Question

Newton’s laws of cooling proposes that the rate of change of temperature is proportional to the temperature difference to the ambient (room) temperature. And can be modelled using the equation: 𝑑𝑇𝑑𝑡=−𝑘(𝑇−𝑇𝑎) This can also be written as: 𝑑𝑇𝑇−𝑇𝑎=−𝑘 𝑑𝑡

Where: 𝑇=𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑇𝑎=𝐴𝑚𝑏𝑖𝑒𝑛𝑡 (𝑟𝑜𝑜𝑚) 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑘=𝐴 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 a) Integrate both sides of the equation and show that the temperature difference is given by: (𝑇−𝑇𝑎)=𝐶𝑜𝑒−𝑘𝑡

Answer 1

GIVEN:

T= temperature of material

T_a= temperature of room or surrounding

k=cooling constant

solution:

𝑑𝑇/𝑑𝑡=−𝑘(𝑇−𝑇𝑎),

𝑑𝑇/(𝑇−𝑇𝑎)=−𝑘𝑑𝑡.

We have to integrate both sides:

$\int \cfrac{dT}{T-T_a} = -\int kdt$

$ln\,|T - T_a| = -kt + ln\,C_0$

$T−T a ​ =e^{ −kt+lnC_ 0} ​$

taking anti (ln)

$T−T a ​ =e^{ lnC 0} ​ e ^{ −kt}$

hence by base change property,

$\boxed{T - T_a = C_0\,e^{-kt}}$

hence we proved.