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Thermal Properties Of Matter

Class 11th Physics Part Ii Bihar Board Solution
Exercise
  1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively.…
  2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B.…
  3. The electrical resistance in ohms of a certain thermometer varies with temperature…
  4. Answer the following: (a) The triple-point of water is a standard fixed point in modern…
  5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following…
  6. A steel tape 1m long is correctly calibrated for a temperature of 27.0°C. The length of a…
  7. A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer…
  8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What…
  9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid…
  10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same…
  11. The coefficient of volume expansion of glycerin is 49 × 10-5 K-1. What is the fractional…
  12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0…
  13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then…
  14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C…
  15. Given below are observations on molar specific heats at room temperature of some common…
  16. Answer the following questions based on the P-T phase diagram of carbon dioxide: (a) At…
  17. Answer the following questions based on the P - T phase diagram of CO2: (a) CO2 at 1 atm…
  18. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers…
  19. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of…
  20. A brass boiler has a base area of 0.15 m^2 and thickness 1.0 cm. It boils water at the…
  21. a body with large reflectivity is a poor emitter Explain why:
  22. a brass tumbler feels much colder than a wooden tray on a chilly day Explain why:…
  23. an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body…
  24. the earth without its atmosphere would be inhospitably cold Explain why:…
  25. heating systems based on circulation of steam are more efficient in warming a building…
  26. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60…

Exercise
Question 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.


Answer:

Given,

Triple point of Neon (TNeon) = 23.57 K


Triple point of Carbon dioxide (TCarbon dioxide) = 216.55 K


We know,


Tc = Tk – 273.15


Where, Tc is temperature in Celsius scale


Tk is temperature in Kelvin scale


Also,


Tf = (9/5)Tc + 32


Where, Tf is temperature in Fahrenheit scale


For Neon:


Tk = 24.57 K


∴ Tc = Tk – 273.15


⇒ Tc = 23.57 – 273.15


⇒ Tc = -248.58 °C


And Tf = (9/5)Tc + 32


⇒ Tf = (9/5)(-248.58) + 32


⇒ Tf = -447.44 + 32


⇒ Tf = -415.44°F


For Carbon dioxide:


Tk = 216.55 K


∴ Tc = Tk – 273.15


⇒ Tc = 216.55 – 273.15


⇒ Tc = -56.6 °C


And Tf = (9/5)Tc + 32


⇒ Tf = (9/5)(-56.6) + 32


⇒ Tf = -101.88 + 32


⇒ Tf = -69.88 °F


NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.



Question 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?


Answer:

Given,

Triple point of water in absolute scale A, TA = 200A


Triple point of water in absolute scale B, TB = 350B


We know that the triple point of water in Kelvin scale (TK) is 273.15K.


So, 200A = 273.15


⇒ A = 273.15/200 …………(1)


Also, 350B = 273.15


⇒ B = 273.15/350 …………(2)


Dividing equation (1) by (2),we get


A/B = (273.15/200)/(273.15/350)


⇒ A/B = 350/200


⇒ A/B = 7/4


⇒ 4A = 7B


Hence, A and B are related by the relation 4A = 7B.


NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.



Question 3.

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

R = Ro [1 + α(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?


Answer:

Given,

Triple point of water, T0 = 273.16 K


Resistance at triple point, R0 = 101.6 Ω


Normal melting point of lead, T = 600.5 K


Resistance at normal melting point of lead, R = 165.5 Ω


Also,


R = R0[1+α(T-T0)]


Where, R0 is the initial resistance


Ris the final resistance


T0 is the initial temperature


T is the final temperature


∴ 165.5 Ω = (101.6 Ω)[1 + α(600.5 K - 273.16 K)]


⇒ 1+α(327.34) = 165.5/101.6


⇒ 1+327.34α = 1.629


⇒ 327.34α = 0.629


⇒ α = 0.629/327.34


⇒ α = 1.92 × 10-3 K-1


For R = 123.4 Ω,


123.4 Ω = (101.6 Ω)[1 + (1.92 × 10-3 K-1)(T - 273.16 K)]


⇒ 1 + (1.92 × 10-3 K-1)(T - 273.16 K) = 123.4/101.6


⇒ 1 + (1.92 × 10-3 K-1)(T - 273.16 K) = 1.214


⇒ (1.92 × 10-3 K-1)(T - 273.16 K) = 0.214


⇒ T - 273.16 K = 0.214/(1.92 × 10-3 K-1)


⇒ T – 273.16 K = 111.75 K


⇒ T = 111.75 K + 273.16 K


⇒ T = 384.91 K



Question 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

tc = 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?


Answer:

(a) The melting point of ice and boiling point of water are temperature and pressure dependent but the triple point of water is not. Hence, the triple-point of water is a standard fixed point in modern thermometry.


(b) The absolute zero temperature or 0K is the other fixed point on the Kelvin scale.


(c) The temperature 273.16 K is the triple point of water. The temperature 273.15 K which is the melting point of ice in Kelvin scale corresponds to 0°C in Celsius scale. Hence, the relation has 273.15 in it and not 273.16.


(d) Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale.


The relation between both the temperatures is as follows:


(TF - 32)/180 = (TK - 273.15)/100 …………(1)


Let TF’ be the temperature on Fahrenheit scale and TK’ be the temperature on absolute scale.


The relation between both the temperatures is as follows:


(TF’ - 32)/180 = (TK’ - 273.15)/100 …………(2)


It is given that TK’ – TK = 1 K


Subtracting equation (1) from equation (2), we get


(TF’ – TF)/180 = (TK’ - TK)/100 = 1/100


⇒ TF’ – TF = (1×180)/100


⇒ TF’ = 9/5


Triple point of water = 273.16 K


∴ Triple point of water on absolute scale, T = 273.16×(9/5)


⇒ T = 491.69


NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.



Question 5.

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:


(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?


Answer:

(a) We know that the triple point of water, T0 = 273.16 K.


At this temperature, pressure in thermometer A,


PA = 1.250 × 105 Pa


Let T be the normal melting point of sulphur.


At this temperature, pressure in thermometer A,


P = 1.797 × 105 Pa


According to Charles’ law,


PA/T0 = P/T


∴ T = (PT0)/PA


⇒ T = (1.797×105 Pa × 273.16 K)/(1.250×105 Pa)


⇒ T = 392.69 K


Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer A is 392.69 K.


At triple point T0 = 273.16 K, the pressure in thermometer B,


PB = 0.200×105 Pa


At temperature T, the pressure in thermometer B,


P’ = 0.287 × 105 Pa


According to Charles’ law,


PB/T0 = P’/T


⇒ (0.200 × 105 Pa)/(273.16 K)= (0.287 × 105 Pa)/T


∴ T = [(0.287 × 105 Pa)/(0.200 × 105 Pa)] × 273.16 K


⇒ T = 391.98 K


Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer B is 391.98 K.


(b) The oxygen and hydrogen gases used in thermometers A and B respectively are not ideal gases. These gases have different physical behaviours. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as ideal gases and will show same characteristics.


NOTE: Ideal gases are those gases which follow ideal gas equation exactly.



Question 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.


Answer:

Given,

Length of the steel tape at temperature T = 27°C = 27 + 273.15 K = 300.15 K, L = 1 m


Coefficient of linear expansion of steel, α = 1.20 × 10-5 K-1


At temperature T1 = 45°C = 45 + 273.15 K = 318.15 K, the length of the steel rod,


L1 = 63 cm = 0.63m


Let L2 be the actual length of the steel rod and L2’ be the length measured by the steel tape at 45°C.


We know,


L2’ = L + αL(T1 - T)


∴ L2’ = 1 m + (1.20 × 10-5 K-1)(1 m)(318.15 K – 300.15 K)


⇒ L2’ = 1 + 0.000216


⇒ L2’ = 1.000216 m = 100.0216 cm


Hence, the actual length of the steel rod at 45°C is given by


L2 = (L2’/L)×L1


⇒ L2 = [(100.0216 cm)/(100 cm)](63 cm) = 63.0136 cm Therefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.



Question 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: αsteel = 1.20 × 10–5 K–1.


Answer:

Given,

Initial temperature, T1 = 27°C = 27 + 273.15 = 300.15 K


Outer diameter of the steel shaft at T1, D1 = 8.70 cm


Diameter of the central hole in the wheel at T1, d1 = 8.69 cm


Coefficient of linear expansion of steel, αsteel = 1.20 × 10-5 K-1


Let the temperature be T2 after the shaft is cooled using ‘dry ice’.


The wheel will slip on the shaft, if the change in diameter,


Δd = 8.69 – 8.70 = – 0.01 cm


Temperature T2 can be calculated as follows.


Δd = D1αsteel(T2 – T1)


⇒ -0.01 cm = 8.70 cm × 1.20 × 10-5 K-1 × (T2 – 300.15 K)


⇒ (T2 – 300.15 K) = -95.78 K


⇒ T2 = 204.37 K = 204.37 – 273.15 = -68.78°C


Therefore, the wheel will slip on the shaft when the temperature of the shaft is –68.78°C.



Question 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.


Answer:

Given,

Initial temperature, T1 = 27.0°C = 27 + 273.15 = 300.15 K


Diameter of the hole at T1, d1 = 4.24 cm


Final temperature, T2 = 227°C = 227 + 273.15 = 500.15 K


Co-efficient of linear expansion of copper, αCu = 1.70 × 10-5 K-1


Let the diameter of the hole at T2 be d2.


If the co-efficient of superficial expansion is β, and the change in temperature is ΔT, then


Change in area (∆A)/Original area (A) = β∆T


∆A/A = [(πd22/4) - (πd12/4)]/(πd12/4)]


⇒ ∆A/A = (d22 - d12)/d12


But β = 2α


∴ (d22 - d12)/d12 = 2α∆T


⇒ (d22/d12) -1 = 2α(T2 – T1)


⇒ d22/(4.24 cm)2 – 1 = 2 × (1.7 × 10-5 K-1) (500.15 K – 300.15 K)


⇒ d22/(17.98 cm2) = 68 × 10-4 + 1


⇒ d22 = 17.98 cm2 × 1.0068


⇒ d22 = 18.1022 cm2


⇒ d2 = 4.2546 cm


Change in diameter = d2– d1 = 4.2546 cm – 4.24 cm =0.0146 cm Hence, the diameter increases by 0.0146 cm.



Question 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? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.


Answer:

Given,

Initial temperature, T1 = 27°C = 27 + 273.15 K = 300.15 K


Length of the brass wire at T1, L = 1.8 m


Final temperature, T2 = –39°C = -39 + 273.15 K = 234.15 K


Diameter of the wire, d = 2.0 mm = 2 × 10-3 m


Co-efficient of linear expansion of brass, α = 2.0 × 10–5 K–1


Young’s modulus of brass, Y = 0.91 × 1011 Pa


Let the tension developed in the wire be T.


Young’s modulus is given by


Y = Stress/Strain


⇒ Y = (F/A)/(∆L/L)


⇒ ∆L = FL/AY


So, for tension T,


ΔL = TL/AY


Where, T = Tension developed in the wire


A = Area of cross-section of the wire


ΔL = Change in the length


But, ΔL = αL(T2 – T1)


∴ αL(T2 – T1) = TL/[π(d/2)2 × Y]


⇒ T = απ(T2 – T1)Y(d/2)2


⇒ T = (2.0 × 10–5 K–1)×3.14×(234.15 K - 300.15 K)×( 0.91 × 1011 Pa)×( 2 × 10-3 m/2)2


⇒ T = -377.37 N


The negative sign indicates that the direction of tension is inwards.



Question 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).


Answer:

Given,

Initial temperature, T1 = 40°C = 40 + 273.15 K = 313.15K


Final temperature, T2 = 250°C = 250 + 273.15 = 523.15 K


Length of the brass rod at T1, L1 = 50 cm = 0.5 m


Diameter of the brass rod at T1, d1 = 3.0 mm = 3 × 10-3 m


Length of the steel rod at T2, L2 = 50 cm = 0.5 m


Diameter of the steel rod at T2, d2 = 3.0 mm = 3 × 10-3 m


Coefficient of linear expansion of brass, α1 = 2.0 × 10–5 K–1 Coefficient of linear expansion of steel, α2 = 1.2 × 10–5 K–1


Change in temperature, ΔT = T2 – T1 = 523.15 K-313.15K=210 K


For the expansion in the brass rod,


Change in length (∆L1)/Original length (L1) = α1ΔT


⇒ ΔL1 = α1L1ΔT


⇒ ΔL1 = (2.0 × 10–5 K–1)×(0.5 m)×(210 K)


⇒ ΔL1 = 0.0021 m = 0.21 cm


For the expansion in the steel rod,


Change in length (∆L2)/Original length (L2) = α2ΔT


⇒ ∆L2 = α2L2ΔT


⇒ ∆L2 = (1.2 × 10–5 K–1)×(0.5 m)×(210 K)


⇒ ∆L2 = 0.00126 m = 0.126 cm


Total change in length, ΔL = ∆L1 + ∆L2


⇒ ΔL = 0.21 cm + 0.126 cm


⇒ ∆L= 0.336 cm


No thermal stress develops at the junction since both the ends of the rod expand freely.



Question 11.

The coefficient of volume expansion of glycerin is 49 × 10–5 K–1. What is the fractional change in its density for a 30°C rise in temperature?


Answer:

Given,

Coefficient of volume expansion of glycerin, αV = 49 × 10–5 K–1


Rise in temperature, ΔT = 30°C = 30 K


Fractional change in its volume = ΔV/V


We know,


ΔV/V = αVΔT


⇒ VT2 - VT1 = VT1αVΔT


⇒ (m/ρT2)-(m/ρT1) = (m/ρT1VΔT


⇒ [(1/ρT2)-(1/ρT1)]/(1/ρT1) = αVΔT


⇒ (ρT1 - ρT2)/ρT2 = αVΔT


Where, m is the mass of glycerin


ρT1 is the initial density at T1


ρT2 is the initial density at T2


T1 - ρT2)/ρT2 is the fractional change in density


So, fractional change in density of glycerine = 49 × 10–5 K–1 × 30 K


= 1.47 × 10-2



Question 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.


Answer:

Given,

Power of the drilling machine, P = 10 kW = 10 × 103 W


Mass of the aluminium block, m = 8.0 kg = 8 × 103 g


Machine operation time, t = 2.5 min = 2.5×60=150 s Specific heat of aluminium, c = 0.91 J g–1 K–1


Let the rise in the temperature of the block after drilling be ΔT.


Total energy of the drilling machine, E = Pt


⇒ E = 10 × 103 × 150


⇒ E = 1.5 × 106 J


It is given that only 50% of the power is useful.


So, Useful energy, ∆Q = (50/100) × E


⇒ ΔQ = 7.5 × 105 J


But ∆Q = mc∆T


∴ ∆T = ∆Q/mc = (7.5 × 105 J)/(8×103 g × 0.91 J g–1 K–1) = 103 K Hence, the rise in the temperature of the block is 103 K in 2.5 minutes of drilling.


NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.



Question 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).


Answer:

Given,

Mass of the copper block, m = 2.5 kg = 2500 g


Rise in the temperature of the copper block, ΔT = 500 °C


Specific heat of copper, C = 0.39 J g–1 °C–1


Heat of fusion of water, L = 335 J g–1


The maximum heat that the copper block can lose, Q = mCΔT


⇒ Q = 2500 g × 0.39 J g–1 °C–1 × 500 °C


⇒ Q = 487500 J


Let m’ be the amount of ice that melts when the copper block is placed on the ice block.


So, the heat gained by the melted ice, Q = m’L


∴ m’ = Q/L


⇒ m’ = 487500/335


⇒ m’ = 1455.22 g = 1.455 kg


Hence, the maximum amount of ice that can melt is 1.455 kg.


NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.



Question 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 cm3 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?


Answer:

Given,

Mass of the metal, m = 0.20 kg = 200 g


Initial temperature of the metal, T1 = 150°C


Final temperature of the metal, T2 = 40°C


Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g Volume of water, V = 150 cm3


Mass of water at temperature T = 27°C, M = 150 × 1 = 150 g


Fall in the temperature of the metal, ΔT = T1 – T2


⇒ ΔT = 150 – 40


⇒ ΔT = 110°C


Specific heat of water, Cw = 4.186 J g-1 °C-1


Let the specific heat of the metal be C.


Heat lost by the metal, θ = mCΔT … (1)


Rise in the temperature of the water and calorimeter system, ΔT′’ = 40 – 27


⇒ ΔT’’ = 13°C


Heat gained by the water and calorimeter system, Δθ′′ = m1CwΔT’ ⇒ Δθ′′= (M+m′)CwΔT’ … (2)


We know that,


Heat lost by the metal = Heat gained by the water and calorimeter system


So, mCΔT = (M+m’)CwΔT’


⇒ 200 g × C × 110 °C = (150 g + 25 g) × 4.186 J g-1 °C-1 × 13 °C


∴ C = (175 × 4.186 × 13)/(110 × 200)


⇒ C = 0.43 J g-1 K-1


If some heat is lost to the surroundings, then the value of specific heat of the metal will be smaller than its actual value.


NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.



Question 15.

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


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 somewhat larger (than the rest) value for chlorine?


Answer:

The gases given in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom. Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the degrees of freedom. 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 heat of a diatomic gas is equal to (5/2)R (where, R = Ideal gas law constant = 1.98 cal mol-1 K-1)

So, molar specific heat = (5/2) × 1.98 = 4.95 cal mol-1 K-1


With the exception of chlorine, all the observations in the given table agree with this value of molar specific heat. This is because of the fact that at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.


NOTE: Molecular degrees of freedom refer to the number of ways a molecule in the gas phase may move, rotate, or vibrate in space.


The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.



Question 16.

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 CO2 co-exist in equilibrium?

(b) What is the effect of decrease of pressure on the fusion and boiling point of

CO2?

(c) What are the critical temperature and pressure for CO2? What is their significance?

(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?


Answer:

The P-T phase diagram of CO2 is as follows:


(a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the triple point, i.e., the point corresponding to the temperature value of - 56.7° C and pressure value of 5.1 atm.


(b) With the decrease in pressure, both the fusion and boiling point of carbon dioxide will decrease.


(c) For carbon dioxide, the critical temperature is 31°C and critical pressure is 72.9 atm. If the temperature of carbon dioxide is more than 31° C, it cannot be liquified irrespective of the applied pressure.


(d) Carbon dioxide will be a vapour at 70°C under 1 atm, a solid at -6°C under 10 atm and a liquid at 15°C under 56 atm.



Question 17.

Answer the following questions based on the P – T phase diagram of CO2:

(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally.

Does it go through a liquid phase?

(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?

(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.

(d) CO2 is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?


Answer:

The P-T phase diagram of CO2 is as follows:


(a) Since the temperature -60° C lies to the left of 56.7° C on the curve i.e., it lies in the region vapour and solid phase, carbon dioxide will condense directly into the solid without becoming liquid.


(b) Since the pressure 4 atm is less than 5.1 atm the carbon dioxide will condense directly into solid without becoming liquid.


(c) When a solid CO2 at 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temperature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the Temperature-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of CO2 at 10 atm.


(d) Since 70° C is higher than the critical temperature of CO2, the CO2 gas cannot be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will deviate from its perfect gas behaviour with increasing pressure.



Question 18.

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 min, 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.


Answer:

Given,

Mass of the child, m = 30kg


Initial temperature of the body of the child, T1 = 101°F


Final temperature of the body of the child, T2 = 98°F


Change in temperature, ΔT = 101 – 98 = 3°F


⇒ ΔT = [3 × (5/9)] °C


⇒ ΔT = 1.667 °C


Time taken to reduce the temperature, t = 20 min


Mass of the child, m = 30 kg = 30 × 103 g


Specific heat of the human body = Specific heat of water (c) = 1000 cal kg-1 °C-1


Latent heat of evaporation of water, L = 580 cal g–1


The heat lost by the child is given as


∆θ = mc∆T


⇒ ∆θ = 30 kg × 1000 cal kg-1 °C-1 × 1.667 °C = 50000 cal


Let m1 be the mass of the water evaporated from the child’s body in 20 min.


Loss of heat through water is given by


∆θ = m1L


∴ m1 = ∆θ/L


⇒ m1 = (50000 cal)/(580 cal g–1) = 86.2 g


Hence, average rate of extra evaporation caused by the drug, Rave = m1/t


⇒ Rave = (86.2 g)/(200 min)


⇒ Rave = 4.3 g/min.


NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.



Question 19.

A ‘thermacole’ icebox is a cheap and 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]


Answer:

Given,

Side of the given cubical ice box, s = 30 cm = 0.3 m


Thickness of the ice box, l = 5.0 cm = 0.05 m


Mass of ice kept in the ice box, m = 4 kg


Time gap, t = 6 h = 6 × 60 × 60 s = 21600 s


Outside temperature, T = 45°C


Coefficient of thermal conductivity of thermacole,


K = 0.01 J s–1 m–1 K–1


Heat of fusion of water, L = 335 × 103 J kg–1


Let m’ be the total amount of ice that melts in 6 h.


The amount of heat lost by the food, θ = KA(T - 0)t/l


Where, A = Surface area of the box = 6×s×2 = 6×0.3×2 = 0.54 m3


∴ θ = [(0.01 Js–1m–1K–1)×(0.54m3)×(45°C)×(21600s)]/(0.05 m)


⇒ θ = 104976 J


But θ = m'L


∴ m’ = θ/L = 104976/(335 × 103 J kg-1)


⇒ m’ = 0.313 g


Mass of ice left = 4 – 0.313 = 3.687 kg


Hence, the amount of ice remaining after 6 h is 3.687 kg.


NOTE: Heat of fusion is the amount of heat supplied to a specific quantity of the substance to change its state from a solid to a liquid at constant pressure.



Question 20.

A brass boiler has a base area of 0.15 m2 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 × 103 J kg–1.


Answer:

Given,

Base area of the boiler, A = 0.15 m2


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 of brass, K = 109 J s–1 m–1 K–1


Heat of vaporisation, L = 2256 × 103 J kg–1


The amount of heat flowing into water through the brass base of the boiler is given by


θ = KA(T1 - T2)t/l ....(1)


where, T1 is the temperature of the flame in contact with the boiler


T2 is the boiling point of water (= 100°C)


Heat required for boiling the water is given by


θ = mL … (2)


Equating equations (1) and (2), we get


mL = KA(T1 - T2)t/l


⇒ T1 - T2 = mLl/KAt


⇒ T1 - T2 = (6 kg × 2256 × 103 J kg-1× 0.01 m)/(109 J s–1 m–1 K–1× 0.15 m2× 60 s)


⇒ T1 - T2 = 137.98°C


⇒ T1 = 137.98°C + 100°C


⇒ T1 = 237.98°C


So, the temperature of the part of the flame in contact with the boiler is 237.98°C.


NOTE: Heat of vaporisation is the amount of heat supplied to a specific quantity of the substance to change its state from a liquid to a gas at constant pressure.



Question 21.

Explain why:

a body with large reflectivity is a poor emitter


Answer:

A body with a large reflectivity is a poor absorber of light radiations since it doesn’t absorb energy. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.



Question 22.

Explain why:

a brass tumbler feels much colder than a wooden tray on a chilly day


Answer:

Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler. Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool. Thus, a brass tumbler feels colder than a wooden tray on a chilly day.



Question 23.

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 calibrated for an ideal black body radiation gives too low a value for temperature of a red-hot iron piece kept in the open. Black body radiation equation is given by E = σ (T4 - T04)


Where, E = Energy radiation


T = Temperature of optical pyrometer


To = Temperature of open space


σ = Constant


Hence, an increase in the temperature of open space reduces the radiation energy. When the same piece of iron is placed in a furnace, the radiation energy, E = σ T4.



Question 24.

Explain why:

the earth without its atmosphere would be inhospitably cold


Answer:

In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface. So, without its atmosphere, earth would be inhospitably cold.



Question 25.

Explain why:

heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water


Answer:

A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).



Question 26.

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.


Answer:

According to Newton’s law of cooling,

(-dT/T) = K(T - T0)


⇒ dT/(T - T0) = -Kdt ………….(1)


Where, T is the temperature of the body


T0 is the temperature of the surroundings = 20°C


K is a constant


Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s


Integrating equation (1), we get






………….(1)


Let t’ be the time it takes to cool from 60 °C to 30 °C. Then,







⇒ t’ = 2 × 300


⇒ t’ = 600 s


⇒ t’ = 600/60 = 10 min


Hence, it takes 10 min to cool the body from 60 °C to 30 °C.


NOTE: Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).