Question
A liquid in a beaker has temperature $$\theta \left( t \right)$$ at time $$t$$ and $${{\theta _0}}$$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $${\log _e}\left( {\theta - {\theta _0}} \right)$$ and $$t$$ is:
A.
B.
C.
D.
Answer :
Solution :
Newton's law of cooling
$$\eqalign{ & \frac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right) \cr & \Rightarrow \,\,\frac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}} = kdt \cr} $$
Integrating
$$ \Rightarrow \,\,\log \left( {\theta - {\theta _0}} \right) = - kt + c$$
Which represents an equation of straight line.
Thus the option (A) is correct.
Newton's law of cooling
$$\eqalign{ & \frac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right) \cr & \Rightarrow \,\,\frac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}} = kdt \cr} $$
Integrating
$$ \Rightarrow \,\,\log \left( {\theta - {\theta _0}} \right) = - kt + c$$
Which represents an equation of straight line.
Thus the option (A) is correct.
