State Newton's II law of motion.

Explanation

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 decereases 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)=(msd)/(dt)" ...(3)"`
From Newton.s law of cooling `(dQ)/(dt) prop -(T-T_(s))`
`(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`
Integrating equation (5) on both sides,
`int(dT)/(T-T_(s))=-int(a)/(ms)dt `
`ln(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)`
Here `b_(2)=e^(b_(1)) = "Constant"`