Setu Bridge of Pathway

 Chapter 10 Thermal Properties of Matter

Introduction :

• Ø  Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference.
• Ø  The SI unit of heat energy transferred is expressed in joule (J).
  In CGS system, unit of heat is calorie and kilocalorie (kcal).
• Ø  1 cal = 4.186 J and 1 kcal = 1000 cal = 4186 J.
• Ø  Temperature of a substance is a physical quantity that 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.
• Ø   A branch of science which deals with the measurement of the temperature of a substance is known as thermometry.
• Ø  A device used to measure the temperature of a body is called a thermometer.
• Ø   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 the freezing point and boiling point of water at standard pressure.
• Ø   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.
• Ø  Kelvin scale is the most commonly used temperature scale in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero. The freezing and boiling points of water on this scale are 273.15 K and 373.15 K, respectively. 

Relationship between Different Temperature Scales

• The relationship between the three temperature scales is given in the table below:

• 
  
  
  
  _____________________________________________________________________

• What is Ideal Gas Law?

  The ideal gas law, also known as the general gas equation, is an equation of the state of a hypothetical ideal gas. Although the ideal gas law has several limitations, it is a good approximation of the behaviour of many gases under many conditions. Benoit Paul Émile Clapeyron stated the ideal gas law in 1834 as a combination of the empirical Charles’s law, Boyle’s Law, Avogadro’s law, and Gay-Lussac’s law.

• The empirical form of ideal gas law is given by:

  PV=nRT where, P is the pressure. V is the volume. n is the amount of substance. R is the ideal gas constant. ( R = 8.31 J/K.mol, or R = 0.082 L.atm/K.mol )

• 

  ______________________________________________________________________________

  THERMAL EXPANSION IN METALS

  It is a well-known phenomenon now that substances expand on heating and contract on cooling. If you heat a body, it alters its dimensions. Depending on the shape of the body.

  1.      The expansion can occur in length in which case it is called Linear Expansion.

  Δl / l = α_{l .}ΔT

  where,

  • l is the initial length of the solid
  • Δl is the change in length.
  • α_l length expansion coefficient
  • ΔT is the temperature difference

   

  2.      If we take a square tile and heat it, the expansion will be on two fronts, length and breadth, here it is called Area Expansion.

                               ΔA / A = α_{l .}ΔT

  where,

  • A is the initial area of the solid
  • ΔA is the change in the area.
  • α_A Area expansion coefficient
  • ΔT is the temperature difference

   

  3.      If we take a cube and heat it, all its sides expand and now the body experiences an increase in the overall volume due to this and it is called Volume Expansion.

                         ΔV / V = α_{l .}ΔT

  where,

  • V is the initial volume of the solid
  • ΔV is the change in volume.
  • α_V volume expansion coefficient
  • ΔT is the temperature difference

  Thermal Expansion of Solids Examples

  Ø  Metal hot water heating pipes should not be used in long straight lengths

  Ø  Metal framed windows need rubber spacers

  Ø  Large structures and mega constructions such as railways, and bridges need expansion joints in the structures to avoid subkink

  Ø  Thermometers are another example of an application of thermal expansion

• ___________________________________________________________________

• Coefficient of Thermal Expansion Table

  The value of the coefficient of thermal expansion of different solids is very important to choose the right material. The table given below lists the name of the material along with the corresponding coefficient of thermal expansion values.

• 

  Example 11.2 A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27 °C. To what temperature should the ring be heated to fit the rim of the wheel?( iron = 1.2 10^–5 K^–1.)

  Answer

  Given, T₁ = 27 °C

  L_T1 = 5.231 m

  L_T2 = 5.243 m

  So,

  L_T2 =L_T1 [1+α_l (T₂–T₁)]

  5.243 m = 5.231 m [1 + 1.2010^–5 K^–1 (T₂–27 °C)]

  or T₂ = 218 °C.

  ______________________________________________________________________________

• WHAT IS THE SPECIFIC HEAT CAPACITY FORMULA?

   

  Ø  Specific heat is the heat energy required to change the temperature of one unit mass of a substance of a constant volume by 1 °C. 

  Ø  The specific heat capacity of a substance is the amount of energy needed to change the temperature by 1 unit of material of 1 kg mass.

  Ø  The SI unit of specific heat and specific heat capacity is J/Kg.

   

  Thermal Capacity Formula

  The thermal capacity or heat capacity is a physical property of matter or substance. We define this property as the amount of heat supplied to a given mass of a material to generate a change of unit temperature.

   

  The thermal capacity SI unit is Joule per Kelvin or J/Kg. Besides this, the heat capacity formula or the thermal capacity formula is:

  
  
   Here,

  Delta Q = It is the amount of heat that must be added to the object of mass  ‘M” to raise its temperature by delta T.

  Also, the value of the above parameter varies considerably depending on the initial temperature {ΔT} T of the object and the pressure {ΔP}P applied to it.

   

  Molar Heat Capacity Formula

  Molar heat capacity is the amount of heat needed for the temperature rise of a given substance by 1 ⁰C. The molar heat capacity formula is given by:

  C_m = C/n

  Here,

  C_m = molar heat capacity

  C = heat capacity

  n = number of moles

  We see that “n” is the number of moles of the sample. The number of moles can be determined by the following formula:

  
  
   Specific Heat Formula 

  The specific heat capacity formula is:

   

  Q=mcΔt

                  Or,

  c=Q / mΔt

   

  Here,

  Q  is the heat energy

  m = mass in Kg

  c = specific heat capacity, and

  t is the temperature change in Kelvin

  Also, the change in temperature is given by:

  Δ T = (T_f – T_i)

  Where T_f is the final temperature and T_i is the initial temperature in K.

   Unit of Specific Heat Capacity

  The unit of specific heat capacity is J/Kg. K, or J/Kg °C. 

   The dimensional Formula Of Specific Heat Capacity Is   =M⁰L²T⁻²K⁻¹

   _______________________________________________________________________

   WHAT IS CALORIMETRY?

   

  Ø  It is a way of measuring the amount of heat that is either released or absorbed during any chemical reaction that takes place. 

  Ø  When a body at a higher temperature is brought in contact with another body at a lower temperature, there is a transfer of heat from the body with the higher temperature that is gained by the body with a lower temperature. No heat escapes or is lost in the surroundings.

   Heat Gained = Heat Lost

• CHANGE OF STATE

  Ø  Depending on temperature and pressure, all matter can exist in a solid, liquid or gaseous state. These states or forms of matter are also called the phases of matter.

   

  Ø  The change of state from solid to liquid is called melting and from liquid to solid is called freezing. It is observed that the temperature remains constant until the entire amount of the solid substance melts. That is, both the solid and the liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid.

   

  Ø  The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each other is called its melting point. The change of state from liquid to vapour (or gas) is called vaporisation. It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour.

   

  Ø  The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point. The change from solid state to vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime.

   

  Ø  Let us take a quick look at these thermal properties of matter:

• When there is a change of state from solid to liquid, it is known as melting.
• The change of liquid to solid is called freezing.
• When the liquid state changes to vapor or gas, it is called vaporization/evaporation.
• Change of state of a gas into a liquid is called condensation

___________________________________________________________________________

Latent Heat

 

Ø  Latent heat is defined as the heat or energy that is absorbed or released during a phase change of a substance. It could either be from a gas to a liquid or liquid to solid and vice versa.

Ø  Latent heat is related to a heat property called enthalpy.

Ø   It is denoted by L and its SI unit is J/kg.

 

L = Q / m

Where, Q = Heat released or consumed

             m = mass of substance

 

Ø  There are two types of latent heat.

(i). Latent Heat of melting

It is a amount of heat which is required to change of phase from solid to liquid for unit mass at constant temperature. Ex- Latent heat of melting of ice is 3.33 x 10⁵ J/kg.

 

(ii). Latent Heat of Vaporization

It is a amount of heat which is required to change of phase from liquid to vapor for unit mass at constant temperature. Ex- Latent heat of vaporization of water is 22.6 x 10⁵ J/kg.

 

 

  Example :

When 0.15 kg of ice at 0∘C is mixed with 0.30 kg of water at 50∘C in a container, the resulting temperature of the mixture is 6.7∘C. Calculate the latent heat of fusion of ice. (Swater=4186 J kg–1K−1)

_________________________________________________________________

Heat Transfer

There are three mechanisms of heat transfer which name is given as- conduction, convection and radiation. Conduction occurs within a body or between two bodies in contact. Convection depends on motion of mass from one region of space to another. Radiation is heat transfer by electromagnetic radiation, such as sunshine, with no need for matter to be present in the space between bodies.

(i). Conduction

Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference. Suppose, one end of a metallic rod is put in a flame, the other end of the rod will soon be so hot that you cannot hold it by your bare hands.

Here, heat transfer takes place by conduction from the hot end of the rod through its different parts to the other end. Gases are poor thermal conductors, while liquids have conductivities intermediate between solids and gases.

(ii). Convection

Convection is a mode of heat transfer by actual motion of matter. It is possible only in fluids. Convection can be natural or forced. In natural convection, gravity plays an important part. When a fluid is heated from below, the hot part expands and, therefore, becomes less dense. Because of buoyancy, it rises and the upper colder part replaces it. This again gets heated, rises up and is replaced by the relatively colder part of the fluid. The process goes on.

In forced convection, material is forced to move by a pump or by some other physical means. The common examples of forced convection systems are forced-air heating systems in home.

(iii). Radiation

Radiation is the transfer of heat by electromagnetic waves such as visible light, infrared, and ultraviolet rays. Everyone has felt the warmth of the sun’s radiation and intense heat from a charcoal grill or the glowing coals in a fireplace. Most of the heat from these bodies reaches you not by conduction or convection in the intervening air but by radiation. This heat transfer would occur even if there were nothing but vacuum between you and the source of heat.

________________________________________________________________

Newton’s Law of Cooling

Ø  According to Newton’s law of cooling, "The rate of loss of heat of a body is directly proportional to the excess of the temperature of the body with respect to the surroundings".

Ø  Formula :

T(t) = T_s + (T_o – T_s) e^-kt

Where,

o    t = time,

o    T(t) = Temperature of the given body at time t

o    T_s = Surrounding temperature

o    T_o = Initial temperature of the body

o    k = Constant

 

Ø  Limitations of Newton’s Law of Cooling

o    The difference in temperature between the body and surroundings must be small

o    The loss of heat from the body should be by radiation only

o    The major limitation of Newton’s law of cooling is that the temperature of the surroundings must remain constant during the cooling of the body

Example: The oil is heated to 70^oC. It cools to 50^oC after 6 minutes. Calculate the time taken by the oil to cool from 50^oC to 40^oC given the surrounding temperature T_s = 25^oC.

Solution:

Given:

The temperature of oil after 6 min, T(t) = 50^oC,

• T_s = 25^oC,
• T_o = 70^oC,
• t = 6 min

On substituting the given data in Newton’s law of cooling formula, we get,

T(t) = T_s + (T_s – T_o) e^-kt

[T(t) – T_s]/[T_o – T_s] = e^-kt

-kt ln = [ln T(t) – T_s]/T_o – T_s

-kt = [ln 50 – 25]/70 – 25 = ln 0.555

k = – (-0.555/6) = 0.092

If T(t) = 45^oC (average temperature as the temperature decreases from 50^oC to 40^oC)

Time taken is -kt ln e = [ln T(t) – T_s]/[T_o – T_s]

-(0.092) t = ln 45 – 25/[70 – 25]

-0.092 t = -0.597

t = -0.597/-0.092 = 6.489 min.