$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

The temperature of a body $\theta$ is slightly more than the temperature of the surrounding $\theta_0$. Its rate of cooling $(R)$ versus the temperature of the body $(\theta)$ is plotted. Its shape would be:

$A$ can is taken out from a refrigerator at $0^{\circ}C$. The atmospheric temperature is $25^{\circ}C$. If $t_1$ is the time taken to heat from $0^{\circ}C$ to $5^{\circ}C$ and $t_2$ is the time taken to heat from $10^{\circ}C$ to $15^{\circ}C$,then:

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5 \, s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

$A$ liquid is filled in a vessel which is kept in a room with a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at the rate of $60 \; cal/sec$. What will be the rate of loss of heat when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$?

$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)