NCERT Solutions for Class 11 Physics Chapter 11 - Thermal Properties of Matter
11.1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Ans. The relation between Kelvin scale and Celsius
scale is :
T_{k} = T_{c} + 273.5
T_{c} = T_{k} – 273.5
For neon, T_{k} = 24.57
So, T_{c} = 24.57 – 273.15
= – 248.58°C
For CO_{2}, T_{k} = 216.55 K,
So, T_{c} = 216.55 – 273.15
= – 56.60°C
Also the relation between Kelvin scale and Fahrenheit scale is
$$\frac{(\text{T}_\text{k}-273.15)}{100}=\frac{(\text{T}_\text{F}-32)}{180}\\T_k=\frac{9}{5}(T_k-273.15)+32$$
Now for neon,
T_{k} = 24.57 K,
$$T_F=\frac{9}{5}[24.57-273.15]+32$$
= – 415.44°C
For CO_{2},T_{k} = 216.55 K,
$$T_F=\frac{9}{5}[216.55-273.15]+32$$
= – 69.88°F.
11.2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?
Ans. Triple point of water is
T = 273.16 K
Since, the absolute scales measure the triple
point as 200 A and 350 B.
200 A = 350 B
= 273.16 K
$$lA=\frac{273.16}{200}k\\and\space 1B=\frac{273.16}{200}k$$
If T_{A} and T_{B} are the temperatures on the two scales, then
$$\frac{273.16}{200}T_A=\frac{273.16}{350}T_B\\So, T_A=\frac{200}{300}T_B=\frac{4}{7}T_B$$
Therefore, the required relation is TA : TB = 4/7
11.3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = R_{o} [1 + a (T – To )]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Ans. Given, the triple point of water
T = 273.16 K and R = 101.6 Ω
$$R = R_0 [1 + 5 × 10^{–3} (T – T_0)]\\(Given) ...(A)\\101.6 = R_0 [1 + 5 × 10^{–3} (273.16 – T_0)]\\...(i)$$
At temperature
T = 600.5 K,
R = 165.5 Ω
165.5 = R_{0} [1 + 5 × 10^{–3} (600.5 – T_{0})] ...(ii)
Dividing eq. (i) by eq. (ii)
$$\frac{101.6}{165.5}=\frac{[1+5×10^{-3}(273.16-T_0]}{[1+5×10^{-3}(600.5-T_0)]}$$
T_{0} = – 49.3 K
Put this value in eq. (i) we get R_{0} = 38.9 Ω
For R = 123.4 Ω, using eq. (A)
123.4 Ω = 38.9 [1 + 5 × 10^{–3} (T+ 49.3)]
T = 384.8 K
11.4. 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°C and 100°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 K. What is the other fixed point on this (Kelvin)scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc 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 ?
Ans. (a) Triple point of water has unique value i.e., 273.16 K. The melting point of ice and boiling point of water do not have unique values and change with change in pressure.
(b) The other fixed point is absolute zero of temperature.
(c) On Celsius scale 0°C corresponds to the melting ponit of ice at normal pressure and the value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
(d) The Fahrenheit scale and Absolute scale are related as
$$\frac{T_F-32}{180}=\frac{T_k-273}{100}.....(i)$$
For another set of temperatures T_{F}′ and T′_{K}
$$\frac{T_F'-32}{180}=\frac{T_k'-273}{100}....(ii)$$
Subtracting (i) from (ii)
$$\frac{T_F'-T_F}{180}=\frac{T'_K-T_K}{100}\\T_F-T_F=\frac{9}{5}(T_K'-T_K)×10\\=\frac{9}{5}×1K=\frac{9}{5}K$$
Triple point of water
= 273.16 K,
Triple ponit of water on the absolute scale is
$$=\frac{9}{5}×273.16=491.688$$
11.5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
Temperature | Press thermometer A | Pressure thermometer B |
Triple-point of water | 1.250 × 10^{5} Pa | 0.200 × 10^{5} Pa |
Normal melting point of sulphur | 1.797 × 10^{5} Pa | 0.287 × 10^{5} 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 ?
Ans. (a) For thermometer A,
P_{tr} = 1.250 × 10^{5}
Pa, Ttr = 273.16 K
P = 1.797 × 10^{5}
Pa, T = ?
According to Charles law, we have
$$\frac{P_tr}{T_tr}=\frac{p}{T}\\T=P.\frac{T_{tr}}{P_{tr}}\\=\frac{1.797×10^5×273.16}{1.250×10^5}$$
T = 392.69 K
Therefore, the absolute temperature of normal melting point of sulphur as read by thermometer A is 392.69 K. For thermometer B
$$P_{tr} = 0.200 × 10^5 Pa,\\T_{tr} = 273.16 K, P = 0.287 × 10^5, T = ?\\T=\frac{0.287×10^5}{0.200×10^5}×273.16\\= 391.98 K$$
Therefore, the absolute temperature of normal melting point of sulphur as read by thermometer B is 392.69 K.
(b) The discrepancy between the two readings is due to the fact that gases (O_{2} and H_{2}) are not perfectly ideal gases. To reduce the discrepancy, the readings should be taken at low pressure so that the gases could show perfect behaviour.
11.6. A steel tape 1m long is correctly calibrated for a temperature of 27.0° C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0° 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° C? Coefficient of linear expansion of steel = 1.20 × 10^{–5} K^{–1}.
Ans. Length of the steel tape at 27°C is 100 cm, i.e.,
L = 100 cm
T_{1} = 27° C = 300 kIts length at 45°C is
L′ = L + ΔL = L + αL ΔT
T2 = 45° C = 45 + 273 = 318 K
Then, L′ = 100 + 1.2 × 10^{–5} × 100 × (318 – 300)
= 100.0216 cm
Length of 63 cm measured by this tape at 45° C will actually be
$$L"= \frac{(100.0216)}{100}×63=63.0136\space cm.$$
11.7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 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 : a_{steel} = 1.20 × 10^{–5} K^{–1}.
Ans. Given, a_{steel} = 1.20 × 10^{–5} K. Here T_{1} = 27° C = 27
+ 273 = 300 K
Length at temperature T_{1}K = l_{T}_{1} = 8.70 cm
Length at temperature T2K = lT_{2} = 8.69 cm
Change in length = l_{T}_{2} – l_{T1} = l_{T1} a(T_{2} – T_{1})
or 8.69 – 8.70 = 8.70 × (1.20 × 10^{–5}) × (T_{2} – 300)
$$or\space T_2-300=\frac{0.01}{8.70×1.2×10^{-5}}\\or\space T_2-300=\frac{0.01}{8.70×1.2×10^{-5}}\\T_2=300-95.8=204.2 k=-68.8\degree C$$
11.8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0° C. What is the change in the diameter of the hole when the
sheet is heated to 227° C? Coefficient of linear expansion of copper = 1.70 × 10^{–5} K^{–1}.
Ans. Here L = Diameter of hole = 4.24 cm
ΔT = 227 – 27 = 200° C
= 200 + 273 = 473 K
and α = 1.70 × 10–5 K
Change in diameter,
ΔL = αL ΔT
= 1.70 × 10^{–5} × 4.24 × 473
= 1.44 × 10^{–2} cm
11.9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of – 39° C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 × 10^{–5} K^{–1}; Young’s modulus of brass = 0.91 × 10^{11 }Pa.
Ans. Here, L = 1.8 cm, T = 27° C, T – DT = – 39°C,
DT = T – (T – DT)
= 27 + 36 = 66°C
Also a = 2 × 10^{–5} K^{–1}
Change in length L = aL DT
= 2 × 10–5 × 1.8 × 66
= 2.376 × 10^{–3} m
$$Strain\space produced=\frac{2.376×10^{-3}m}{1.8}=10^{-3}\\ Now\space \frac{stress}{Strain}= Young's modulus × Strain\\or Stress = 0.91 × 10^11 × 1.32 × 10^–3= 1.2 × 10^8 Pa\\Now, Diameter of wire,\\ d = 2 mm = 2 × 10^–3 m\\Area\space of\space the\space wire =\frac{\pi d^2}{4}= \frac{3.14×(2×10^{-3})^2}{4}= 3.14×10^{-6}\\=3.77×10^2 N$$
11.10. A brass rod of length 50 cm and diameter 3.0 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°C, if the original lengths are at 40.0° 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 × 10–5 K–1, steel = 1.2 × 10–5 K–1 ).
Ans. $$Given, l_{brass} = l_{steel} = 50 cm\\d_{brass} = d_{steel} = 3 mm = 0.3 cm\\\Delta l_{brass} = ?, \Delta l_{steel} = ?\\Now \Delta T = 250 – 40 = 210° C\\\alpha{brass} = 2 × 10^{–5} K^{–1}\\\alpha{steel} = 1.2 × 10^{–5} K^{–1}\\\Delta l_{brass} = \alpha_{brass} × l_{brass} × \Delta T\\= 2 × 10^{–5} × 50 cm × 210\\ = 0.21 cm\\And \Delta l_{steel} = \alpha {steel} × l_{steel} × \Delta T\\= 1.2 × 10^{–5} × 50 cm × 210\\ = 0.126 cm = 0.13 cm (app)\\Total\space change\space in\space length,\\\Delta l = \Delta l_{brass} + \Delta l_{steel}\\ = 0.21 cm + 0.13 cm = 0.34 cm\\$$
Since, the rod expands freely from both the ends, not thermal stress is developed at the junction.
11.11. The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30° C rise in temperature ?
Ans. Given, g = 49 × 10^{–5} °C^{–1}
ΔT = 30° C
Fractional change in density,
$$\frac{\Delta p}{p}=\gamma\Delta T\space \begin{bmatrix}\frac{\Delta V}{V}= \frac{\Delta p}{p} \end{bmatrix}\\=49×10^{-5} K^{-1}×30 \degree C\\=1.47×10^{-2}$$
11.12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 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 J g^{–1} K^{–1}.
Ans. Given Power, P = 10 kW = 10 × 10^{3} W = 10^{4} W Time, t = 2.5 min = 2.5 × 60 s = 150 s Specific heat of aluminium
C = 0.91 Jg^{–1} K^{–1}
= 0.91 × 10^{3}
Jkg^{–1} K^{–1}
Energy used, E = Pt = 10^{4 }× 150 s = 1.5 × 10^{6 }J
It is given that only 50% of the power is useful.
$$Useful energy Q= \begin{pmatrix}\frac{50}{100}\end{pmatrix}×1.5×10^6 J\\=7.5×10^5J\\Let \Delta T= increase\space in\space temperature\\Using Q = mC\Delta T\\so \Delta T =\frac {Q}{mC}\\=\frac {7.5×10^5 J}{8 kg}×0.91×10^3 J kg^{-1o}C^{-1}\\=103 \degree C$$
Therefore, in 2.5 minutes of drilling the rise in
temperature is 103°C
11.13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°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 J g^{–1} K^{–1}; heat of fusion of water = 335 J g^{–1}).
Ans. Let m be the mass of ice melted when hot copper
block is placed over ice block.
Given, C_{cu} = 0.39 J g^{–1} K^{–1}
= 0.39 × 10^{3} J kg^{–1} K^{–1}
Heat lost by copper block
= Heat gained by ice
m_{cu}CΔT = mL [ΔT = Rise in temperature]
$$m=\frac{m_{cu}C\Delta T}{L}\\=\frac{2.5×10^3×0.39×500}{335}\\=1455.2 g=1.455\space kg\\Hence,\space the\space maximum\space amount\space of\space ice\space that\space can\space melt\space is\space 1.455\space kg.$$
11.14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm^{3} of water at 27 °C. The final temperature is 40 °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 ?
Ans.
$$Let C be the specific heat of the metal (in cal g^{–1} °C^{–1})\\Heat lost by metallic block = mC\Delta T\\= 200 × C × (150 – 40) cal\\= 22000 C cal\\Given, volume of water V = 150 cm^3\\Mass of water, M = V × r = 150 g\\Water equivalent, W = 0.025 kg = 25 g\\Specific heat of water, C′ = 1 cal g^{–1}° C^{–1}\\Heat gained by water and calorimeter system\\= (M + W)C′ \Delta T′\\= (150 + 25) × 1 × (40 – 27)\\ = 175 × 13 = 2275 cal\\By principle of calorimetry,\\Heat lost by metallic block\\= Heat gained by calorimeter system\\22000 C cal = 2275 cal\\C=\frac{2275}{22000}=0.1 C cal g^{-1}\degree C^{-1}\\=0.43 J g^{-1} K^{-1}$$
If some heat is lost to the surroundings, then the value of C will be smaller than the acutal value.
11.15. Given below are observations on molar specific heats at room temperature of some common gases.
Gas | Molar specific heat (Cv) (cal mo1^{–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, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the some what larger (than the rest) value for chlorine ?
Ans. The gases listed in the table are diatomic. Besides the translational degree of freeedom, they have vibrational as well as rotational motion degrees of freedom. Therefore, to increase the temperature of these gases heat must be supplied. This increase the average energy of all mods of motion. Hence, the molar
specific heat of diatomic gases is more than that of monatomic gases. If only rotational mode of motion is considered, then the molar specific of a diatomic gas
$$=\frac{5}{2}R=\begin{pmatrix}\frac{5}{2}\end{pmatrix}×1.98\\=4.95\space cal\space mol^{-1}k^{-1}$$
With the exception of chlorine all the observations in the given table agree with
$$\begin{pmatrix}\frac{5}{2}\end{pmatrix}R$$
[R is called gas constant].
This is because at room temperature, chlorine also has vibrational modes at motion besides rotational and translational modes of motion. This is the reason that chlorine has somewhat larger value of molar specific heat.
11.16. A child running a temperature of 101 °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 °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 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 cal g^{–1}.
Ans. Fall in temperature = DT = 101 – 98 = 3° F
$$But\space 1\degree F = \begin{pmatrix}\frac{5}{9}\end{pmatrix}\degree C\\\Delta T=3×\begin{pmatrix}\frac{5}{9}\end{pmatrix}\degree C =\begin{pmatrix}\frac{5}{3}\end{pmatrix}\degree C$$
Time taken to reduce the temperature
t = 20 mins
Mass of the child, m = 30 kg
Specific heat of human body
= specific heat of water
= c = 1000 cal/kg/°C
Latent heat of evaporation of water,
L = 580 cal/g
Now heat lost by the child = H = mcΔT
=\frac{30×1000×5}{3}
=50000 cal
Let m′ be the mass of the water evoporated from
the child's body in 20 mins, then m′L = H
$$m'=\frac{H}{L}=\frac{5×10^4}{580}=86.2\space g$$
Average rate of extra evaporation caused by
drug
$$=\frac{m'}{t}=\frac{86.2g}{20\space min}=4.3/min$$
11.17. 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 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m^{–1} K^{–1}. [Heat of fusion of water = 335 × 103 J kg^{–1}]
Ans. Here length of each side,
l = 30 cm = 0.3 m
Thickness of each side,
Δx = 5 cm = 0.05 m
Total surface area through which heat enters
into the box,
A = 6l^{2 }= 6 × 0.3 × 0.3 = 0.54 m^{2}.
Temperature difference
ΔT = (45 – 0)°C = 45°C
Thermal conductivity,
K = 0.01 Js^{–1} m^{–1} k^{–1}
Time Δt = 6 hrs = 6 × 60 × 60 s
Latent heat of fusion,
L = 335 × 10^{3 }J/kg
Let m be the mass of ice melted in this time, then
$$\Delta Q=mL=KA \begin{pmatrix}\frac{\Delta T}{\Delta x}\end{pmatrix} \Delta t\\m= KA \begin{pmatrix}\frac{\Delta T}{\Delta x}\end{pmatrix} \Delta t/ L\\= 0.01×0.54× \begin{pmatrix}\frac{45}{0.54}\end{pmatrix}×\begin{pmatrix}\frac{6×60×60}{335×10^3}\end{pmatrix}$$
= 0.313 kg
Mass of ice left = (4 – 0.313) kg = 3.687 kg
Hence, the amount of ice remaining after 6 hrs is 3.687 kg
11.18. A brass boiler has a base area of 0.15 m^{2 }and thickness 1.0 cm. It boils water at the rate of 6.0 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 J s^{–1} m^{–1} K^{–1} ; Heat of vaporisation of water = 2256 × 10^{3 }J kg^{–1}.
Ans. Given, Base area of the boiler,
A = 0.15 m^{2}
Thickness of the boiler
l = 1.0 cm = 0.01 m
Boiling rate of water,
R = 6.0 kg/min
Mass m = 6 kg
Time, t = 1 min = 60 s
Thermal conductivity
K = 109 Js^{–1} m^{–1} K^{–1}
Heat of vaporisation of water,
L = 2256 J/kg
The amount of heat flowing into water through the brass base of the boiler is given by
$$Q=\frac{[KA(T_1-T_2)t]}{l}.....(i)$$
Where,
T^{1} = Temperature of the flame in contact with the boiler.
T^{2} = Boiling point of water = 100°C
Heat required for boiling the water,
Q = mL ...(ii)
Equating equations (i) and (ii),
$$\frac{[KA(T_1-T_2)t]}{l}=mL\\Or, (T_1-T_2)=\frac{mLl}{KAt}\\=\begin{pmatrix}\frac{6×2256×1000×0.01}{109×0.15×60}\end{pmatrix}\\=137.98\degree C$$
Therefore, the temperature of the part of the flame in contact with the boiler is 137.98 °C
11.19. 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
Ans. (a) According to Kirchhoff’s law of black body radiations , good emitters are good absorbers and bad emitters are bad absorber. A body with large reflectivity is poor absorber of heat and consequently, it is also a poor emitter.
(b) Brass is a good conductor of heat, while wood is a bad conductor. When we touch the brass tumbler on a chilly day, heat starts flowing from our body to the tumbler and we feel it cold. However, when the wooden tray is touched, heat does not flow from our hands to the tray and we do not feel cold.
(c) An optical pyrometer is based on the principle that the brightness of a glowing surface of a body depends upon its temperature. Therefore, if the temperature of the body is less than 6000°C, the image formed by the optical pyrometer is not brilliant and we do not get the reliable result. It is for this reason that the pyrometer gives a too low value for the temperature of red hot iron in the open.
(d) The lower layer of Earth's atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus, the heat radiations received by the earth from the Sun during the day time are kept trapped by the atmosphere. If there were no atmosphere, whole of heat radiated by the earth during night would escape to the universe leaving it under the intense cold.
(e) Steam at a temperature of 100°C has more heat (because its latent heat is 540 cal /g) compared to the same amount of water at the same temperature. Therefore, heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
11.20. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Ans. According to Newton's law of cooling,
Time taken 't' to cool a body from temperature T_{1} to T_{2} is
$$t=\frac{1}{K},\frac{In[T_1-T_0]}{[T_2-T_0]}where,$$
Temperature of the surroundings = T_0 = 20° C, K is constant
Temperature of the body falls from 80° C to 50°
C in time t′ = 5 min = 300 s
$$5=\frac{2.303}{K}\begin{bmatrix}log\frac{[80-20]}{[50-20]}\end{bmatrix}\\K=\begin{bmatrix}\frac{2.303}{5}\end{bmatrix}log \begin{pmatrix}\frac{6}{3}\end{pmatrix}=\begin{bmatrix}\frac{2.303}{5}\end{bmatrix}×.3010\\=0.1386\space min^{-1}\\$$
In the second case T_{1} = 60°C, T_{2} = 30° C, hence time
$$t=\begin{bmatrix}\frac{2.303}{0.1386}\end{bmatrix}\begin{bmatrix}log\frac{[60-20]}{[30-20]}\end{bmatrix}\\=\begin{bmatrix}\frac{2.303}{0.1386}\end{bmatrix}log\space 4= \begin{pmatrix}\frac{2.303×0.6021}{0.1386}\end{pmatrix}\\=\frac{1.3866363}{0.1386}=\frac{1.3866}{0.1386}= 10\space mins$$
Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.
11.21. Answer the following questions based on the P–T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO_{2 }co-exist in equilibrium ?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO_{2} ?
(c) What are the critical temperature and pressure for CO_{2} ? What is their significance ?
(d) Is CO_{2} solid, liquid or gas at (a) –70 °C under 1 atm, (b) – 60°C under 10 atm, (c) 15°C under 56 atm?
Ans. (a) Liquid and vapour phases coexist with solid phase at the triple point temperature – 56.6 °C and a pressure of 5.11 atm.
(b) The boiling point as well as freezing point decrease with decrease in pressure.
(c) The critical temperature is 31.1°C and critical pressure is 73.0 atm.
(d) (a) It is vapour. (b) It is solid and (c) It is liquid.
11.22. Answer the following questions based on the P – T phase diagram of CO2:
(a) CO_{2} at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?
(b) What happens when CO_{2} at 4 atm pressure is cooled from room temperature at constant pressure ?
(c) Describe qualitatively the changes in a given mass of solid CO_{2} at 10 atm pressure and temperature – 65 °C as it is heated up to room temperature at constant pressure.
(d) CO_{2} is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?
Ans. Refer to above diagram ,
(a) Since, – 60 °C lies to left of – 56.6 °C on the curve, the vapour will condense directly into the solid.
(b) Since, 4 atm pressure is less than 5.11 atm, the vapour will condense directly into solid without entering into the liquid region.
(c) When the solid CO_{2} at – 65 °C is heated at 10 atm pressure, It is first converted into a liquid. A further increase in its temperature brings it into the vapour phase. If a horizontal line at P= 10 atm is drawn parallel to the T axis, then the points of intersection of this line with the fusion and vaporization curve give the fusion and boiling points at 10 atm.
(d) Above 31.1 °C, the gas cannot be liquified. Therefore, on being compressed isothermally at 70 °C, there will be no transition to the liquid region. However,
the gas will depart more and more from its perfect gas behaviour with the increase in pressure.
NCERT Solutions for Class 11 Physics Chapter 11 Free PDF Download
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