Answer to Question #175165 in Calculus for mia

Question #175165

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: 𝑑𝑇𝑑𝑑=βˆ’π‘˜(π‘‡βˆ’π‘‡π‘Ž) This can also be written as: π‘‘π‘‡π‘‡βˆ’π‘‡π‘Ž=βˆ’π‘˜ 𝑑𝑑

Where: 𝑇=π‘‡π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘šπ‘Žπ‘‘π‘’π‘Ÿπ‘–π‘Žπ‘™ π‘‡π‘Ž=π΄π‘šπ‘π‘–π‘’π‘›π‘‘ (π‘Ÿπ‘œπ‘œπ‘š) π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘˜=𝐴 π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ a) Integrate both sides of the equation and show that the temperature difference is given by: (π‘‡βˆ’π‘‡π‘Ž)=πΆπ‘œπ‘’βˆ’π‘˜π‘‘


1
Expert's answer
2021-03-25T13:04:13-0400

GIVEN:

T= temperature of material

Ta= temperature of room or surrounding

k=cooling constant


solution:

𝑑𝑇/𝑑𝑑=βˆ’π‘˜(π‘‡βˆ’π‘‡π‘Ž),

𝑑𝑇/(π‘‡βˆ’π‘‡π‘Ž)=βˆ’π‘˜π‘‘π‘‘.


We have to integrate both sides:


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


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


"T\u2212T \na\n\u200b\t\n =e^{ \n\u2212kt+lnC_ \n0}\n\u200b"


taking anti (ln)


"T\u2212T \na\n\u200b\t\n =e^{ \nlnC \n0}\n\u200b\t\n \n e ^{\n\u2212kt}"


hence by base change property,


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



hence we proved.


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