$A$ hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses $60 \%$ and $80 \%$ of the maximum heat it can lose in time $t_1$ and $t_2$ respectively. The ratio $t_2 / t_1$ will be

An ordinary body cools from $4 \theta$ to $3 \theta$ in $t$ minutes. The temperature of that body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature is $\theta$)

State Newton's law of cooling and derive its mathematical equation.

Hot water cools from $60^oC$ to $50^oC$ in the first $10$ minutes and to $42^oC$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$.

$A$ container with $1 \ kg$ of water is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received from sunlight is $700 \ W \ m^{-2}$ and it is absorbed by the water over an effective area of $0.05 \ m^2$. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $^{\circ}C$) in the temperature of water and the surroundings after a long time will be. . . . . . . . (Ignore the effect of the container, and take the constant for Newton's law of cooling $k = 0.001 \ s^{-1}$, Heat capacity of water $s = 4200 \ J \ kg^{-1} \ K^{-1}$)

$A$ cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes when the room temperature is $20^{\circ} C$. The time taken by the similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at the same room temperature is $:$