Class 11 Physics Chapter 11 Thermal Properties of Matter Notes

Thermal Properties of Matter

TEMPERATURE

It is a relative measure or indication of hotness or coldness.

Scales of Temperature

There are three different scales of temperature.

(a)  The Celsius Scale:

This scale was devised by Anders Celsius in the year 1710. The interval between the lower fixed point and the upper fixed points is divided into 100 equal parts. Each division of the sale is called one degree centigrade or one degree Celsius (1oC). At normal pressure, the melting point of ice is 0oC. This is the lower fixed point of the Celsius scale. At normal pressure, the boiling point of water is 100oC. This is the upper fixed point of the Celsius scale.

(b) The Fahrenheit Scale.

This scale was devised by Gabriel Fahrenheit in the year 1717. The interval between the lower and the upper fixed points is divided into 180 equal parts. Each division of this scale is called one degree Fahrenheit (1oF). On this scale, the melting point of ice at normal pressure is 32oF. This is the lower fixed point. The boiling point of water at normal pressure is taken as 212oF. This is the upper fixed point.

(c)  The Reaumer Scale

This scale was devised by R. A. Reaumer in the year 1730.
The interval between the lower and the upper fixed point is divided into 80 equal parts. Each division is called one degree Reaumer (1oR). on this scale, the melting point of ice at normal pressure 0oR. This is lower fixed point. The boiling point of water at normal pressure is 80oR. This is the upper fixed point.

Conversion of temperature from one scale to another

In order to convert temperature from one scale to another, following relation is used.

Temperatureononescale−LowerfixedpointUpperfixedpoint−Lowerfixedpoint=Temperatureonotherscale−LowerfixedpointUpperfixedpoint−Lowerfixedpoint
⸫   C−0100−0=F−32212−32=R−080−0
or  C10=F−32180=R80

Equilibrium state of energy thermodynamic system is completely described by specific values of some macroscopic variables or state variables. The relation between the state variables is called the equation of state. The thermodynamic state variables are of two types (i) extensive and (ii) intensive.
Extensive variables depend on the quantities of system e.g. volume, mass etc.
Intensive variables are independent of quantity of system e.g. pressure, density, etc.

HEAT

It is the form of energy transferred between two or more systems or a system and its surrounding by virtue of temperature difference.

IDEAL GAS EQUATION

Combining first four laws (i.e. Boyle’s law, Charle’s Law, Gay – Lussac’s Law and Avagadro’s law) we get one single equation for an ideal gas, i.e.

PV = nRT        for n moles of gas

or PVnT = R = constant

Here, R = universal gas constant = 8.31 J/mol-K  = 2 calorie/mol-K

For 1 mole of a gas  n = 1

i.e.  PV = RT

or PVT=R = constant

THERMAL EXPANSION

Experiments show that most of bodies increase their volume upon heating. The extent of expansion of various bodies is characterized by the temperature coefficient of expansion, or simply the coefficient of expansion. While considering solid which retain their shape during temperature variations, the distinction is made between (a) a change in their linear dimensions (viz. the dimensions in a certain direction), i.e. linear expansion, and (b) a change in the volume of a body, i.e. cubic expansion.

The coefficient of linear expansion is the quantity α equal to the fraction of the initial length by which a body taken at 0oC has elongated as a result of heating it by 1oC (or by 1 K):

α=(lt−lo)/lot

where l­o is the initial length at 0oC and lt is the length at a temperature t. From this expansion, we can find

lt=lo(1+αt)

The dimensions of α are K-1  (or   oC-1).
The coefficient of cubic expansion is the quantity γ equal to the fraction of the initial volume by which the volume of a body taken at 0oC has increased upon heating it by 1oC (or by 1 K):

γ=(Vt−Vo)/Vot,

where Vo is the volume of a body at 0oC and Vt is its volume at a temperature t. From this equation, we obtain

Vt=Vo(1+γt)

The quantity γ has also the dimensions of K-1 (or  oC-1).
The coefficient of cubic expansion is about three times larger than the coefficient of linear expansion:

γ=3α

The coefficient of cubic expansion γ for liquids are somewhat higher than for solid bodies, ranging between 10-3 and 10-4 K-1.
What obeys the general laws of thermal expansion only at a temperature above 4 oC. From 0 oC to 4 oC, water contracts rather than expands. At 4 oC, water occupies the smallest volume, i.e. it has the highest density. At the bottom of deep lakes, there is denser water in winter, which remains the temperature of 4 oC even after the upper layer has been frozen.

Application 1

The lengths l1i = 100 m of iron wire and l1c = 100 m of copper wire are marked off at t1 = 20 oC. What is the difference in lengths of the wires at t2 = 60 oC? The coefficients of linear expansion for iron and copper are α1 = 1.2 x 10-5 K-1 and αc = 1.7 x 10-5 K-1.

Solution:

l1i=lo(1+αit1) and  l2i=loi(1+αit2)

The elongation of the iron wires is

l2i−l1i=loiαi(t2−t1).

Substituting  loi=l1i/(1+fit1), we find the elongation of the iron and copper wires

l2i−l1i=l1iαi(t2−t1)/(1+αit1)             (1)
l2c−l1c=l1cαc(t2−t1)/(1+αct1)             (2)

Subtracting (1) from (2) and considering that l1i=l1c=l1, we obtain

l2c−l2i=l1(αc−αi)(t2−t1)(1+αct1)(1−αit1)=19.9mm

For low values of temperature t, when αt<1, it is not necessary to reduce l1 and l2 to lo1 and l­o2 at t = 0 oC. To a sufficiently high degree of accuracy, we an assume that Δl=lαΔt. Under this assumption, the problem can be solved in a simpler way:

Δli=l1iαi(t2−t1),Δlc=l1cαc(t2−t1)

Consequently, since l1i=l1c=l1, we have

Δl=Δlc−Δli=l1(t2−t1)(αc−αi)=20mm

It can been seen that the deviation from a more exact value of 19.9 mm amounts to 0.1 mm, i.e. the relative error Δ=0.1/19.9=0.5%.

Application 2

A solid body floats in a liquid at a temperature t = 0o C and is completely submerged in it at 50o C. What fraction d of volume of the body is submerged in the liquid at 0o C if  γs = 0.3 x 10-5 K-2 and of the liquid, γ1 = 8.0 x 10-5 K-1?

Solution:

In both the cases the weight of the body will be balanced by the force of buoyancy on it.

At  to = 0 oC, the buoyancy is Fb=δVoρog            (1)

where Vo is the volume of the body and ρo is the density of the liquid at to = 0oC. At t = 50 oC, the volume of the body becomes V=Vo(1+γst) and the density of the liquid is ρ1=ρo/(1+γ1t). The buoyancy in this case is

Fb=Voρog(1+γst)/(1+γ1t)                (2)

Equating the right-hand sides of equation (1) and (2), we get

δ=(1+γst)/(1+γ1t)=96%

Specific Heat and Heat Capacity

If a quantity of heat Q produces a change in temperature DT in a body, its heat capacity is defined as
Heat capacity  C=QΔT
The SI unit of heat capacity is JK-1.
The quantity of heat Q required to produce a change in temperature DT is also proportional to the mass m of the sample.
Q=mCΔT
where C is called the specific heat of the substance.
Specific heat may be defined as the heat capacity per unit mass.
C=Heatcapacitymass
It is sometimes convenient, especially in the case with gases, to deal with the number of moles n of a substance rather than its mass. Then,
Q=nCmΔT
where Cm is the molar specific heat, measured in J/mol-K  (or cal/mol-K)
Cm = Mc

The specific heat of a substance usually varies with temperature.

The specific heat changes abruptly when the substance transforms from solid to liquid, or from liquid to gas. It also depends on the conditions under which the heat is supplied. For example, the specific heat of a gas kept at constant pressure Cp is different from its specific heat at constant volume Cv. For air, Cv = 0.17 cal/g-K and Cp = 0.24 cal/g-K For solids and liquids the difference is generally small, and in practice Cp is usually measured.