Newton's Law Of Cooling - Practice Questions & MCQ

According to Newton's Law of Cooling   

$$
\frac{d \theta}{d t} \alpha\left(\theta-\theta_0\right)
$$

or we can say that $\frac{d \theta}{d t}=k\left(\theta-\theta_0\right)$
where
k is the proportionality constant

$$
R=\frac{d \theta}{d t}=\text { Rate of cooling }
$$

$\theta=$ Temperature of the body
$\theta_0=$ Temperature of the surrounding
Using the above formula we can plot various curves
1.The curve between $\log \left(\theta-\theta_0\right)$ Vs time $(t)$

As $\log _e\left(\theta-\theta_0\right)=-k t+c$

So the graph will be 

2. The curve between Temperature of body and time i.e $\theta$ Vs $t$ As $\theta-\theta_0=A e^{-k_t}$

So the graph will be 

3. The curve between rate of Cooling and body temperature I.e $R=\frac{d \theta}{d t}$ vs $\theta$

$$
R=\frac{d \theta}{d t}=K\left(\theta-\theta_0\right)=K \theta-K \theta_0
$$
 

So the graph will be 

4. The curve between the Rate of Cooling (R) and the Temperature difference between body and Surrounding 

$\begin{aligned} & \quad R=\frac{d \theta}{d t} \quad \text { Vs } \quad\left(\theta-\theta_0\right) \\ & \text {As } \quad R \propto \left(\theta-\theta_0\right)\end{aligned}$

So the graph will be