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
θ=θ1 ,t=0
log(θ1-θ0 )=-k(0)+c
C=log(θ1-θ0 ) ------(2)
Substituting (2) in (1), we get
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
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=- k∫dt
log(θ-40) = -kt + logc
log(θ-40/c)=-kt
θ-40/c=e^ -kt
θ-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
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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