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March 2011 - Career Point

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PHYSICS FUNDAMENTAL FOR IIT-JEE<br />

Calorimetry, K.T.G., Heat transfer<br />

KEY CONCEPTS & PROBLEM SOLVING STRATEGY<br />

Calorimetry :<br />

The specific heat capacity of a material is the amount<br />

of heat required to raise the temperature of 1 kg of it<br />

by 1 K. This leads to the relation<br />

Q = ms θ<br />

where Q = heat supplied, m = mass, θ = rise in<br />

temperature.<br />

The relative specific heat capacity of a material is the<br />

ratio of its specific heat capacity to the specific heat<br />

capacity of water (4200 J kg –1 K –1 ).<br />

Heat capacity or thermal capacity of a body is the<br />

amount of heat required to raise its temperature by 1<br />

K. [Unit : J K –1 ]<br />

Thus heat capacity = Q/θ = ms<br />

dθ 1 dQ<br />

Also = ×<br />

dt ms dt<br />

i.e., the rate of heating (or cooling) of a body depends<br />

inversely on its heat capacity.<br />

The water equivalent of a body is that mass of water<br />

which has the same heat capacity as the body itself.<br />

[Unit : g or kg] This is given by<br />

m×s<br />

W =<br />

s w<br />

where m = mass of body, s = specific heat capacity of<br />

the body, s w = specific heat capacity of water.<br />

Principle of Calorimetry : The heat lost by one<br />

system = the heat gained by another system. Or, the<br />

net heat lost or gainsed by an isolated system is zero.<br />

It system with masses m 1 , m 2 , ...., specific heat<br />

capacities s 1 , s 2 , ...., and initial temperatures θ 1 , θ 2 , ....<br />

are mixed and attain an equilibrium temperature θ<br />

then<br />

θ =<br />

Σmsθ<br />

, for equal masses θ =<br />

Σms´<br />

Σsθ<br />

Σs<br />

Newton's law of cooling :<br />

The rate of loss of heat from a body in an<br />

environment of constant temperature is proportional<br />

to the difference between its temperature and that of<br />

the surroundings.<br />

If θ = temperature of the surroundings then<br />

dθ<br />

– ms = C´(θ – θ0 )<br />

dt<br />

where C´ is a constant that depends on the nature and<br />

extent of the surface exposed. Simplifying<br />

dθ C<br />

= –C(θ – θ0 ) where C = = constant<br />

dt<br />

mś<br />

Kinetic theory of gases :<br />

The pressure of an ideal gas is given by p = 3<br />

1 µnC<br />

2<br />

where µ = mass of each molecule, n = number of<br />

molecules per unit volume and C is the root square<br />

speed of molecules.<br />

p = 3<br />

1 ρC<br />

2<br />

or pV = 3<br />

1 mC<br />

2<br />

where ρ is the density of the gas and m = mass of the<br />

gas.<br />

Root Mean Square Speed of Molecules :<br />

This is defined as<br />

2<br />

2<br />

2<br />

C1 + C2<br />

+ C3<br />

+ ... + CN<br />

C =<br />

N<br />

where N = total number of molecules. It can be<br />

obtained through these relations<br />

C =<br />

3p<br />

ρ<br />

=<br />

3RT<br />

M<br />

Total Energy of an ideal gas (E) :<br />

This is equal to the sum of the kinetic energies of all<br />

the molecules. It is assumed that the molecules do not<br />

have any potential energy. This follows from the<br />

assumption that these molecules do not exert any<br />

force on each other.<br />

1<br />

E = mC 2 3 m 3<br />

= RT = pV<br />

2 2 M 2<br />

Thus, the energy per unit mass of gas = 2<br />

1 C<br />

2<br />

The energy per unit volume = 2<br />

3 p<br />

The energy per mole = 2<br />

3 pV = 2<br />

3 RT<br />

2<br />

XtraEdge for IIT-JEE 24 MARCH <strong>2011</strong>

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