A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge(θ-θ0) and t is :

Question

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge(θ-θ0) and t is : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(dθ/dt)=-k(θ-θ0) ∫ limitsθθ0(dθ/θ-θ0)=-k∫ limitst0dt In (θ-θ0) = -kt + C So graph is straight line.

Q. A liquid in a beaker has temperature $θ (t)$ at time $t$ and $θ_{0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $l o g_{e} (θ - θ_{0})$ and $t$ is :

$\frac{d θ}{d t} = - k (θ - θ_{0})$
$θ_{0} \int θ$ $\frac{d θ}{θ - θ _{0}} = - k 0 \int t d t$
$I n (θ - θ_{0}) = - k t + C$
So graph is straight line.

Q. A liquid in a beaker has temperature θ(t) at time t and θ0​ is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge​(θ−θ0​) and t is :

dtdθ​=−k(θ−θ0​) θ0​∫θ​θ−θ0​dθ​=−k0∫t​dt In(θ−θ0​)=−kt+C So graph is straight line.

Q. A liquid in a beaker has temperature $θ (t)$ at time $t$ and $θ_{0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $l o g_{e} (θ - θ_{0})$ and $t$ is :

$\frac{d θ}{d t} = - k (θ - θ_{0})$
$θ_{0} \int θ$ $\frac{d θ}{θ - θ _{0}} = - k 0 \int t d t$
$I n (θ - θ_{0}) = - k t + C$
So graph is straight line.