Question
Instantaneous temperature difference between cooling body and the surroundings obeying Newton's law of cooling is $\theta$. Which of the following represents the variation of $\ln \theta$ with time $t ?$
✓
Answer
$\ln \left(\frac{T_f-T_0}{T-T_0}\right)=K t$
If $\theta$ is the instantaneous temperature than
$\ln \left(\frac{\theta_i-\theta_0}{\theta-\theta_0}\right)=K t$
$\ln \left(\theta_i-\theta_0\right)-\ln \left(\theta-\theta_0\right)=K T \left\{\begin{array}{l}\theta_i \longrightarrow \text { initial temperature } \\ \theta_0 \longrightarrow \text { temperature of surrounding }\end{array}\right\}$
$\ln \left(\theta-\theta_0\right)=-K t+\ln \left(\theta_i-\theta_0\right)$
Comparing to
$y=m x+C$
If $\theta$ is the instantaneous temperature than
$\ln \left(\frac{\theta_i-\theta_0}{\theta-\theta_0}\right)=K t$
$\ln \left(\theta_i-\theta_0\right)-\ln \left(\theta-\theta_0\right)=K T \left\{\begin{array}{l}\theta_i \longrightarrow \text { initial temperature } \\ \theta_0 \longrightarrow \text { temperature of surrounding }\end{array}\right\}$
$\ln \left(\theta-\theta_0\right)=-K t+\ln \left(\theta_i-\theta_0\right)$
Comparing to
$y=m x+C$
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