A hot body placed in the air is cooled according to Newton's law of cooling. The rate of decrease of the temperature being k times the temperature difference from the surroundings, starting from t = 0, In $\frac{ln (x)}{k}$ time, the body will lose half the maximum temperature it can lose. Find the value of $x$ .

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Solution

According to N.L.C.
$\Rightarrow \frac{d T}{d t} = k (T - T_{o}) \Rightarrow \int \frac{d T}{(T_{f} - T_{i})} = \int k d t \Rightarrow ln (T_{f} - T_{i}) = k t {G i v e n : T_{f} = \frac{T_{i}}{2}} \Rightarrow \frac{ln (\frac{- T_{i}}{2})}{k} = t \Rightarrow t = \frac{ln (2)}{k}$

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A hot body placed in the air is cooled according to Newton's law of cooling. The rate of decrease of the temperature being k times the temperature difference from the surroundings, starting from t = 0, In ln(x)k time, the body will lose half the maximum temperature it can lose. Find the value of x.

According to N.L.C.⇒dTdt=k(T−To)⇒∫dT(Tf−Ti)=∫kdt⇒ln(Tf−Ti)=kt{Given:Tf=Ti2}⇒ln(−Ti2)k=t⇒t=ln(2)k

According to N.L.C.
$\Rightarrow \frac{d T}{d t} = k (T - T_{o}) \Rightarrow \int \frac{d T}{(T_{f} - T_{i})} = \int k d t \Rightarrow ln (T_{f} - T_{i}) = k t {G i v e n : T_{f} = \frac{T_{i}}{2}} \Rightarrow \frac{ln (\frac{- T_{i}}{2})}{k} = t \Rightarrow t = \frac{ln (2)}{k}$