First slide

Radiation

Question

Rate of cooling $\frac{dθ}{dt}$ of a body depends upon

Easy

A

Nature of surface of the body
B

Temperature of body and surroundings
C

Mass and Specific heat of the body
D

All the above

Solution

$\frac{dQ}{dt} = \frac{KA (θ - θ_{o})}{l}$
$ms \frac{dθ}{dt} = \frac{KA (θ - θ_{o})}{l}$
$\frac{dθ}{dt} = \frac{KA (θ - θ_{o})}{l}$

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