### Thermal Properties of Matter : Notes and Study Materials -pdf

- Concepts of Thermal Properties of Matter
- Thermal Properties of Matter Master File
- Thermal Properties of Matter Revision Notes
- Thermal Properties of Matter MindMap
- NCERT Solution Thermal Properties of Matter
- NCERT Exemplar Solution Thermal Properties of Matter
- Thermal Properties of Matter: Solved Example 1
- Thermal Properties of Matter: Solved Example 2

### Class 11 Physics Chapter 11 Thermal Properties of matter

Topics and Subtopics in ** Class 11 Physics Chapter 11 Thermal Properties of matter****:**

Section Name | Topic Name |

11 | Thermal Properties of matter |

11.1 | Introduction |

11.2 | Temperature and heat |

11.3 | Measurement of temperature |

11.4 | Ideal-gas equation and absolute temperature |

11.5 | Thermal expansion |

11.6 | Specific heat capacity |

11.7 | Calorimetry |

11.8 | Change of state |

11.9 | Heat transfer |

11.10 | Newton’s law of cooling |

### Thermal Properties of Matter Class 11 Notes Physics Chapter 11

• Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is expressed in joule (J).

In CGS system, unit of heat is calorie and kilocalorie (kcal).

1 cal = 4.186 J and 1 kcal = 1000 cal = 4186 J.

• Temperature of a substance is a physical quantity which measures the degree of hotness or coldness of the substance. The SI unit of temperature is kelvin (K) and °C is a commonly used unit of temperature.

• A branch of science which deals with the measurement of temperature of a substance is known as thermometry. A device used to measure the temperature of a body is called thermometer.

• A thermometer calibrated for a temperature scale is used to measure the value of given temperature on that scale. For the measurement of temperature, two fixed reference points are selected. The two convenient fixed reference points are the ice point and the steam point of water at standard pressure, which are known as freezing point and boiling point of water at standard pressure.

• The two familiar temperature scales are the Fahrenheit temperature scale and the Celsius temperature scale. The ice and steam point have values 32°F and 212°F respectively, on the Fahrenheit scale and 0°C and 100°C on the Celsius scale. On the Fahrenheit scale, there are 180 equal intervals between two reference points, and on the Celsius scale, there are 100.

• If t_{c} and t_{F} are temperature values of a body on Celsius temperature scale and Fahrenheit temperature scale respectively, then the relationship between Fahrenheit and Celsius temperature is given by

• An ideal gas obeys the following law. That is PV = gRT, where P,V and T are the pressure, volume and temperature of the gas respectively, g is the number of moles in an ideal gas and R = 8.31 J mol^{-1} K^{-1} is known as universal gas constant. The equation, PV – gRT is known as ideal gas equation.

• The absolute minimum temperature for an ideal gas, inferred by extrapolating the straight line P – T graph is found to be – 273.15 °C and is designated as absolute zero. Absolute temperature scale (T) and Celsius scale are related by

t° C = T – 273.15

•** Thermal Expansion**

The increase of size of a body due to the increase in the temperature is called thermal expansion. Three types of expansions can take place in solids viz. linear, superficial and volume expansion,

(i) **Linear Expansion:** The increase in the length of a solid on heating is called linear expansion.

If the temperature of a rod of original length l is raised by a small amount Δt, its length increases by Δl. Then the linear expansion is given by

Δl = l ∞ Δt

where a is the coefficient of linear expansion of the given solid. The unit of α is per degree Celsius (°C^{-1}) in the CGS and per kelvin (K^{-1}) in the SI system.

(ii) **Superficial or Area Expansion:** The increase in surface area of the solid on heating is called superficial expansion.

If A_{0} is the area of a solid at 0°C and A( its area at t°C then A_{t} = A_{0}(l + βt)

where β is known as the coefficient of superficial expansion. Unit of β is °C^{-1} or K^{-1}.

(iii) **Volume Expansion:** The increase in volume of the solid on heating is called volume expansion.

The change in the volume of a solid with a change in temperature Δt is given by Δv = V_{γ }Δt

where y is the coefficient of volume expansion.

• The relation among coefficients of linear expansion (α), superficial expansion (β) and volume expansion (γ) is given as

• For a given solid, the three coefficients of expansion α , β, γ are not constant. Their values depend on the temperature range.

• Liquids have volume expansion only. If we do not take into account the expansion of solid container, then the expansion of liquid is called apparent expansion. On the other hand, if we take into account the expansion of solid too, it is referred as the real expansion of liquid. It is found that γ_{r} = γ_{a} + γ_{g}, where γ_{r}= real expansion coefficient of liquid, γ_{a} = apparent expansion coefficient of liquid and γ_{g} = volume expansion coefficient of container vessel (glass).

• Water exhibits an anomalous behaviour. It contracts on heating between 0 °C and 4 °C but expands on heating beyond 4 °C. Thus, specific volume of water is minimum at 4 °C or density of water is maximum at 4 °C. This property of water has an important environmental effect.

• **Thermal Stress**

When a rod is held between two fixed supports and its temperature is increased, the fixed supports do not allow the rod to expand, which results in a stress which is called thermal stress.

Thermal stress in the rod is given by

where Y is the Young’s modulus for the material of the rod, A is the area cross-section of the rod, a is the coefficient of linear expansion and F is the developed force in the rod.

• **Thermal Capacity**

The thermal capacity of a body is the quantity of heat required to raise the temperature of the whole of the body through a unit degree. It is measured in calorie per °C or joule per K.

If Q be the amount of heat needed to produce a change in temperature (Δt) of the substance, then thermal capacity of the substance is given by

Dimensional formula of heat capacity is [ML^{2}T ^{-2}K^{-1} ],

• **Specific Heat Capacity**

The specific heat capacity (also referred to as specific heat) of a substance is the amount of heat required to raise the temperature of a unit mass of substance through 1 °C. It is measured in cal g^{-1}(°C)^{-1} or J kg^{-1} K^{-1}.

The specific heat capacity of a substance is given by

where m is mass of substance and Q is the heat required to change its temperature Δt.

• Molar specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 mole of the substance by 1°C.

It is given by

The unit of molar specific heat capacity is J mole^{-1} K^{-1} in SI system and Cal mol^{-1} °C^{-1 }in CGS system.

The dimensional formula of molar specific heat capacity is [ML^{2}T^{-2} K^{-1} mole^{-1}].

• **Calorimetry**

Calorimetry is concerned with the measurement of heat, the basic apparatus for this purpose being called the calorimeter.

When two bodies at different temperatures are ‘mixed’, heat ‘flows’ from the body at a higher temperature to the one at a lower temperature, until a common ‘equilibrium’ temperature is reached. Assuming this ‘heat exchange’ to be confined to the two bodies alone (i.e, neglecting any heat loss to the surroundings) we have, from the law of energy conservation:

Heat gained by one body = heat lost by the other.

• Transition of matter from one state (solid, liquid and gas) to another is called a change of state.

• The change of state from solid to liquid is called melting and from liquid to solid is called fusion. It is observed that the temperature remains constant until the entire amount of the solid substance melts i.e., both the solid and liquid states of the substance co-exist in thermal equilibrium during the change of state from solid to liquid.

• The temperature at which a solid melts is called its melting point. The value of melting point of a solid is characteristic of the substance and depends on pressure also.

• Melting of ice under increased pressure and refreezing on reducing the pressure is called regelation.

• The change of state from liquid to vapour (or gas) is called vaporisation. The temperature at which the liquid and vapour states of a substance co-exist is called its boiling point.

• The change from solid state to vapour state without passing through the liquid state is called sublimation.

• **The Basic Heat Formula**

The heat Q required to raise the temperature of a mass m of a substance of specific heat capacity s through t degrees is given by

i.e., Heat required = mass x specific heat x change in temperature

• **Latent Heat**

Latent heat of a substance is the amount of heat energy required to change the state of unit mass of the substance from solid to liquid or from liquid to gas/vapour without any change in temperature.

• The latent heat of fusion (L_{f}) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporisation (L_{v}) is the heat per unit mass required to change a substance from liquid to vapour state without change in temperature and pressure.

• **Heat Transfer**

Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation. Conduction usually takes place in solids, convection in liquids and gases, and no medium is required for radiation.

(i) Conduction: According to Maxwell, conduction is the flow of heat through an unequally heated body from places of higher temperature to those of lower temperature. Rate of heat transfer is given by

where K is called Thermal Conductivity and A is area of cross-section.

(ii) Convection: Maxwell defines convection as the flow of heat by the motion of the hot body itself carrying its heat with it.

(iii) Radiation: Radiation is the mode of heat transfer in which heat travels directly from one place to another without the agency of any intervening medium.

• Thermal conductivity is defined as heat energy transferred in unit time from unit area having a unit difference in temperature over unit length. It is expressed in Js^{-1} m^{-1 }°C^{-1} or W^{-1} K^{-1}

• **Thermal Resistance**

The thermal resistance of a body is a measure of its opposition to the flow of heat through it. It is defined as

•** Newton’s Law of Cooling**

Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature of the body and the surroundings, provided the difference in temperature is small, not more than 40 °C.

– ve sign implies that as time passes, temperature T decreases.

When an object at temperature T_{1} is placed in a surrounding of temperature T_{2} the net energy

radiated per second is,

• **Black Body Radiation**

(i) **Emissive Power:** The amount of heat energy rediated per unit area of the surface of a body, per unit time and per unit wavelength range is constant which is called as the ’emissive power’ (e_{λ}) of the given surface, given temperature and wavelength. Its S.I. unit is Js^{-1} m^{-2 }.

(ii) **Absorptive Power :** When any radiation is incident over a surface of a body, a part of it gets reflected, a part of it gets refracted and the rest of it is absorbed by that surface. Therefore, the ‘absorptive power’ of a surface at a given temperature and for a given wavelength is the ratio of the heat energy absorbed by a surface to the total energy incident on it at a certain time. It is represented by (a_{λ}). It has no unit as it is a ratio.

(iii) **Perfect Black Body:** A body is said to be a perfect black body is its absorptivity is 1. It neither reflects nor transmits but absorbs all the thermal radiations incident on it irrespetive of their wavelengths.

(iv) **Wein’s Displacement Law :** This law states that as the temperature increases, the maximum value of the radiant energy emitted by the black body, move towards shorter wavelengths. Wein found that “The product of the peak wavelength ( λ_{m}) and the Kelvin temperature (T) of the black body should remain constant.” λ_{m }x T= b

Where b is constant known as Wein’s constant. Its value is 2.898 x 10^{-3} mk.

(v) **Stefan’s Law :** This law states that the thermal radiations energy emitted per second from the surface of a black body is directly proportional to its surface area A and to the fourth power of its absolute temperature T.

Emission coefficient or degree of blackness of a body is represented by a dimensionless quantity ε, 0 < ε < 1. If ε = 1 then the body is perfectly black body. Hence

Let us consider an object at absolute temperature T and To be the temperature of the surroundings.

• H_{1} = Rate of energy emitted by the body

(vi) The Solar Constant: The average energy emitted from the surface of the sun, absorbed per unit area, per minute by the earth is constant which is called as solar constant which is represented by S whose value is 8.135 jm^{-2} min^{-1}.

Let the earth be moving in a circular path of radius r taking sun as its centre.

Taking sun as perfectly black body, the energy radiated per unit time from the surface of the sun is given by

CBSE Class 11 Physics Chapter-11 Important Questions

**VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)**

**What will be the effect of increasing temperature on (i) angle of contact (ii) surface tension.**Angle of contact increases with increase of temperature while surface tension generally decreases with increase of temperature

Ans.**Explain do why detergents have small angle of contact?**Detergents should have small angle of contact so that they hawe low surface tension and greater ability to Wet a surface. Further as

Ans.

is small Cos will be large so h i.e. penetration will be high.**Why does mercury not Wet glass?**Mercury does not wet glass because of larger cohesive force between Hg-Hg molecules than the adhesive forces between mercury-glass molecules.

Ans.**Why ends of a glass tube become rounded on heating?**When glass is heated, it melts. The surface of this liquid tends to have a minimum area. For a given volume, the surface area is minimum for a sphere. This is why the ends of a glass tube become rounded on heating.

Ans.**State Wein’s displacement law for black body radiation.****State Stefan Boltzmann law.****Name two physical changes that occur on heating a body.**Volume and electrical resistance.

Ans.**Distinguish between heat and temperature.****Which thermometer is more sensitive a mercury or gas thermometer?**Gas thermometer is more sensitive as coefficient of expansion of Gas is more than mercury.

Ans.**Metal disc has a hole in it. What happens to the size of the hole when disc is heated?**Expansion is always outward, therefore the hole size increased on heating.

Ans.**Name a substance that contracts on heating.**Ice

Ans.**A gas is free to expand what will be its specific heat?**Infinity

Ans.**What is Deby’s temperature?**The temperature above which molar heat capacity of a solid substance becomes Constant.

Ans.**What is the absorptive power of a perfectly black body?**One

Ans.**At what temperature does a body stop radiating?**

Ans. At oK.**If Kelvin temperature of an ideal black body is doubled, what will be the effect on energy radiated by it?**

Ans.**In which method of heat transfer does gravity not play any part?**In conduction and radiation

Ans.**Give a plot of Fahrenheit temperature versus Celsius temperature.**

Ans.**Why birds are often seen to swell their feather in winter?**When birds swell their feathers, they trap air in the feather. Air being a poor conductor prevents loss of heat and keeps the bird warm.

Ans.**A brass disc fits snugly in a hole in a steel plate. Should we heat or cool the system to loosen the disc from the hole.**The temp. coefficient of linear expansion for brass is greater than that for teel. On cooling the disc shrinks to a greater extent than the hole, and hence brass disc gets lossened.

Ans.

**2 Marks Questions**

**1. **** 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; heat of fusion of water = 335 J).**

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

Rise in the temperature of the copper block, Δ*θ* = 500°C

Specific heat of copper, *C* = 0.39 J

Heat of fusion of water, *L* = 335 J

The maximum heat the copper block can lose, *Q* = *mC*Δ*θ*

= 2500 0.39 500

= 487500 J

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

The heat gained by the melted ice, *Q* =

Hence, the maximum amount of ice that can melt is 1.45 kg.

**2. **** 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. [Heat of fusion of water =]**

**Ans.**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

Outside temperature, *T*= 45°C

Coefficient of thermal conductivity of thermacole, *K*=

Heat of fusion of water, *L*=

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

The amount of heat lost by the food:

Where,

*A*= Surface area of the box =

But

Mass of ice left = 4 – 0.313 = 3.687 kg

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

**SHORT ANSWER TYPE QUESTIONS (2 MARKS)**

**On a hot day, a car is left in sunlight with all windows closed. Explain why it is considerably warmer than outside, after some time?****Ans.**Glass transmits 50% of heat radiation Coming from a hot source like Sun but does not allow the radiation from moderately hot bodies to pass through it.**Name the suitable thermometers to measure the following temperatures****(a) -100°C (b) 80° C****(c) 780°C (d) 2000°C****Ans.**

(a) Gas thermometer;

(b) Mercury thermometer;

(c) Platinum resistance thermometer;

(d) Radiation pyrometer.**The earth without its atmosphere would be inhospitably cold. Why?****Ans.**Due to green-house effect, the presence of atmosphere prevents heat radiations received by earth to go back. In the absence of atmosphere radiation will go back at night making the temperature very low and inhospitable.**The Coolant used in a nuclear plant should have high specific heat. Why?****Ans.**So, that it absorbs more heat with comparatively small change in temperature and extracts large amount of heat.**A sphere, a cube and a disc made of same material and of equal masses heated to same temperature of 200°C. These bodies are then kept at same lower temperature in the surrounding, which of these will cool (i) fastest, (ii) slowest. Explain.****Ans.**Rate of energy emission is directly proportional to area of surface for a given mass of material. Surface area of sphere is least and that of disc is largest. Therefore cooling of (i) disc is fastest and (ii) sphere is slowest.**Why pendulum clocks generally go faster in winter and slow in Summer?****Why the brake drums of a car are heated when it moves down a hill at constant speed?**

**Ans.**- When the car moves downhill, the decrease in gravitational potential energy is converted into work against force of friction between brake shoe and drum which appears as heat.
- Time period of pendulum =

In winter I becomes shorter so its time period reduces so it goes faster. In summer increases resulting in increase in time period so the clock goes slower.

**The plots of intensity versus wavelength for three blackbodies at temperature T**_{1}, T_{2}and T_{3}respectively are shown.**Arrange the temperature in decreasing order. Justify your answer.****Ans.**

∴ from Wein displacement law

T_{1}> T_{3}> T_{2}**The triple point of water is a standard fixed point in modern thermometry. Why? Why melting point of ice or boiling point of water not used as standard fixed points.****Ans.**The melting point of ice as well as the boiling point of water changes with change in pressure. The presence of impurities also changes the melting and boiling points. However the triple point of water has a unique temperature and is independent of external factors. It is that temperatures at which water, ice & water vapour co-exist that is 273.16K and pressure 0.46 cm of Hg.

**3 Marks Questions**

**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.**Kelvin and Celsius scales are related as:

– 273.15 … (*i*)

Celsius and Fahrenheit scales are related as:

… (*ii*)

For neon:

= 24.57 K

= 24.57– 273.15 = –248.58°C

For carbon dioxide:

= 216.55 K

∴ = 216.55 – 273.15 = –56.60°C

**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?**

**Ans. **Triple point of water on absolute scale A, = 200 A

Triple point of water on absolute scale B, = 350 B

Triple point of water on Kelvin scale, = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.

=

200 A = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.

=

350 B = 273.15

is triple point of water on scale A.

is triple point of water on scale B.

Therefore, the ratio is given as 4: 7.

**3. **** 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 =.**

**Ans. **Length of the steel tape at temperature *T* = 27°C, *l* = 1 m = 100 cm

At temperature = 45°C, the length of the steel rod, = 63 cm

Coefficient of linear expansion of steel, *α* =

Let be the actual length of the steel rod and *l*‘ be the length of the steel tape at 45°C.

= 100.0216 cm

Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as:

= 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.

**4. **** 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.**

**Ans.**Power of the drilling machine, *P*= 10 kW =

Mass of the aluminum block, *m*= 8.0 kg =

Time for which the machine is used, *t*= 2.5 min = 2.5 60 = 150 s

Specific heat of aluminium, *c*= 0.91 J

Rise in the temperature of the block after drilling = *T*

Total energy of the drilling machine =

=

=

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

Useful energy,

But

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

**5.**** 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?**

**Ans. **Mass of the metal, *m* = 0.20 kg = 200 g

Initial temperature of the metal, = 150°C

Final temperature of the metal, = 40°C

Calorimeter has water equivalent of mass, *m*‘ = 0.025 kg = 25 g

Volume of water, *V* =

Mass (*M*) of water at temperature *T* = 27°C:

150 1 = 150 g

Fall in the temperature of the metal:

Specific heat of water, = 4.186 J/g/°K

Specific heat of the metal = *C*

Heat lost by the metal, … (*i*)

Rise in the temperature of the water and calorimeter system:

Heat gained by the water and calorimeter system:

… (*ii*)

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

*mC*Δ*T* = (*M* + *m*‘) Δ*T*‘

200 C 110 = (150 + 25) 4.186 13

If some heat is lost to the surroundings, then the value of *C* will be smaller than the actual value.

**SHORT ANSWER TYPE QUESTION (3 MARKS)**

**A big size balloon of mass M is held stationary in air with the help of a small block of mass M/2 tied to it by a light string such that both float in mid air. Describe the motion of the balloon and the block when the string is cut. Support your answer with calculations.****Ans.**

When the balloon is held stationary in air, the forces acting on it get balance

Up thrust = Wt. of Balloon + Tension in string

U = Mg + T

For the Small block of mass floating stationary in air

When the string is cut T = 0, the small block begins to fall freely, the balloon rises up with an acceleration “a such that

U — Mg = Ma

in the upward direction.**Describe the different types of thermometers commonly used. Give the relation between temperature on different scales. Give four reasons for using mercury in a thermometer.****Two rods of different metals of coefficient of linear expansion and and initial length I**_{1}and I_{2}respectively are heated to the same temperature, Find relation in and I_{2}such that difference between their lengths remain Constant.**Ans.**

Given that the difference in their length remain constant**Explain the principle of platinum resistance thermometer.****A steel rail of length 5m and area of cross section 40cm**^{2}is prevented from expanding while the temperature rises by 10°C. Given coefficient of linear expansion of steel is 1.2 × 10^{-5}k^{-1}. Explain why space needs to be given for thermal expansion.**Ans.**The Compressive strain

Thermal stress

which corresponds to an external force

A force of this magnitude can easily bend the rails, hence it is important to leave space for thermal expansion.**Define (i) Specific heat capacity (ii) Heat capacity (iii) Molar specific heat capacity at Constant pressure and at Constant Volume and Write their units.****Plot a graph of temperature versus time showing the change in the state of ice on heating and hence explain the process (with reference to latent heat)****What is the effect of pressure on melting point of a substance? What is regelation. Give a practical application of it.****What is the effect of pressure on the boiling point of a liquid. Describe a simple experiment to demonstrate the boiling of H**_{2}O at a temperature much lower than 100°C. Give a practical application of this phenomenon.**State and explains the three modes of transfer of heat. Explain how the loss of heat due to these three modes is minimised in a thermos flask.****Define Coefficient of thermal conductivity. Two metal slabs of same area of cross-section, thickness d**_{1}and d_{2}having thermal conductivities K_{1}and K_{2}respectively are kept in contact. Deduce expression for equivalent thermal Conductivity.**Ans.**Definition of coefficient of thermal conductivity.

In steady state the heat passing in unit time through the rod remain same that is

where k is the coefficient of thermal conductivity

Also T_{1}– T_{2}= (T_{1}-T) + (T-T_{2})

**4 Marks Questions**

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

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

**Ans.**It is given that:

*R* = … (*i*)

Where,

are the initial resistance and temperature respectively

*R* and *T* are the final resistance and temperature respectively

α is a constant

At the triple point of water, = 273.15 K

Resistance of lead, = 101.6

At normal melting point of lead, *T* = 600.5 K

Resistance of lead, *R* = 165.5

Substituting these values in equation (*i*), we get:

For resistance, = 123.4

Where, T is the temperature when the resis tan ce of lead is 123.4

**2.**** 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: =.**

**Ans.**The given temperature, *T* = 27°C can be written in Kelvin as:

27 + 273 = 300 K

Outer diameter of the steel shaft at *T*, = 8.70 cm

Diameter of the central hole in the wheel at *T*, = 8.69 cm

Coefficient of linear expansion of steel, =

After the shaft is cooled using ‘dry ice’, its temperature becomes* *.

The wheel will slip on the shaft, if the change in diameter, Δ*d* = 8.69 – 8.70

= – 0.01 cm

Temperature , can be calculated from the relation:

=

0.01 =

= 95.78

∴ = 204.21 K

= 204.21–273.16

= –68.95°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is -69°C.

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

Gas | Molar specific heat () (cal mol K) |

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 somewhat larger (than the rest) value for chlorine?**

**Ans. **The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).

Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes 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 heat of a diatomic

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

**4.The coefficient of volume expansion of glycerin is. What is the fractional change in its density for a 30 °C rise in temperature?**

**Ans.**Coefficient of volume expansion of glycerin, =

Rise in temperature, = 30°C

Fractional change in its volume =

This change is related with the change in temperature as:

Where,

*m* = Mass of glycerine

= Initial density at

= Final density at

Where,

= Fractional change in density

∴Fractional change in the density of glycerin =

**5 Marks Questions**

**1. 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 – 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)** The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

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

**(c)** The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.

Hence, absolute temperature (Kelvin scale) *T*, is related to temperature *t*c, on Celsius scale as:

= *T* –273.15

**(d)** Let *T*F be the temperature on Fahrenheit scale and be the temperature on absolute scale. Both the temperatures can be related as:

……..(i)

Let be the temperature on Fahrenheit scale and be the temperature on absolute scale. Both the temperatures can be related as:

………..(ii)

It is given that:

Subtracting equation (*i*) from equation (*ii*), we get:

Triple point of water = 273.16 K

∴Triple point of water on absolute scale = = 491.69

**2. Two ideal gas thermometers A and Buse oxygen and hydrogen respectively. The following observations are made:**

Temperature | Pressure thermometer A | Pressure thermometer B |

Triple-point of water | ||

Normal melting point of sulphur |

**(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)** Triple point of water, *T* = 273.16 K.

At this temperature, pressure in thermometer A,

Let be the normal melting point of sulphur.

At this temperature, pressure in thermometer A,

According to Charles’ law, we have the relation:

= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.

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

At temperature, the pressure in thermometer B,

According to Charles’ law, we can write the relation:

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

**(b)** The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. 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 perfect ideal gases.

**3.**** 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 =.**

**Ans. **Initial temperature, = 27.0°C

Diameter of the hole at = 4.24 cm

Final temperature, = 227°C

Diameter of the hole at

Co-efficient of linear expansion of copper,

For co-efficient of superficial expansion, and change in temperature Δ*T*, we have the relation:

But

Change in diameter = = 0.0144 cm

Hence, the diameter increases by cm.

**4. **** 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 =; Young’s modulus of brass =.**

**Ans. **Initial temperature, = 27°C

Length of the brass wire at, *l* = 1.8 m

Final temperature, = –39°C

Diameter of the wire, *d* = 2.0 mm = m

Tension developed in the wire = *F*

Coefficient of linear expansion of brass, *α*=

Young’s modulus of brass, *Y* =

Young’s modulus is given by the relation:

…………..(i)

Where,

*F* = Tension developed in the wire

*A* = Area of cross-section of the wire.

Δ*L* = Change in the length, given by the relation:

Δ*L* = *αL* … (*ii*)

Equating equations (*i*) and (*ii*), we get:

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is.

**5. 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 =, steel =).**

**Ans. **Initial temperature, = 40°C

Final temperature, = 250°C

Change in temperature, = 210°C

Length of the brass rod at = 50 cm

Diameter of the brass rod at = 3.0 mm

Length of the steel rod at = 50 cm

Diameter of the steel rod at = 3.0 mm

Coefficient of linear expansion of brass, =

Coefficient of linear expansion of steel, =

For the expansion in the brass rod, we have:

= 0.2205 cm

For the expansion in the steel rod, we have:

= 0.126 cm

Total change in the lengths of brass and steel,

= 0.2205 + 0.126

= 0.346 cm

Total change in the length of the combined rod = 0.346 cm

Since the rod expands freely from both ends, no thermal stress is developed at the junction.

**6. **** 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-exist in equilibrium?**

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

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

**(d) Is 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)** The *P*–*T* phase diagram for is shown in the following figure.

C is the triple point of the phase diagram. This means that at the temperature and pressure corresponding to this point (i.e., at -56.6°C and 5.11 atm), the solid, liquid, and vaporous phases of co-exist in equilibrium.

**(b)** The fusion and boiling points of decrease with a decrease in pressure.

**(c)** The critical temperature and critical pressure of are 31.1°C and 73 atm respectively. Even if it is compressed to a pressure greater than 73 atm, will not liquefy above the critical temperature.

**(d)** It can be concluded from the *P*–*T* phase diagram of that:

(a) is gaseous at -70°C, under 1 atm pressure

(b) is solid at -60°C, under 10 atm pressure

(c) is liquid at 15°C, under 56 atm pressure

**7.**** Answer the following questions based on the P-T phase diagram of :**

**(a) at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?**

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

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

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

**Ans.(a)** No

**(b)** It condenses to solid directly.

**(c)** The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

**(d)** It departs from ideal gas behaviour as pressure increases.

**Explanation:**

**(a)** The *P*–*T* phase diagram for is shown in the following figure.

At 1 atm pressure and at -60°C, lies to the left of -56.6°C (triple point C). Hence, it lies in the region of vaporous and solid phases.

Thus, C condenses into the solid state directly, without going through the liquid state.

**(b)** At 4 atm pressure, lies below 5.11 atm (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid state directly, without passing through the liquid state.

**(c)** When the temperature of a mass of solid (at 10 atm pressure and at -65°C) is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

**(d)** If is heated to 70°C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70°C is higher than the critical temperature of. It will remain in the vapour state, but will depart from its ideal behaviour as pressure increases.

**8.**** 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 .**

**Ans. **Initial temperature of the body of the child, = 101°F

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

Change in temperature, Δ*T* = °C

Time taken to reduce the temperature, *t* = 20 min

Mass of the child, *m* = 30 kg =

Specific heat of the human body = Specific heat of water = *c*

= 1000 cal/kg/ °C

Latent heat of evaporation of water, *L* = 580 cal

The heat lost by the child is given as:

= 50000 cal

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

Loss of heat through water is given by:

=

∴Average rate of extra evaporation caused by the drug

**9. A brass boiler has a base area of 0.15 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 = ; Heat of vaporisation of water =.**

**Ans. **Base area of the boiler, *A* = 0.15

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*=

Heat of vaporisation, *L*=

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

…………(i)

Where,

= Temperature of the flame in contact with the boiler

= Boiling point of water = 100°C

Heat required for boiling the water:

θ = *mL* … (*ii*)

Equating equations (*i*) and (*ii*), we get:

=

= 137.98°C

Therefore, the temperature of the part of the flame in contact with the boiler is 137.98°C.

**10. 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)** A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.

**(b)** 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.

**(c)** 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:

Where,

*E* = Energy radiation

*T* = Temperature of optical pyrometer

= Temperature of open space

= 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,

**(d)** Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface.

**(e)** 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).

**11.**** 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, we have:

Where,

Temperature of the body = *T*

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 (*i*), we get:

…….(ii)

The temperature of the body falls from 60°C to 30°C in time = *t*‘

Hence, we get:

………….(iii)

Equating equations (*ii*) and (*iii*), we get:

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

**NUMERICALS**

**How much pressure will a man of weight 80 kgf exert on the ground when (i) he is lying and (2) he is standing on his feet. Given area of the body of the man is 0.6 m**^{2}and that of his feet is 80 cm^{2}.**Ans.**(i) When man is lying P = = 1.307 × 10^{3}N/m^{-2}

(ii) When man is standing then A = 2 × 80 cm^{2}= 160 × 10^{-4}m^{2}**The manual of a car instructs the owner to inflate the tyres to a pressure of 200 k pa. (a) What is the recommended gauge pressure? (b) What is the recommended absolute pressure (c) if, after the required inflation of the tyres, the car is driven to a mountain peak where the atmospheric pressure is 10% below that at sea level, what will the tyre gauge read?****Ans.**

(a) Pressure Instructed by manual = P_{g}= 2OOK Pa

(b) Absolute Pressure = 101 kP + 200 kPa = 301 kPa

(c) At mountain peak Pa’ is 10% less

Pa’ = 90 k Pa

If we assume absolute pressure in tyre does not change during driving then P_{g}= P – P’_{a}= 301 – 90 = 211 k Pa

So the tyre will read 211 k Pa, pressure.**Two stars radiate maximum energy at wavelength, 3.6 x 10**^{7}m respectively. What is the ratio of their temperatures?**Ans.**By Wien’s Displacement Law

= 4:3**Find the temperature of 149°F on kelvin scale.****Ans.**

T = 286K**A metal piece of 50 g specific heat 0.6 cal/g°C initially at 120°C is dropped in 1.6 kg of water at 25°C. Find the final temperature or mixture.****Ans.**

C_{2}= 1 cal/gm°c

∴ 50 × 0.6 × (120 – θ) = 1.6 × 103 × 1 × (θ – 25)

Θ = 26.8°C**A iron ring of diameter 5.231 m is to be fixed on a wooden rim of diameter 5.243 m both initially at 27°C. To what temperature should the iron ring be heated so as to fit the rim (Coefficient of linear expansion of iron is 1.2 x 10**^{5}k^{-1}?**Ans.**

5.243 = 5.231[1 + 1.2**×**10^{-5}(T – 300)]

T = 191 + 300 = 491 K = 218°C**100g of ice at 0°C is mixed with 100 g of water at 80°C. The resulting temperature is 6°C. Calculate heat of furion of ice.****Ans.**

m_{1}c_{1}(80 – 6) = m_{2}L + m_{2}c_{2}(6 – 0)

100 × 1 × 74 = 100 L + 100 × 1 × 6

L = (1 × 74) -6

= 68 cal/g**Calculate heat required to covert 3kg of water at 0°C to steam at 100°C Given specific heat capacity of H**_{2}0 = 4186J kg^{-1}k^{-1}and latent heat of stem = 2.256 x 10^{6}J/kg**Ans.**Heat required to convert H_{2}O at 0° to H_{2}O at 100 = m_{1}c_{1}t

= 3 × 4 1 86 × 1OO

= 1255BOO J

Heat required to convert H_{2}O at 100°C to steam at 100°C is = mL

= 3 × 2.256 x 10^{6}

= 676BOOO J

Total heat = 80.238OO J**Given length of steel rod 15 cm; of copper 10 cm. Their thermal conductivities are 50.2 j – s m**^{-1}k^{-1}and 385 j- s – m^{-1}k^{-1}respectively. Area of crossection of steel is double of area of Copper rod?**Ans.**

A_{1}= 2A_{2}

T = 317.43 K = 44.43°C**A body at temperature 94°C cools to 86°C in 2 min. What time will it take to cool from 82°C to 78°C. The temperature of surrounding is 20°C.****Ans.**

…….1

…..2

dividing eq. (1) by eq (2)**A body re-emits all the radiation it receives. Find surface temperature of the body. Energy received per unit area per unit time is 2.835 Watt/m**^{2}and = 5.67 x 10^{-8}W m^{-2}K^{-4}.**Ans.**

= 85 k**At what temperature the resistance of thermometer will be 12% more of its resistance at 0°C (given temperature coefficient of resistance is 2.5 × 10**^{-3}C^{-1}?**Ans.**