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The temperature T of a cooling object drops at a rate that is proportional to the difference T-C, where C is the constant temperature of the surrounding medium. Thus,

dT/dt=-k(t-c), (1)

where, t is the elapsed time. The function that satisfies the above equation is

T = T(t) = ae^-kt +C, (2)

where, a is a constant. It is actually the difference between the temperature of the surrounding medium and the initial temperature of the cooling object. Your mission should you choose to accept it is to Integrate equation (1) to obtain equation (2).

The temperature $T$ of a cooling object drops at a rate that is proportional to the difference $T−C$, where $C$ is the constant temperature of the surrounding medium. Thus, $dtdT =−k(T−C),$ where, $t$ is the elapsed time. The function that satisfies the above equation is $T=T(t)=ae_{−kt}+C$ where, $a$ is a constant. It is actually the difference between the temperature of the surrounding medium and the initial temperature of the cooling object. Your mission should you choose to accept it is to Integrate equation (1) to obtain equation $(2)$ Hint: If you can't figure out direct integration, you may find the derivative of equation (2) to obtain equation (1) as a clue and reverse the process.

The answer is described below properly here Solution:- This is the First part of the Answer

The temperature T of a cooling object drops at a rate that is proportional to the difference T-C, where C is the constant temperature of the surrounding medium.

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