A liquid in a beaker has temperature $ heta(t)$ at time $t$ and $ heta_{0}$ is temperature of surroundings, then according to Newton’s law of cooling the correct graph between $\log_{e}{( heta - heta_{0})}$ and $t$ is

Question

A liquid in a beaker has temperature $	heta(t)$ at time $t$ and $	heta_{0}$ is temperature of surroundings, then according to Newton’s law of cooling the correct graph between $\log_{e}{(	heta - 	heta_{0})}$ and $t$ is

Dr. Abubacker · Accepted Answer

(a) $\frac{d	heta}{dt} = - k(	heta - 	heta_{0})$
$\int_{	heta_{0}}^{	heta}{\frac{d	heta}{	heta - 	heta_{0}} = - k\int_{0}^{t}{dt}}$
$\ln{\left( 	heta - 	heta_{0} ight) = - kt + C}$
So graph is straight line
[image:11_image67.emf]

Q. A liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_{0}$ is temperature of surroundings, then according to Newton’s law of cooling the correct graph between $\log_{e}{(\theta - \theta_{0})}$ and $t$ is

Explanation

Reviewed by Dr. Abubacker

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