If a piece of metal is heated to temperature `theta` and the allowed to cool in a room which is at temperature `theta_0`, the graph between the temperature T of the metal and time t will be closet to

Options

A.
B.
C.
D.

Answer

B

Explanation

According to Newton’s law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of the body and surroundings.
`(dQ)/(dt)=k(T_(2)-T_(1))`... (i)
and `dQ=msdT_(2)`
`therefore(dQ)/(dt)=(msdT_(2))/(dt)` ...(ii)
From Eqs. (i) and (ii) , we get `-(msdT_(2))/(dt)=k(T_(2)-T_(1))`
`rArr(dT_(2))/(T_(2)-T_(1))=-(k)/(ms)dt=k dt[thereforeK=(k)/(ms)]`
On integrating we, get `log_(e)(T_(2)-T_(1))=-Kt+C or T_(2)=T_(1)+C'e^(-kt)`