{K}$ and $216.55\ \text{K}$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Solution. The Kelvin and Celsius scales have the same unit size and differ only in their zero, $t_C = T – 273.15$. The Fahrenheit degree is $5/9$ of a Celsius degree and its zero is offset by 32, so $t_F = \tfrac95\,t_C + 32$.

For neon, $t_C = 24.57 – 273.15 = -248.58\ {}^\circ\text{C}$, and

$$t_F = \tfrac95(-248.58) + 32 = -447.44 + 32 = -415.44\ {}^\circ\text{F}$$

For carbon dioxide, $t_C = 216.55 – 273.15 = -56.60\ {}^\circ\text{C}$, and

$$t_F = \tfrac95(-56.60) + 32 = -101.88 + 32 = -69.88\ {}^\circ\text{F}$$

Question 10.2

Two absolute scales $A$ and $B$ have triple points of water defined to be $200\ \text{A}$ and $350\ \text{B}$. What is the relation between $T_A$ and $T_B$?

Solution. Both scales are absolute, so both have their zero at absolute zero, and any temperature on either is proportional to the same physical quantity. The ratio of a temperature to the triple point is therefore the same on both scales:

$$\frac{T_A}{200} = \frac{T_B}{350} \quad\Rightarrow\quad T_A = \frac{200}{350}\,T_B = \frac47\,T_B$$

A check: at the triple point, $T_B = 350$ gives $T_A = 200$, as it should.

Question 10.3

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law

$$R = R_o\,[1 + \alpha\,(T – T_o)]$$

The resistance is $101.6\ \Omega$ at the triple-point of water $273.16\ \text{K}$, and $165.5\ \Omega$ at the normal melting point of lead ($600.5\ \text{K}$). What is the temperature when the resistance is $123.4\ \Omega$?

Solution. The law has two unknowns, $R_o$ and $\alpha$, and the two calibration points fix them. Take the triple point as the reference, so $T_o = 273.16\ \text{K}$ and $R_o = 101.6\ \Omega$. The melting point of lead then gives $\alpha$:

$$165.5 = 101.6\,[1 + \alpha(600.5 – 273.16)] \quad\Rightarrow\quad \alpha = \frac{63.9}{101.6\times327.34} = 1.921\times10^{-3}\ \text{K}^{-1}$$

Now turn the law round for $R = 123.4\ \Omega$:

$$T – 273.16 = \frac{123.4 – 101.6}{101.6\,\alpha} = \frac{21.8}{101.6\times1.921\times10^{-3}} = 111.7\ \text{K}$$

so $T = 384.8\ \text{K}$.

Question 10.4

Answer the following:

(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0\ {}^\circ\text{C}$ and $100\ {}^\circ\text{C}$ respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number $273.16\ \text{K}$. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) $T$ is related to the temperature $t_c$ on the Celsius scale by $t_c = T – 273.15$. Why do we have $273.15$ in this relation, and not $273.16$?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Question 10.4 (a)

Solution. A fixed point is useful only if it can be reproduced exactly, anywhere. The melting point of ice and the boiling point of water both shift with pressure — water boils well below $100\ {}^\circ\text{C}$ on a mountain — and with dissolved impurities. At the triple point, ice, liquid water and water vapour coexist, and that can happen at only one pressure ($611\ \text{Pa}$) and one temperature. Nothing has to be specified or controlled, so it is the same in every laboratory.

Question 10.4 (b)

Solution. An absolute scale needs only one fixed point to set its unit size, because its zero is not a choice: it is absolute zero, where the pressure of an ideal gas thermometer would fall to zero. So the second “fixed point” of the Kelvin scale is absolute zero itself, $0\ \text{K}$.

Question 10.4 (c)

Solution. The triple point of water is not $0\ {}^\circ\text{C}$ but $0.01\ {}^\circ\text{C}$. The Kelvin scale assigns it $273.16\ \text{K}$, so the ordinary melting point of ice, $0.01\ \text{K}$ lower, is $273.15\ \text{K}$. That is the temperature at which $t_c = 0$, and so it is the number subtracted in $t_c = T – 273.15$.

Question 10.4 (d)

Solution. On such a scale (the Rankine scale) the zero is absolute zero, as on the Kelvin scale, but each degree is a Fahrenheit degree, $5/9$ of a kelvin. The same temperature is therefore $9/5$ times the Kelvin number:

$$T = \tfrac95\times273.16 = 491.69$$

Question 10.5

Two ideal gas thermometers $A$ and $B$ use oxygen and hydrogen respectively. The following observations are made. At the triple-point of water, the pressure in thermometer $A$ is $1.250\times10^5\ \text{Pa}$ and in thermometer $B$ is $0.200\times10^5\ \text{Pa}$. At the normal melting point of sulphur, the pressure in $A$ is $1.797\times10^5\ \text{Pa}$ and in $B$ is $0.287\times10^5\ \text{Pa}$.

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers $A$ and $B$?

(b) What do you think is the reason behind the slight difference in answers of thermometers $A$ and $B$? (The thermometers are not faulty.) What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Question 10.5 (a)

Solution. A constant-volume gas thermometer reads temperature as proportional to pressure, calibrated so that the triple point is $273.16\ \text{K}$. So $T = 273.16\times P/P_{tr}$:

$$T_A = 273.16\times\frac{1.797}{1.250} = 392.69\ \text{K}, \qquad T_B = 273.16\times\frac{0.287}{0.200} = 391.98\ \text{K}$$

Question 10.5 (b)

Solution. The proportionality $T \propto P$ is exact only for an ideal gas. Real oxygen and hydrogen deviate slightly from ideal behaviour, and by different amounts, so the two thermometers disagree a little. Every gas becomes more nearly ideal as it is made more dilute. The remedy is to repeat the measurement with less and less gas in each thermometer (lower and lower triple-point pressure), plot the temperature read against that pressure, and extrapolate to zero pressure. In that limit both gases behave ideally and both thermometers give the same reading.

Question 10.6

A steel tape $1\ \text{m}$ long is correctly calibrated for a temperature of $27.0\ {}^\circ\text{C}$. The length of a steel rod measured by this tape is found to be $63.0\ \text{cm}$ on a hot day when the temperature is $45.0\ {}^\circ\text{C}$. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is $27.0\ {}^\circ\text{C}$? Coefficient of linear expansion of steel $= 1.20\times10^{-5}\ \text{K}^{-1}$.

Solution. At $45.0\ {}^\circ\text{C}$ the tape has expanded, so each centimetre marked on it is really a little longer than a centimetre, by the factor $(1 + \alpha\,\Delta T)$ with $\Delta T = 18.0\ \text{K}$. A reading of $63.0\ \text{cm}$ therefore stands for

$$l = 63.0\,(1 + 1.20\times10^{-5}\times18.0) = 63.0 + 0.0136 = 63.0136\ \text{cm}$$

On a $27.0\ {}^\circ\text{C}$ day the tape reads true. The rod is also steel, so it contracts from $45.0$ to $27.0\ {}^\circ\text{C}$ by exactly the same factor that the tape did, and its length is back to $63.0\ \text{cm}$.

The correction, $0.0136\ \text{cm}$, is smaller than the last significant figure of the data, so to three significant figures the rod is $63.0\ \text{cm}$ long on both days.

Question 10.7

A large steel wheel is to be fitted on to a shaft of the same material. At $27\ {}^\circ\text{C}$, the outer diameter of the shaft is $8.70\ \text{cm}$ and the diameter of the central hole in the wheel is $8.69\ \text{cm}$. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: $\alpha_{\text{steel}} = 1.20\times10^{-5}\ \text{K}^{-1}$.

Solution. Only the shaft is cooled, and the wheel slips on once the shaft’s diameter has shrunk to the $8.69\ \text{cm}$ of the hole. A diameter is a length, so it contracts according to $\Delta d = d\,\alpha\,\Delta T$:

$$\Delta T = \frac{\Delta d}{d\,\alpha} = \frac{-0.01}{8.70\times1.20\times10^{-5}} = -95.8\ \text{K}$$

The shaft must be cooled to $27 – 95.8 = -68.8\ {}^\circ\text{C}$, within the reach of dry ice (which sublimes at about $-78\ {}^\circ\text{C}$).

Question 10.8

A hole is drilled in a copper sheet. The diameter of the hole is $4.24\ \text{cm}$ at $27.0\ {}^\circ\text{C}$. What is the change in the diameter of the hole when the sheet is heated to $227\ {}^\circ\text{C}$? Coefficient of linear expansion of copper $= 1.70\times10^{-5}\ \text{K}^{-1}$.

Solution. When a sheet is heated, every length in it grows by the same fraction, as though the sheet were a photograph being enlarged. That includes the circle of copper that would fill the hole, so the hole grows exactly as a copper disc of the same size would. With $\Delta T = 200\ \text{K}$:

$$\Delta d = d\,\alpha\,\Delta T = 4.24\times1.70\times10^{-5}\times200 = 1.44\times10^{-2}\ \text{cm}$$

Question 10.9

A brass wire $1.8\ \text{m}$ long at $27\ {}^\circ\text{C}$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39\ {}^\circ\text{C}$, what is the tension developed in the wire, if its diameter is $2.0\ \text{mm}$? Co-efficient of linear expansion of brass $= 2.0\times10^{-5}\ \text{K}^{-1}$; Young’s modulus of brass $= 0.91\times10^{11}\ \text{Pa}$.

Solution. Left free, the wire would contract by the fraction $\alpha\,\Delta T$. The rigid supports hold it at its original length, which is the same as letting it contract and then stretching it back by that fraction. So the strain in the wire is $\alpha\,\Delta T$, with $\Delta T = 27 – (-39) = 66\ \text{K}$, and the length $1.8\ \text{m}$ drops out:

$$\text{strain} = 2.0\times10^{-5}\times66 = 1.32\times10^{-3}$$

Young’s modulus turns strain into stress, and stress times the cross-section gives the tension. The radius is $1.0\times10^{-3}\ \text{m}$:

$$F = Y\,(\alpha\,\Delta T)\,\pi r^2 = (0.91\times10^{11})(1.32\times10^{-3})\,\pi(1.0\times10^{-3})^2 = 3.8\times10^2\ \text{N}$$

Question 10.10

A brass rod of length $50\ \text{cm}$ and diameter $3.0\ \text{mm}$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250\ {}^\circ\text{C}$, if the original lengths are at $40.0\ {}^\circ\text{C}$? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $= 2.0\times10^{-5}\ \text{K}^{-1}$, steel $= 1.2\times10^{-5}\ \text{K}^{-1}$).

Solution. The ends are free, so each rod simply expands by its own $l\,\alpha\,\Delta T$, and the combined rod grows by the sum. With $l = 50\ \text{cm}$ and $\Delta T = 250 – 40 = 210\ \text{K}$:

$$\Delta l_{\text{brass}} = 50\times2.0\times10^{-5}\times210 = 0.21\ \text{cm}, \qquad \Delta l_{\text{steel}} = 50\times1.2\times10^{-5}\times210 = 0.126\ \text{cm}$$

$$\Delta l = 0.21 + 0.126 = 0.336 \approx 0.34\ \text{cm}$$

The diameter plays no part, because a length change does not depend on the cross-section. Thermal stress arises only when something prevents a body from expanding, as the supports did in Question 10.9. Here nothing restrains either rod along its length, so no stress develops at the junction. (Strictly, the junction face tries to expand sideways by different amounts in the two metals, but the question, like the book, treats the rods as expanding freely.)

Question 10.11

The coefficient of volume expansion of glycerine is $49\times10^{-5}\ \text{K}^{-1}$. What is the fractional change in its density for a $30\ {}^\circ\text{C}$ rise in temperature?

Solution. A fixed mass keeps its mass as it expands, so its density $\rho = m/V$ falls as its volume rises. For small changes, a fractional rise in $V$ is the same fractional fall in $\rho$:

$$\frac{\Delta\rho}{\rho} \approx -\frac{\Delta V}{V} = -\alpha_V\,\Delta T = -49\times10^{-5}\times30 = -1.47\times10^{-2}$$

Worked without the approximation, $\rho’ = \rho/(1 + \alpha_V\Delta T)$ gives $-0.0147/1.0147 = -1.45\times10^{-2}$; both round to $1.5\times10^{-2}$.

Question 10.12

A $10\ \text{kW}$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0\ \text{kg}$. How much is the rise in temperature of the block in $2.5$ minutes, assuming $50\%$ of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $= 0.91\ \text{J g}^{-1}\,\text{K}^{-1}$.

Solution. The work done by the drill ends up as heat, and half of it goes into the block: $5\ \text{kW}$ for $150\ \text{s}$.

$$Q = 5.0\times10^3\times150 = 7.5\times10^5\ \text{J}$$

In SI units the specific heat is $910\ \text{J kg}^{-1}\,\text{K}^{-1}$, and $Q = ms\,\Delta T$ gives

$$\Delta T = \frac{Q}{ms} = \frac{7.5\times10^5}{8.0\times910} = 103\ \text{K}$$

Question 10.13

A copper block of mass $2.5\ \text{kg}$ is heated in a furnace to a temperature of $500\ {}^\circ\text{C}$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $= 0.39\ \text{J g}^{-1}\,\text{K}^{-1}$; heat of fusion of water $= 335\ \text{J g}^{-1}$.)

Solution. The most ice melts when all the heat the copper gives up goes into melting, and the copper can cool no further than the $0\ {}^\circ\text{C}$ of the ice it sits on. The heat released in cooling from $500$ to $0\ {}^\circ\text{C}$ is

$$Q = ms\,\Delta T = 2500\times0.39\times500 = 4.875\times10^5\ \text{J}$$

and each gram of ice needs $335\ \text{J}$ to melt:

$$m_{\text{ice}} = \frac{4.875\times10^5}{335} = 1.46\times10^3\ \text{g}$$

Question 10.14

In an experiment on the specific heat of a metal, a $0.20\ \text{kg}$ block of the metal at $150\ {}^\circ\text{C}$ is dropped in a copper calorimeter (of water equivalent $0.025\ \text{kg}$) containing $150\ \text{cm}^3$ of water at $27\ {}^\circ\text{C}$. The final temperature is $40\ {}^\circ\text{C}$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Solution. The water equivalent of the calorimeter is the mass of water that would take up the same heat, so the calorimeter behaves like an extra $25\ \text{g}$ of water. With $150\ \text{g}$ of water ($150\ \text{cm}^3$), the cold side is $175\ \text{g}$ of water warming by $13\ \text{K}$. Taking $s_w = 4.186\ \text{J g}^{-1}\,\text{K}^{-1}$:

$$\text{heat gained} = 175\times4.186\times13 = 9.52\times10^3\ \text{J}$$

The metal cools from $150$ to $40\ {}^\circ\text{C}$, by $110\ \text{K}$, and gives up exactly this heat if none is lost:

$$200\times s\times110 = 9.52\times10^3 \quad\Rightarrow\quad s = 0.43\ \text{J g}^{-1}\,\text{K}^{-1}$$

If some heat escapes to the surroundings, the metal actually gives out more heat than the water and calorimeter record. The calculation credits the metal with only the heat that was recorded, so it underestimates the metal’s specific heat.

Question 10.15

Given below are observations on molar specific heats at room temperature of some common gases. The molar specific heat $C_v$, in $\text{cal mol}^{-1}\,\text{K}^{-1}$, is: hydrogen $4.87$, nitrogen $4.97$, oxygen $5.02$, nitric oxide $4.99$, carbon monoxide $5.01$, chlorine $6.17$.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92\ \text{cal/mol K}$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Solution. Heating a gas at constant volume raises the average energy of its molecules, and the specific heat measures how much energy that takes per degree. The energy is shared among every independent way a molecule can store it. A monatomic molecule can only move — three translational modes — and $C_v = \tfrac32R \approx 3\ \text{cal mol}^{-1}\,\text{K}^{-1}$, as the question’s $2.92$ reflects.

Every gas in the table is diatomic. A dumbbell-shaped molecule can also rotate about two axes perpendicular to its length, so it has five modes, and heat must be supplied to all of them. That raises $C_v$ to $\tfrac52R \approx 4.97\ \text{cal mol}^{-1}\,\text{K}^{-1}$, which matches hydrogen, nitrogen, oxygen, nitric oxide and carbon monoxide closely.

Chlorine’s $6.17$ is well above $\tfrac52R$. The extra heat must be going into a further mode: vibration of the two atoms along the bond. Chlorine atoms are heavy and the bond is comparatively weak, so its vibration is slow enough to be partly excited at room temperature, whereas in the lighter gases it is not. (Full vibration would give $\tfrac72R \approx 6.95$, and chlorine is on its way there.)

Question 10.16

A child running a temperature of $101\ {}^\circ\text{F}$ is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to $98\ {}^\circ\text{F}$ in 20 minutes, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30\ \text{kg}$. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580\ \text{cal g}^{-1}$.

Solution. A drop of $3\ {}^\circ\text{F}$ is a drop of $3\times\tfrac59 = \tfrac53\ {}^\circ\text{C}$. With the body’s specific heat taken as that of water, $1\ \text{cal g}^{-1}\,\text{K}^{-1}$, the heat the body loses is

$$Q = ms\,\Delta T = 30\,000\times1\times\tfrac53 = 5.0\times10^4\ \text{cal}$$

All of it goes into evaporating sweat, at $580\ \text{cal}$ per gram:

$$m = \frac{5.0\times10^4}{580} = 86.2\ \text{g}\ \text{in 20 min}, \qquad \frac{86.2}{20} = 4.3\ \text{g/min}$$

Question 10.17

A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side $30\ \text{cm}$ has a thickness of $5.0\ \text{cm}$. If $4.0\ \text{kg}$ of ice is put in the box, estimate the amount of ice remaining after $6\ \text{h}$. The outside temperature is $45\ {}^\circ\text{C}$, and co-efficient of thermal conductivity of thermacole is $0.01\ \text{J s}^{-1}\,\text{m}^{-1}\,\text{K}^{-1}$. [Heat of fusion of water $= 335\times10^3\ \text{J kg}^{-1}$]

Solution. Heat leaks in by conduction through all six walls. Inside, the melting ice holds the temperature at $0\ {}^\circ\text{C}$, so the difference across every wall stays $45\ \text{K}$ and the flow is steady. The six faces of a $30\ \text{cm}$ cube have a total area $6\times(0.30)^2 = 0.54\ \text{m}^2$; since this is an estimate, the small difference between the inner and outer surface areas is ignored. The rate of heat flow is

$$H = \frac{KA\,\Delta T}{L} = \frac{0.01\times0.54\times45}{0.050} = 4.86\ \text{W}$$

In $6\ \text{h} = 21\,600\ \text{s}$ the box takes in $4.86\times21\,600 = 1.05\times10^5\ \text{J}$, which melts

$$m = \frac{1.05\times10^5}{335\times10^3} = 0.31\ \text{kg}$$

leaving $4.0 – 0.31 = 3.69\ \text{kg}$.

Question 10.18

A brass boiler has a base area of $0.15\ \text{m}^2$ and thickness $1.0\ \text{cm}$. It boils water at the rate of $6.0\ \text{kg/min}$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $= 109\ \text{J s}^{-1}\,\text{m}^{-1}\,\text{K}^{-1}$; Heat of vaporisation of water $= 2256\times10^3\ \text{J kg}^{-1}$.

Solution. In the steady state, every joule that boils water has first been conducted through the brass base. So the rate of conduction equals the rate at which latent heat is taken up:

$$H = \frac{6.0\ \text{kg}}{60\ \text{s}}\times2256\times10^3\ \text{J kg}^{-1} = 2.256\times10^5\ \text{W}$$

The top face of the base is in contact with boiling water, at $100\ {}^\circ\text{C}$; the bottom face is at the flame temperature $T$. The conduction equation gives the temperature difference needed to drive this flow:

$$T – 100 = \frac{HL}{KA} = \frac{(2.256\times10^5)(0.010)}{109\times0.15} = 138\ \text{K}$$

so $T = 100 + 138 = 238\ {}^\circ\text{C}$.

Question 10.19

Explain why: (a) a body with large reflectivity is a poor emitter; (b) a brass tumbler feels much colder than a wooden tray on a chilly day; (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace; (d) the earth without its atmosphere would be inhospitably cold; (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Question 10.19 (a)

Solution. Radiation falling on a body is partly reflected and the rest absorbed, so a body that reflects most of it absorbs little. Kirchhoff’s law says that, at a given temperature and wavelength, a body’s power to emit is in proportion to its power to absorb: a good absorber is a good emitter. A strong reflector is therefore a poor absorber, and so a poor emitter.

Question 10.19 (b)

Solution. On a chilly day both objects are at the same temperature, below that of the hand. What the hand feels is how fast it loses heat. Brass is a good conductor: it carries heat away from the skin quickly and spreads it through the tumbler, so the skin cools fast and feels cold. Wood is a poor conductor, so only a thin layer at the surface warms up, heat leaves the hand slowly, and the tray feels less cold.

Question 10.19 (c)

Solution. The pyrometer infers temperature from the radiation it receives, assuming it comes from a perfect black body. Iron in the open is not a black body: its emissivity is less than 1, so at a given temperature it radiates less than a black body would, and the pyrometer reads that as a lower temperature. Inside a furnace the iron is surrounded by walls at the same temperature, and the radiation there is in equilibrium; what leaves the iron’s surface is its own emission plus the wall radiation it reflects, which together make up exactly black-body radiation. So the reading is correct.

Question 10.19 (d)

Solution. The earth absorbs sunlight and re-radiates the energy as infrared. Gases in the atmosphere, chiefly water vapour and carbon dioxide, absorb much of that infrared and radiate part of it back to the surface — the greenhouse effect. The atmosphere also carries heat by convection and moderates the swing between day and night. Without it the earth would radiate its heat straight into space, especially at night, and its average surface temperature would be far lower (about $-18\ {}^\circ\text{C}$ rather than about $15\ {}^\circ\text{C}$).

Question 10.19 (e)

Solution. Hot water gives up heat only by cooling, about $4.2\ \text{kJ}$ per kilogram for each degree. Steam at $100\ {}^\circ\text{C}$ first gives up its latent heat as it condenses, $2256\ \text{kJ}$ per kilogram, and only then cools as water. So a kilogram of steam delivers several hundred times as much heat to the radiators as a kilogram of water cooling through a few degrees, and far less mass has to be circulated.

Question 10.20

A body cools from $80\ {}^\circ\text{C}$ to $50\ {}^\circ\text{C}$ in 5 minutes. Calculate the time it takes to cool from $60\ {}^\circ\text{C}$ to $30\ {}^\circ\text{C}$. The temperature of the surroundings is $20\ {}^\circ\text{C}$.

Solution. Newton’s law of cooling says the rate of cooling is proportional to how far the body is above its surroundings. Over each interval, the book’s method (as in its Example 10.8) takes the average rate of cooling to be $k$ times the excess of the average temperature over the surroundings:

$$\frac{\text{fall in temperature}}{\text{time}} = k\,(T_{\text{av}} – T_s)$$

The first interval fixes $k$. The body falls $30\ {}^\circ\text{C}$ in $5\ \text{min}$ at an average of $65\ {}^\circ\text{C}$, which is $45\ {}^\circ\text{C}$ above the surroundings:

$$\frac{30}{5} = k\times45 \quad\Rightarrow\quad k = \frac{2}{15}\ \text{min}^{-1}$$

In the second interval it again falls $30\ {}^\circ\text{C}$, but at an average of $45\ {}^\circ\text{C}$, only $25\ {}^\circ\text{C}$ above the surroundings, so it cools more slowly:

$$\frac{30}{t} = \frac{2}{15}\times25 \quad\Rightarrow\quad t = 9\ \text{min}$$

A note on the approximation. Averaging the temperature is accurate when the fall is small compared with the excess, as in Example 10.8. Here the falls are large, and integrating the law exactly, as the chapter does to get $\ln(T – T_s) = -kt + c$, gives a slightly different time. The excess halves from $60$ to $30\ {}^\circ\text{C}$ in the first interval, so it halves every $5\ \text{min}$; in the second it falls from $40$ to $10\ {}^\circ\text{C}$, a quarter, which takes two halvings, or $10\ \text{min}$. The answer the book expects is the $9\ \text{min}$ of its averaging method.

Common mistakes

• Question 10.2: turning the ratio upside down. $T_A/200 = T_B/350$ gives $T_A = \tfrac47 T_B$. Writing $T_A = \tfrac74 T_B$ comes from matching the numbers 200 and 350 to the wrong side; checking at the triple point ($T_B = 350$ must give $T_A = 200$) catches it at once.
• Question 10.6: expecting the rod to be a different length at $27\ {}^\circ\text{C}$. The rod is steel, like the tape, so the tape’s error and the rod’s expansion are the same fraction. Students who take the $63.0\ \text{cm}$ reading as the rod’s true length at $45.0\ {}^\circ\text{C}$ and then shrink it to $27.0\ {}^\circ\text{C}$ get $62.99\ \text{cm}$; the reading was short by exactly the amount the rod then contracts, so the two cancel.
• Question 10.8: thinking the hole gets smaller. It feels as if the metal should expand into the hole. It does not: heating enlarges every length in the sheet, including the hole’s diameter, by the same fraction.
• Question 10.9: putting the length of the wire into the answer. The tension depends on the strain $\alpha\,\Delta T$, not on the change in length, so the $1.8\ \text{m}$ cancels. Using the diameter instead of the radius in $\pi r^2$ is the other common slip, and it quadruples the force.
• Question 10.11: giving a positive answer. The volume rises, so the density falls. The fractional change is $-1.47\times10^{-2}$; losing the sign reverses the physics.
• Question 10.14: answering “greater”. Heat lost to the surroundings makes the water’s gain less than the metal’s loss. The calculation uses the water’s gain, so it gives a value smaller than the true specific heat.
• Question 10.16: treating $3\ {}^\circ\text{F}$ as $3\ {}^\circ\text{C}$. A Fahrenheit degree is only $5/9$ of a Celsius degree. Forgetting the conversion inflates the answer to $7.8\ \text{g/min}$.
• Question 10.18: giving $138\ {}^\circ\text{C}$ as the flame temperature. Conduction gives the temperature difference across the base. The top of the base is at the $100\ {}^\circ\text{C}$ of boiling water, so the flame is at $238\ {}^\circ\text{C}$. Using $6.0\ \text{kg}$ per second instead of per minute is the other trap.

Practise next

• Chapter 11, Thermodynamics — heat and work as two ways of changing a system’s energy, and the first law that connects them.
• Chapter 12, Kinetic Theory — the equipartition of energy behind Question 10.15, and why diatomic gases have $C_v = \tfrac52R$.
• Chapter 8, Mechanical Properties of Solids — Young’s modulus, stress and strain, which Question 10.9 uses to turn a prevented contraction into a tension.