MHT-CET : Physics Entrance Exam

### MHT - CET : Physics - Radiation Page 2

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5.

Ritchie's experiment

To verify Kirchhoff's law:
P, Q, R are metal cylinders with same cross sectional area
A.
One side - lamp black
One side - polished

• P and Q connected by a U-tube which contains coloured liquid.
• Third cylinder R is kept between P and Q such that they are coaxial.
• Black face of R faces polished face of P.
• Polished face of R faces black face of Q.
• Temperature is raised by pouring hot water in R.
• No change observed in the manometer level.
Þ P and Q absorb equal quantities of heat from R.
• E, Eb are emissive powers of polished and black faces respectively.
a is the coefficient of absorption of the polished face
• Heat incident on P per sec from R
Q = k A Eb     A: area of cross-section of face.
Heat absorbed by P per second =
a Q = a k A Eb
• Heat emitted from the polished face of R = E
Heat incident by Q from R =
k A E
• Heat absorbed by P = Heat absorbed by Q
a K A Eb = k A E
 \ a = E = e Eb

6

The rate of emission of heat from unit area of a perfectly black body is directly proportional to the fourth power of its absolute temperature.

• Eb µ T4
• Eb = s T4   s: Stefan's constant = 5.67 × 10-8 J/m2 s °K
• Stefan's law gives the quantity of radiant energy emitted by the body.
• If A is the total surface area of the body, the total quantity of radiant energy emitted per unit time by the body is A s T4
• Stefan's law is applicable only to perfectly black bodies
• For a body of emissivity 'e',
 E = e \ E = Eb e = s T4 e Eb

 \ dQ = s A e T4 d t
• When a black body is kept in a surrounding, the rate of emission of heat per unit area by a black body is given by
Eb = s (T4 - T04)
T = Temperature of the black body
T0 = Temperature of the surrounding (T > T0)

7

Newton's Law of Cooling

The rate of loss of heat from a body is directly proportional to the excess temperature of the body over the surroundings, provided the excess is small.

 dQ µ (q - q0), dQ = K (q - q0), d t d t

where, K is a constant
(
q - q0) is small difference of temperature between the body and its surroundings.

• Rate of loss of heat from a body
 dQ = m ´ s ´ dq d t d t
• where, m = mass of the body

s = specific heat of its material
 dq = rate of fall of its temperature d t
 \ m s dq = K (q - q0) d t
 \ dq = K = K (q - q0) d t ms
• Newton's law of cooling may also be stated as:
The rate of fall of temperature of a body is directly proportional to the excess temperature of the body over the surroundings, provided the excess is small.

8

It is the correction to be applied to the observed temperature of the body in order to compensate the change in its temperature due to the loss or gain of heat by radiation.
It is equal to the difference between the observed temperature of the body and the temperature that would have been reached in the absence of radiation.

Remember: Radiation correction is given by D q = (q - )
where,
q is the maximum temperature reached in time t

 q¢ is the temperature attained by mixture on cooling for time t . 2

\ Corrected temperature = q + D q

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