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Newtons law of cooling Btech M1

NEWTON'S* LAW OF COOLING


Statement: The rate of change of the temperature of a body is proportIonal to the difference between temperature of the body and temperature of the surrounding medium.

=> rate of change of the temperature of a body temperature of the body - temperature of the surrounding medium.

Let θ be the temperature of the body at time (t) and
θ0 be the temperature of its surrounding medium (usually air).

By the Newton's law of cooling,

we have,

dθ/dt   (θ-θ0 ) 

dθ/dt =-k (θ-θ0 )         (where k is a positive constant)    

1/(θ-θ0 )dθ=- k dt (Variables Separable)


Integrating, we get


          ∫ dθ/(θ-θ0 )=- k dt
log(θ-θ0 )=-kt+c                                         .......  (1)                                 

If initially 0 = 0, is the temperature of the body at time t = 0, then (1) gives

             Î¸=θ1   ,t=0

        log(θ1-θ0 )=-k(0)+c

                      C=log(θ1-θ0 )                                        ------(2)

Substituting (2) in (1), we get

log(θ-θ0) = -kt +log (θ1-θ0)

log(θ-θ0)-log (θ1-θ0)= -kt


                 log(θ-θ0)/(θ1-θ0)= -kt

                      
(θ-θ0)/(θ1-θ0)=e^ -kt

(θ-θ0)=(θ1-θ0)e^ -kt


                            θ=θ0+(θ1-θ0)e^ -kt .......(3)

.'.eq(3)which gives the temperature of the body at any time t.

Example : A body is originally at 80°C and cools down to 60°C in 20 minutes. If the temperature of the air is 40°C, find the temperature of the body after 40 minutes.

Solution :
Let O be the temperature of the body at time t.

By Newton's law of cooling, we have


dθ/dt =-k (θ-θ0 )  [ where θ0, is the temperature of the air]


dθ/dt =-k (θ-40 ) [ θ0 = 40 (given)]

dθ/(θ-40 )dθ=- k dt -------(1)


Integrating on both sides, we get


dθ/(θ-40 )dθ=- kdt

log(θ-40) = -kt + logc

log(θ-40/c)=-kt

                  θ-40/c=e^ -kt

                   Î¸-40=ce^ -kt -------(2)

When t=0,0 = 80°C and when t = 20,0 = 60°C

Substituting in (2), we get c = 40 and 20 =ce^ -20k

20/40=ce^ -20k

k=1/20log2


                      θ-40=40e^ -(1/20log2)t

When t = 40, θ=?

θ-40=40e^ -(1/20log2)40 -------(3)

θ =40+40e^-2log2

i.e., θ = 40 + 40 e^log (1/4)

θ = 40+40.1/4

θ=50°C

.'. Hence the temperature of the body after 40 minutes is 50°C


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