temperature of a body theta is slightly more than the the temperature of surrouding theta'. Its rate of cooling R versus theta is plotted. draw the graph.

Open in App

Solution

Dear Student,

According to newton law of cooling rate of cooling is depend on the temperature difference between the surrounding and the body.
$- \frac{d θ}{d t} = k (θ - θ_{s}) θ_{s} : i t i s t h e t e m p e r a t u r e o f s u r r o u n d i n g . k i s a c o n s \tan t, - \frac{d θ}{d t} i s t h e r a t e o f c o o l i n g . a n d θ i s t h e t e m p e r a t u r e o f b o d y . R = k (θ - θ_{s}) t h i s i s t h e s i m p l e l i n e a r e q n i f w e p l o t t h e g r a p h w i t h θ t h e n w e g o t s t r a i g h t l i n e w i t h s l o p e k a n d i n t e r c e p t a t x a x i s a t θ = θ_{s} . R e g a r d s$

Suggest Corrections

Similar questions

Temperature of a body $θ$ is slightly more than the temperature of the surrounding $θ_{0}$ . Its rate of cooling $(R)$ versus temperature of body $(θ)$ is plotted. Its shape would be

Temperature of a body is slightly more than the temperature of the surrounding ₀. Its rate of cooling (R) versus temperature of body() is plotted. Its shape would be