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

Q. 10.1 The triple points of neon and carbon dioxide are $24.57K$ and $216.55K$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

The relation between the Kelvin and Celsius scales is T _K = T _C + 273.15

Triple Point of Neon in Kelvin T _K = 24.57 K

Triple Point of Neon in Celcius T _C = T _K -273.15 = 24.57 -273.15 =-248.58 ^o C

Triple Point of carbon dioxide in Kelvin T _K = 216.55 K

Triple Point of carbon dioxide in Celcius T _C = T _K -273.15 = 216.55 - 273.15 = -56.60 ^o C

The relation between Celsius and Fahrenheit scales is $T_F=\frac{9}{5}{T}_C+32$

Triple Point of Neon in Fahrenheit is

$T_F=\frac{9}{5}\times (-248.58)+32$

T _F = -415.44 ^o C

Triple Point of carbon dioxide in Fahrenheit is

$T_F=\frac{9}{5}\times (-56.60)+32$

T _F = -69.88 ^o C

Q. 10.2 Two absolute scales $A$ and $B$ have triple points of water defined to be $200A$ and $350B$ . What is the relation between $T_A$ and $T_B$ ?

Answer:

200 A = 273 K

$T_K=\frac{273}{200}{T}_A$

350 B = 273 K

$T_K=\frac{273}{350}{T}_B$

Equating T _K from the above two equations, we have

$\frac{273}{200}{T}_A=\frac{273}{350}{T}_B$

$T_A=\frac{4}{7}{T}_B$

Q. 10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:$R=R_0[1+\alpha (T-T_0)]$ The resistance is $101.6\mathrm{\Omega }$ at the triple-point of water $273.16K,$ and $165.5\mathrm{\Omega }$ at the normal melting point of lead $(600.5K).$ What is the temperature when the resistance is $123.4\mathrm{\Omega }$?

Answer:

$R=R_0[1+\alpha (T-T_0)]$

R ₀ = $101.6\mathrm{\Omega }$

T ₀ = $273.16K$

R = $165.5\mathrm{\Omega }$

T = $600.5K$

Putting the above values in the given equation, we have

$\alpha =\frac{165.5-101.6}{600.5-27316}$

$\alpha =1.92\times {10}^{-2}{K}^{-1}$

For R = $123.4\mathrm{\Omega }$

$T=T_0+\frac{1}{\alpha }(\frac{R}{R_0}-1)$

$T=273.16+\frac{1}{1.92\times {10}^{-2}}(\frac{123.4}{101.6}-1)$

$T=384.75K$

Q. 10.4 (a) Answer the following:
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)?

Answer:

Unlike the melting point of ice and the boiling point of water, the triple point of water has a fixed value of 273.16 K. The melting point of ice and the boiling point of water vary with pressure.

Q. 10.4 (b) Answer the following :

There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^{\circ }C$ and ${100}^{\circ }C$ 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 $273.16K$ . What is the other fixed point on this (Kelvin) scale?

Answer:

The other fixed point on the Kelvin scale is 0 K. 0K is the absolute zero

Q. 10.4 (c) Answer the following :

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 $273.16?$

Answer:

This is because 0 °C on the Celsius scale corresponds to the melting point at standard pressure is equal to 273.15 K, whereas 273.16 K is the triple point of water. The triple point of water is 0.01^oC, not 0 °C

Q. 10.4 (d) Answer the following :

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?

Answer:

Let at a certain temperature, the readings on the Fahrenheit and Kelvin Scales be T _F and T _K, respectively

$T_F-32=\frac{9}{5}(T_K-273)$ $(i)$

Let at another temperature, the reading on the Fahrenheit and Kelvin Scales be T' _F and T' _K respectively

$T_F^{\prime }-32=\frac{9}{5}(T_K^{\prime }-273)$ $(ii)$

Subtracting equation (ii) from (i)

$T_F-T_F^{\prime }=\frac{9}{5}(T_K-T_K^{\prime })$

For T _K - T' _K = 1 K,

T _F - T' _F = 9/5

Therefore, corresponding to 273.16 K, the absolute scale whose unit interval size is equal to that of the Fahrenheit scale

$T_F=\frac{9}{5}\times 273.16$

$T_F=491.688$

Q. 10.5 (a) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

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

Answer:

$PV=nRT$

$\frac{P}{T}=\frac{nR}{V}$

As the moles of oxygen and hydrogen inside the thermometers and the volume occupied by the gases remain constant, P/T would remain constant.

The triple point of water(T ₁ ) = 273.16 K

Pressure in thermometer A at a temperature equal to the triple point of water (P ₁ ) = $1.250\times {10}^{5}Pa$

Pressure in thermometer A at a temperature equal to the Normal melting point of sulphur (P ₂ ) = $1.797\times {10}^{5}Pa$

The normal melting point of sulphur, as read by thermometer A, T_2, would be given as

$T_2=\frac{P_2{T}_1}{P_1}$

$T_2=\frac{1.797\times {10}^5\times 273.16}{1.250\times {10}^5}$

$T_2=392.69K$

Pressure in thermometer B at a temperature equal to the triple point of water (P₁' ) = $0.200\times {10}^{5}Pa$

Pressure in thermometer B at a temperature equal to the Normal melting point of sulphur (P₂' ) = $0.287\times {10}^{5}Pa$

The normal melting point of sulphur, as read by thermometer B, T₂' would be given as

$T_2^{\prime }=\frac{P_2^{\prime }{T}_1}{P_1^{\prime }}$

$T_2^{\prime }=\frac{0.287\times {10}^5\times 273.16}{0.200\times {10}^5}$

$T_2^{\prime }=391.98K$

Q. 10.5 (b) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

What do you think is the reason behind the slight difference in the 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?

Answer:

The slight difference in the answers of thermometers A and B occurs because the gases used in the thermometers are not ideal gases. To reduce this discrepancy, the experiments should be carried out at low pressures where the behaviour of real gases tends closer to that of ideal gases.

Q. 10.6 A steel tape 1m long is correctly calibrated for a temperature of ${27.0}^{\circ }C.$ The length of a steel rod measured by this tape is found to be $63.0cm$ on a hot day when the temperature is ${45.0}^{\circ }C.$ 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 ${27.0}^{\circ }C?$ Coefficient of linear expansion of steel $=1.20\times {10}^{-5}{K}^{-1}.$

Answer:

At 27 ^oC, the 63 cm ($l_1$) mark on the steel tape would be measuring exactly 63 cm as the tape is calibrated at 27 ^o C

Coefficient of linear expansion of steel $=1.20\times {10}^{-5}{K}^{-1}.$

Actual length when the scale is giving a reading of 63 cm at 45 ^oC is $l_2$

$l_2=l_1(1+\alpha \mathrm{\Delta }T)=63\times (1+1.20\times {10}^{-5}\times (45-27))=63.013608cm$

The actual length of the steel rod on a day when the temperature is 45 °C is 63.013608 cm.

Length of the same steel rod on a day when the temperature is 63°C.

Q. 10.7 A large steel wheel is to be fitted on to a shaft of the same material. At ${27}^{\circ }C$ , the outer diameter of the shaft is $8.70cm$ and the diameter of the central hole in the wheel is $8.69cm$ . 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 : ${\alpha }_{steel}=1.20\times {10}^{-5}{K}^{-1}$ .

Answer:

Diameter of the steel shaft at 27 ^o C (T ₁ ) d ₁ = 8.70 cm

The diameter of the central hole in the wheel, d₂ = 8.69 cm

Coefficient of linear expansion of the steel ${\alpha }_{steel}=1.20\times {10}^{-5}{K}^{-1}$.

The wheel will slip on the shaft when the diameter of the steel shaft becomes equal to the diameter of the central hole in the wheel.

Let this happen at temperature T

$d_2=d_1(1+\alpha (T-T_1))8.69=8.7(1+1.2\times {10}^5(T-27))$

$T=-68.79{}^{o}C$

Q. 10.8 A hole is drilled in a copper sheet. The diameter of the hole is $4.24cm$ at ${27.0}^{\circ }C.$ What is the change in the diameter of the hole when the sheet is heated to ${227}^{\circ }C?$ Coefficient of linear expansion of copper $=1.70\times {10}^{-5}{K}^{-1}.$

Answer:

Coefficient of linear expansion of copper $\alpha$ $=1.70\times {10}^{-5}{K}^{-1}.$

Coefficient of superficial expansion of copper is $\beta$

$\beta =2\alpha$

$\beta =2\times 1.7\times {10}^{-5}$

$\beta =3.4\times {10}^{-5}{K}^{-1}$

Diameter of the hole at 27 ^o C (d ₁ ) = 4.24 cm

Area of the hole at 227 ^o C is

$A_2=A_1(1+\beta \mathrm{\Delta }T)$

$A_2=\pi {\left(\frac{4.24}{2}\right)}^2(1+3.4\times {10}^{-5}(227-27))$
$A_2=14.215c{m}^2$

Let the diameter at 227 ^oC be d ₂

$\pi {\left(\frac{d_2}{4}\right)}^2=14.215$

$d_2=4.254cm$

Change in diameter is d ₂ -d ₁ = 4.24 -4.254 = 0.014 cm.

Q. 10.9 A brass wire $1.8m$ long at ${27}^{\circ }C$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-{39}^{\circ }C,$ what is the tension developed in the wire if its diameter is $2.0mm?$ Co-efficient of linear $expansionofbrass=2.0\times {10}^{-5}{K}^{-1};$ $Youn{g}^{\prime }smodulusofbrass=0.91\times {10}^{11}Pa.$

Answer:

Young's Modulus of Brass, $Y=0.91\times {10}^{11}$

Coefficient of linear expansion of Brass, $\alpha =2.0\times {10}^{-5}{K}^{-1}$

The diameter of the given brass wire, d = 2.0 mm

Length of the given brass wire, $l$ = 1.8 m

Initial Temperature T ₁ = 27 ^o C

Final Temperature T ₂ = -39 ^o C

$Y=\frac{Stress}{Strain}$

$Y=\frac{\frac{F}{A}}{\frac{\mathrm{\Delta }l}{l}}$

$F=\frac{\mathrm{\Delta }lAY}{l}$

$F=\frac{(l\alpha \mathrm{\Delta }T)AY}{l}$

$F=\alpha \mathrm{\Delta }TAY$

$F=\alpha \mathrm{\Delta }TY\pi \frac{d^2}{4}$

$F=\frac{2.0\times {10}^{-5}\times (-39-27)\times 0.91\times {10}^{11}\times \pi \times (2\times {10}^{-3}{)}^2}{4}$

$F=-378N$

The tension developed in the wire is 378 N. The negative sign signifies that this tension is inward.

Q. 10.10 A brass rod of length $50cm$ and diameter $3.0mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at ${250}^{\circ }C,$ if the original lengths are at ${40.0}^{\circ }C?$ Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear $expansionofbrass=2.0\times {10}^{-5}{K}^{-1},steel=1.2\times {10}^{-5}{K}^{-1}$ ).

Answer:

Length of the rods $l$ = 50 cm

Coefficient of linear expansion of brass, ${\alpha }_b=2\times {10}^{-5}{K}^{-1}$

Coefficient of linear expansion of steel, ${\alpha }_s=1.2\times {10}^{-5}{K}^{-1}$

Initial Temperature T ₁ = 40.0 ^o C

Final Temperature T ₂ = 250 ^o C

Change in the length of the brass rod is

$\mathrm{\Delta }{l}_b=l_t{\alpha }_b\mathrm{\Delta }T$

$\mathrm{\Delta }{l}_b=50\times 2.0\times {10}^{-5}\times (250-40)$

$\mathrm{\Delta }{l}_b=0.21cm$

A change in the length of the steel rod is

$\mathrm{\Delta }{l}_s=l_s{\alpha }_s\mathrm{\Delta }T$

$\mathrm{\Delta }{l}_s=50\times 1.2\times {10}^{-5}\times (250-40)$

$\mathrm{\Delta }{l}_s=0.126cm$

Change in the length of the combined rod is

$\mathrm{\Delta }l=\mathrm{\Delta }{l}_s+\mathrm{\Delta }{l}_b$

$\mathrm{\Delta }l=0.126+0.21$

$\mathrm{\Delta }l=0.336cm$

Q. 10.11 The coefficient of volume expansion of glycerine is $49\times {10}^{-5}{K}^{-1}.$ What is the fractional change in its density for a ${30}^{\circ }C$ rise in temperature?

Answer:

Coefficient of volume expansion of glycerine is $\gamma =49\times {10}^{-5}{K}^{-1}$

Let the initial volume and mass of a certain amount of glycerine be V and m, respectively.

Initial density is

$\rho =\frac{m}{V}$

Change in volume for a 30 ^oC rise in temperature will be

$\mathrm{\Delta }V=V(\gamma \mathrm{\Delta }T)$

$\mathrm{\Delta }V=V(49\times {10}^{-5}\times 30)$

$\mathrm{\Delta }V=0.0147V$

Final Density is

${\rho }^{\prime }=\frac{m}{V+\mathrm{\Delta }V}$

${\rho }^{\prime }=\frac{m}{1.0147V}$

${\rho }^{\prime }=\frac{0.986m}{V}$

Fractional Change in density is

$\frac{{\rho }^{\prime }-\rho }{\rho }=\frac{\frac{0.986m}{v}-\frac{m}{v}}{\frac{m}{v}}=-0.014$

The negative sign signifies that with an increase in temperature, density will decrease.

Q. 10.12 A $10kW$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0kg.$ How much is the rise in temperature of the block in $2.5$ minutes, assuming ${50}^o{/}_o$ of power is used up in heating the machine itself or lost to the surrounding. $Specificheatofaluminium=0.91J{g}^{-1}{K}^{-1}.$

Answer:

Power of the drilling machine, P = 10 kW

Time. t = 2.5 min

Total energy dissipated, E, is

$E=Pt$

$E=10\times {10}^3\times 2.5\times 60$

$E=1.5\times {10}^{6}J$

The thermal energy absorbed by the aluminium block is

$\mathrm{\Delta }Q=\frac{E}{2}$

$\mathrm{\Delta }Q=\frac{1.5\times {10}^6}{2}$

$\mathrm{\Delta }Q=7.5\times {10}^{5}J$

Mass of the aluminium block, m = 8.0 kg

Specific heat of aluminium, c = 0.91 J g ^-1 K ^-1

Let the rise in temperature be $\mathrm{\Delta }T$

$mc\mathrm{\Delta }T=\mathrm{\Delta }Q$

$\mathrm{\Delta }T=\frac{\mathrm{\Delta }Q}{mc}$

$\mathrm{\Delta }T=\frac{7.5\times {10}^5}{0.91\times 8\times {10}^3}$

$\mathrm{\Delta }T=103.02{}^{o}C$

Q. 10.13 A copper block of mass $2.5kg$ is heated in a furnace to a temperature of ${500}^{\circ }C$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $=0.39J{g}^{-1}{K}^{-1};$ heat of fusion of water $=335J{g}^{-1}$ ).

Answer:

Mass of copper block m = 2.5 kg

Initial Temperature of the copper block, T ₁ = 500 ^o C

Final Temperature of Copper block, T ₂ = 0 ^o C

Specific heat of copper, c = 0.39 J g ^-1 K ^-1

Thermal Energy released by the copper block is $\mathrm{\Delta }Q$

$\mathrm{\Delta }Q=mc\mathrm{\Delta }T$

$\mathrm{\Delta }Q=2.5\times {10}^3\times 0.39\times 500$

$\mathrm{\Delta }Q=487500J$

Latent heat of fusion of water, L = 335 J g ^-1

The amount of ice that can melt is

$w=\frac{\mathrm{\Delta }Q}{L}$

$w=\frac{487500}{335}$

$w=1455.22g$

1.455 kg of ice can melt using the heat released by the copper block.

Q. 10.14 In an experiment on the specific heat of a metal, a $0.20kg$ block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent $0.025kg$ ) containing $150c{m}^3$ of water at ${27}^{\circ }C.$ The final temperature is ${40}^{\circ }C.$ 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?

Answer:

Let the specific heat of the metal be c.

Mass of metal block m = 200 g

Initial Temperature of the metal block = 150 ^o C

Final Temperature of the metal block = 40 ^o C

The heat released by the block is

$\mathrm{\Delta }Q=mc\mathrm{\Delta }T$

$\mathrm{\Delta }Q=200\times c\times (150-40)$

$\mathrm{\Delta }Q=22000c$

Initial Temperature of the calorimeter and water = 27 °C

Final Temperature of the calorimeter and water = 40 °C

Amount of water = 150 cm

Mass of water = 150 g

Water equivalent of calorimeter = 25 g

Specific heat of water = 4.186 J g ^-1 K ^-1

Heat absorbed by the Calorimeter and water is $\mathrm{\Delta }{Q}^{\prime }$

$\mathrm{\Delta }{Q}^{\prime }=(150+25)\times 4.186\times (40-27)$

$\mathrm{\Delta }{Q}^{\prime }=9523.15J$

The heat absorbed by the Calorimeter and water is equal to the heat released by the block

$\mathrm{\Delta }Q=\mathrm{\Delta }{Q}^{\prime }22000c=9523.15c=0.433J{g}^{-1}{K}^{-1}$

The above value would be less than the actual value since some heat must have been lost to the surroundings as well, which we haven't accounted for.

Q. 10.15 Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat (C_v)

$(calmo{l}^{-1}{K}^{-1})$

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, the molar specific heat of a monatomic gas is $2.92cal/molK.$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Answer:

Monoatomic gases have only a translational degree of freedom, but diatomic gases have rotational degrees of freedom as well. The temperature increases with an increase in the spontaneity of motion in all degrees. Therefore, to increase the temperature of diatomic gases, more energy is required than that required to increase the temperature of monoatomic gases by the same value, owing to higher degrees of freedom in diatomic gases.

If we only consider rotational modes of freedom, the molar specific heat of the diatomic gases would be given as

$c=\frac{fR}{2}$

$c=\frac{5}{2}\times 1.92$

$c=4.95calmo{l}^{-1}{K}^{-1}$

The number of degrees of freedom = 5 (3 translational and 2 rotational)

The values given in the table are more or less in accordance with the above calculated ones. The larger deviation from the calculated value in the case of chlorine is because of the presence of vibrational motion as well.

Q. 10.16 A child running a temperature of ${101}^{\circ }F$ 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 ${98}^{\circ }F$ 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 $30kg.$ . 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 $580cal{g}^{-1}$ .

Answer:

Initial Temperature of the boy = 101°F

Final Temperature of the boy = 98°F

Change in Temperature is

$\mathrm{\Delta }T=3{}^{o}F$

$\mathrm{\Delta }T=3\times \frac{5}{9}$

$\mathrm{\Delta }T=1.67{}^{o}C$

Mass of the child is m = 30 kg

Specific heat of human body = 1000 cal kg ^{-1 °C -1}

Heat released is $\mathrm{\Delta }Q$

$\mathrm{\Delta }Q=mc\mathrm{\Delta }T$

$\mathrm{\Delta }Q=30\times 1000\times 1.67$

$\mathrm{\Delta }Q=50000cal$

Latent heat of evaporation of water = 580 cal g ^-1

The amount of heat lost by the boy has been absorbed by the water.

Let the mass of water which has evaporated be m'

$\mathrm{\Delta }Q=m^{\prime }L$

$m^{\prime }=\frac{Q}{L}$

$m^{\prime }=\frac{50000}{580}$

$m^{\prime }=86.2g$
Time in which the water has evaporated, t = 20 min.

The rate of evaporation is m'/t

$\frac{m^{\prime }}{t}=\frac{86.2}{20}$

$\frac{m^{\prime }}{t}=4.31gmi{n}^{-1}$

Q. 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 cm has a thickness of $5.0cm.$ If $4.0kg$ of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is ${45}^{\circ }C.$ and co-efficient of thermal conductivity of thermacole is $0.01J{s}^{-1}{m}^{-1}{K}^{-1}.$ $[Heatoffusionofwater=335\times 103Jk{g}^{-1}]$

Answer:

Side of the box s = 30 cm

Area available for conduction A

A = 6s ²

A=6(30) ²

A=5400 cm ² = 0.54 m ²

Temperature difference = 45 ^o C

Coefficient of thermal conductivity of thermacole is k = 0.01 J s ^-1 m ^-1 K ^-1

Width of the box is d = 5 cm

Heat absorbed by the box in 6 hours is $\mathrm{\Delta }Q$

$\mathrm{\Delta }Q=\frac{kA\mathrm{\Delta }T}{l}$

$\mathrm{\Delta }Q=\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}$

$\mathrm{\Delta }Q=104976J$

The heat of fusion of water is $L=335\times {10}^{3}Jk{g}^{-1}$

The amount of ice which has melted is m'

$m^{\prime }=\frac{\mathrm{\Delta }Q}{L}$

$m^{\prime }=\frac{104976}{335\times {10}^3}$

$m^{\prime }=0.313kg$

Amount of ice left after 6 hours = 4 - 0.313 = 3.687 kg

Q. 10.18 A brass boiler has a base area of $0.15m^2$ and thickness $1.0cm.$ It boils water at the rate of $6.0kg/min$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $=109J{s}^{-1}{m}^{-1}{K}^{-1};$ Heat of vaporisation of water $=2256\times 103Jk{g}^{-1}.$

Answer:

The rate at which water boils, R = 6.0 kg min ^-1

The heat of vaporisation of water, $L=2256\times {10}^{3}Jk{g}^{-1}$

The rate at which heat enters the boiler

$\frac{dQ}{dT}=RL$

$\frac{dQ}{dT}=\frac{6\times 2256\times {10}^3}{60}$

$\frac{dQ}{dT}=2.256\times {10}^{5}J{s}^{-1}$

The base area of the boiler, A = 0.15 m ²

Thickness, $l$ = 1.0 cm

Thermal conductivity of brass $=109J{s}^{-1}{m}^{-1}{K}^{-1};$

The temperature inside the boiler = Boiling point of water = 100 ^o C

Let the temperature of the flame in contact with the boiler be T

The amount of heat flowing into the boiler is

$\begin{array}{r}\frac{dQ}{dt}=\frac{KA\mathrm{\Delta }T}{l}\\ 2.256\times {10}^5=\frac{109\times 0.15\times (T-100)}{1\times {10}^{-2}}\\ T-100=137.98\\ T={237.98}^{\circ }\mathrm{C}\end{array}$

The temperature of the flame in contact with the boiler is 237.98 ^o C

Q. 10.19 (a) Explain whya body with large reflectivity is a poor emitter

Answer:

A body with a large reflectivity is a poor absorber. As we know, a body which is a poor absorber will also be a poor emitter. Therefore, a body with large reflectivity is a poor emitter.

Q .10.19 (b) Explain whya brass tumbler feels much colder than a wooden tray on a chilly day

Answer:

Brass is a good conductor of heat. Therefore, once someone touches brass, the heat from their body flows into it, and it feels cold. In the case of a wooden tray, no such conduction of heat from the body takes place, as wood is a very poor conductor of heat.

Q. 10.19 (c) Explain why 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

Answer:

An optical pyrometer relates the brightness of a glowing body to its temperature. In the open, because of other sources of light, the sensor in the optical pyrometer does not detect the true brightness of a red hot piece of iron and thus does not predict its temperature correctly, whereas in the furnace, the piece of iron is the only source of light and the sensor detects its brightness correctly thus giving the correct value of the temperature.

Q. 10.19 (d) Explain why the Earth without its atmosphere would be inhospitably cold

Answer:

The sun's rays contain infrared radiation. These are reflected back by the lower part of the atmosphere after being reflected by the surface of the Earth and are trapped inside the atmosphere, thus maintaining the Earth's temperature at a hospitable level. Without these rays being trapped, the temperature of the Earth would go down severely, and thus the Earth without its atmosphere would be inhospitably cold.

Q. 10.19 (e) Explain why heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water

Answer:

Heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water because the same amount of steam at 100 °C contains more energy available for heat dissipation than the same amount of water at 100 °C in the form of latent heat of vaporisation.

Q. 10.20 A body cools from ${80}^{\circ }C$ to ${50}^{\circ }C$ in 5 minutes. Calculate the time it takes to cool from ${60}^{\circ }C$ to ${30}^{\circ }C.$ The temperature of the surroundings is ${20}^{\circ }C.$

Answer:

Let a body initially be at temperature T ₁

Let its final Temperature be T ₂

Let the surrounding temperature be T ₀

Let the temperature change over time t.

According to Newton's Law of Cooling

$-\frac{dT}{dt}=K(T-T_0)$

$\frac{dT}{T-T_0}=-Kdt$

${\int }_{T_1}^{T_2}\frac{dT}{T-T_0}=-K{\int }_0^{t}dt$

${[ln(T-T_0)]}_{T_1}^{T_2}=-K[t{]}_0^t$

$ln\left(\frac{T_2-T_0}{T_1-T_0}\right)=-Kt$

where K is a constant.

We have been given that the body cools from 80 ^o C to 50 ^o C in 5 minutes when the surrounding temperature is 20 ^o C.

T ₂ = 50 ^o C

T ₁ = 80 ^o C

T ₀ = 20 ^o C

t = 5 min = 300 s.

$ln\left(\frac{50-20}{80-20}\right)=-K\times 300K=\frac{ln(2)}{300}$

For T ₁ = 60 ^o C and T ₂ = 30 ^o C we have

$ln\left(\frac{30-20}{60-20}\right)=-\frac{ln2}{300}t$

$t=\frac{ln(4)\times 300}{ln(2)}$

$t=\frac{2ln(2)\times 300}{ln(2)}$

$t=600s$

$t=10min$

The body will take 10 minutes to cool from 60 ^o C to 30 ^o C at the surrounding temperature of 20 ^o C.