Question

Newton’s law of cooling states that the temperature of an object changes at a rate proportionalto the difference between the temperature of the object itself and the temperatureof its surroundings (the ambient air temperature in most cases). Suppose that the ambienttemperature is 70◦F and that the rate constant is 0.05 (min)−1.Write a differential equationfor the temperature of the object at any time. Note that the differential equation is thesame whether the temperature of the object is above or below the ambient temperature.

Answer 1

Answer:

$\frac{dT}{dt} = -0.05\;min^{-1}  (T-70\ºF)$

Step-by-step explanation:

Hi!

Lets call:

T = temprature of the object

T₀ = temperature of surroundings

t = time

The rate of change of T is its derivative with respecto to time. If T > T₀, the object looses heat, so T decreases. Then, being k > 0:

$\frac{dT}{dt} = -k(T-T_0)$

In this case T₀ = 70ºF and k = 0.05/min. Then the differential equation is:

$\frac{dT}{dt} = -0.05\;min^{-1}  (T-70\ºF)$