Question

Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.49787 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes).

Answer 1

Answer:

$T = 40 + e^{3.68t} e^{2.99}$

Explanation:

The differential equation for given  is given as

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

integrating above equation we have

ln(T-T_s) = -kt + C

At t = 0 , T(0) =  60

$ln(60- 40) = -k\times 0 + C$

2.99 = C

 At t =1 , T(1) = 40.49887

$ln(40.49787 - 40) = -k\times 1 + 2.99$

- k = -3.687

So we have

$T- 40 = e^{3.68t + 2.99}$

$T = 40 + e^{3.68t} e^{2.99}$