Answer:
[tex]T=T_o+C'e^{-kt}[/tex]
Explanation:
Newton's law of cooling
Rate of cooling
[tex]\dfrac{dT}{dt}\alpha(T-T_o)[/tex]
Where To is the surrounding temperature
T is object temperature at any time t
now by removing proportionality sign
[tex]\dfrac{dT}{dt}=-k(T-T_o)[/tex]
Now by separating variables
[tex]\dfrac{dT}{(T-T_o)}=-k\ dt[/tex]
[tex]\int \dfrac{dT}{(T-T_o)}=-\int k\ dt[/tex]
So
[tex]\ln (T-T_o)=- k\ t +C[/tex]
Where C is constant
[tex]T=T_o+C'e^{-kt}[/tex]
C' is also a constant and it can be find by using boundary conditions.