It is defined to this law, if the temperature T of the object is not very different from that of the environment T0, then rate of cooling -dT/dt is proportional to the temperature difference between them. T0 solve it let us consider that, T = T0 + ΔT
So that T4 = (T0 + ΔT)4 = To4 = (1 +ΔT/To ) ≈ To4 = (1 +4ΔT/To ) (from binomial expansion)
(T4 - ) = 4To3 (ΔT) or (T4 - To4) ∝ ΔT (as T0 = constant)
Now, we have already given that rate of cooling, (-dT/dt) ∝(T4 - To4)
and here we have define that, (T4 - To4) ∝ ΔT, if the temperature difference is very small.
Thus, rate of cooling, (-dT/dt) ∝ ΔT or -dθ/dt ∝ Δθ
as dT = dθ or ΔT = Δθ
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Variation of temperature of a body according to Newton's law: Consider a system has a temperature qi at time t = 0. It is placed in an environment whose temperature is q0. We are interested in searching the temperature of the object at time t, assuming Newton's law of cooling to hold good or by taking that the temperature difference is very small. As per that law, rate of cooling µ temperature difference
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or 
Here 
θ = θo +(θi - θo)e-αt
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From that expression we take that θ = θi at t = 0 and θ = θ0 at t = ∞, i.e. temperature of the object varies exponentially with time from qi to θ0 (<θi). The temperature versus time relation is as given in diagram.
Note: If the system cools by radiation from θ1 to θ2 in time t, then giving the approximation
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(-dθ/dt) = θ1 -θ2/t and θ = θav =(θ1 +θ2)/2
The equation (-dθ/dt) = α(θ -θo) converts into θ1 -θ2/t = α{(θ1+ θo)/2-θo}
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