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Explain in detail newton's law of coolin...

Explain in detail newton's law of cooling .

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Newton.s law of cooling: Newton.s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
`(dQ)/(dt) prop (T-T_(s))` ...(1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
`T_(s)= "Temperature of the surrounding"`
From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T . Let `T_(s)` be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT ..(2)
Dividing both sides of equation (2) by dt
`(dQ)/(dt)= (msdT)/(dt)` ...(3)
From Newton.s law of cooling
`(dQ)/(dt) prop-(T-T_(s))`
`(dQ)/(dt)=-a(T-T_(s))` ....(4)
where a is some positive constant.
From equation (3) and (4)
`-a(T-T_(s))=ms(dT)/(dt)`
`(dT)/(T-T_(s))=-(a)/(ms)dt` ...(5)
Integrating equstion (5) on both sides,
`int_(0)^(infty)(dT)/(T-T_(s))=-int_(0)^(1)(a)/(ms)dt`
`"In"(T-T_(s))=-(a)/(ms)t+b_(1)`
Where `b_(1)` is the constant of integration. Taking exponential both sides, we get
`T=T_(s)+b_(2)e^((-a)/(ms)t)` ...(6)
Here `b_(2)=e^(b_(1))="Constant"`
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