−du/dt=k(u−s)5/4
is the differential equation for law of natural cooling.
Now, as degree of the differential equation is 1, thus it is linear.
∫u0udu/(u−s)5/4=−∫0tkdt
(u−s)−1/4−(u0−s)−1/4=−4kt
u=s+((u0−s)−1/4−4kt)−4
is the solution for the above differential equation.
Comments
Dear Raj, please use the panel for submitting new questions.
According to Newton’s law of cooling, the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is C o 290 and the substance cools from C o 370 to C o 330 in 10 minutes, find when the temperature will be C o 295