Question 208: Properties of Matter

Question

A liquid in a beaker has temperature $$	heta \left( t ight)$$ at time $$t$$ and $${	heta _0}$$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $${\log _e}\left( {	heta  - {	heta _0}} ight)$$ and $$t$$ is :

Accepted Answer

Newton's law of cooling
$${{d	heta } \over {dt}} =  - k\left( {	heta  - {	heta _0}} ight)$$
$$ \Rightarrow {{d	heta } \over {\left( {	heta  - {	heta _0}} ight)}} =  - kdt$$
Intergrating
$$ \Rightarrow \log \left( {	heta  - {	heta _0}} ight) =  - kt + c$$
Which represents an equation of straight line.
Thus the option $$(a)$$ is correct.

Answer

Answer

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Answer

Properties of Matter

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 :

Choose the correct answer: