Establish a relationship between temperature difference of system and surroundings with time taken by the system to cool .

Explanation

Newton.s law of cooling
The rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings , for small temperature difference .
The rate of loss also depends upon
(i) Area of exposed surface of the body
(ii) Nature of surface of the body .
Consider a body of mass m and specific capacity s at temperature T Let `T_(0)` be the temperature of the surroundings .
Then ,according to Newton .s law of cooling , we have
Rate of loss of heat `prop` temperature difference between body and surroundings
`(dQ)/(dt)prop(T-T_(o))`
`(dQ)/(dt)=k(T-T_(o))` ...... (i)
If temperature of body decreases by dT in time dt . Then ,amount of heat lost by in time dt is given by
dQ=msdT
dividing by dt both sides , we get
`(dQ)/(dt)=ms(dT)/(dt)` ......(ii)
From (i) and (ii) `-ms(dt)/(Dt)=k(T-T_(o))rArr(dT)/(T-T_(o))=(-k)/(ms)dt`
`(dT)/(T-T_(o))=-K.dt` , where k. `=(k)/(ms)`
On integrating
`log_(e)(T-T_(o))=-k.t+C`
`T=T_(o)+Ce^(-kt)`
where `C^(n)=e^(c )`.
Equation (iii) is required relationship between temperature difference of system and surroundings with time taken by system to cool .