Answer:
[tex]\dfrac{dT}{dt}=-0.4(T_o-84)[/tex]
Explanation:
dT = Change in temperature
dt = Time interval
k = constant of proportionality = 0.4
[tex]T_o[/tex] = Temperature of object
[tex]T_A[/tex] = Temperature of ambiance = 84°C
From Newton's law of cooling we have the expression
[tex]\dfrac{dT}{dt}=-k(T_o-T_A)[/tex]
So, the differential equation is
[tex]\mathbf{\dfrac{dT}{dt}=-0.4(T_o-84)}[/tex]