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  • If a piece of metal is heated to temperature   and then allowed to cool in a room which is at temperature <img alt=

If a piece of metal is heated to temperature  \theta and then allowed to cool in a room which is at temperature \theta_{0}, the graph between the temperature T of the metal and time t will be closed to :

 

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

According to Newton's law of cooling, the rate of loss of heat is directly proportional to the difference oi temperature of the body and surroundings.

\frac{d Q}{d t}=k\left(T_{2}-T_{1}\right)$--------(1)\\ and $\quad d Q=m s d T$\\ $\therefore$ $\frac{d Q}{d t}=\frac{m s d T}{d t}$ $\ldots(\mathrm{ii})$

From Eqs. (i) and (ii), we get $$ \begin{aligned} -\frac{m s d T}{d t} &=k\left(T_{2}-T_{1}\right) \\ \frac{d T}{T_{2}-T_{1}} &=-\frac{k}{m s} d t=-K d t \quad\left[\therefore K=\frac{k}{m s}\right] \end{aligned} $$

\\ On \ integrating\ we\ get\ \\ $\log _{e}\left(T_{2}-T_{1}\right)=-K t+C$ \\ Or\ $T_{2}=T_{1}+C^{\prime} e^{-K t} \quad\left[\text { where }, C^{\prime}=e^{C}\right]$\\ This\ variation\ is\ correctly\ represented \ in \ option $(\mathrm{d})$

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