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Question 86 of 126

Q.The temperature T of a cooling object drops at a rate proportional to the difference (T−S)(T-S), where S is constant temperature of surrounding medium. If initially T=150°CT=150°C, find the temperature of the cooling object at any time 'tt'.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2016Subjective· 6mImportance★★★★★
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Set up the differential equation from Newton's law of cooling, separate variables, integrate, and apply the initial condition T(0)=150T(0)=150.

1. Set up the differential equation. The rate of cooling is proportional to (T−S)(T-S):

dTdt=−k(T−S),k>0 constant\frac{dT}{dt}=-k(T-S),\qquad k>0 \text{ constant}

2. Separate variables.

dTT−S=−k dt\frac{dT}{T-S}=-k\,dt

3. Integrate both sides.

∫dTT−S=∫−k dt\int\frac{dT}{T-S}=\int -k\,dt

ln⁡∣T−S∣=−kt+C1\ln|T-S|=-kt+C_1

4. Exponentiate.

T−S=eC1e−kt=Ae−kt,A=eC1T-S=e^{C_1}e^{-kt}=Ae^{-kt},\qquad A=e^{C_1}

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