$A$ cup of tea cools from $65.5^{\circ} C$ to $62.5^{\circ} C$ in one minute in a room at $22.5^{\circ} C$. How long will it take to cool from $46.5^{\circ} C$ to $40.5^{\circ} C$ in the same room?

$A$ body cools in $7$ minutes from $60^{\circ}C$ to $40^{\circ}C$. What time (in minutes) does it take to cool from $40^{\circ}C$ to $28^{\circ}C$ if the surrounding temperature is $10^{\circ}C$? Assume Newton's Law of cooling holds.

$A$ suspended sphere has density $d$ and specific heat $s$. The radius of the sphere is $r$. The temperature difference between the sphere and the surroundings $(\Delta \theta)$ is very small. If the temperature of the surroundings is $\theta_0$,then the rate of cooling of the sphere will be .......

Assuming Newton's law of cooling to be valid,a body at temperature $50^{\circ} C$ in a surrounding of temperature $20^{\circ} C$ achieves a steady state with the help of a $100 \ W$ heater. If the same body has a temperature of $35^{\circ} C$ in the same surrounding,then the power of the heater required to maintain a steady state is ........ $W$.

$A$ 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

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. If the temperature of the surroundings is $10^{\circ} C$,then after the next $6$ minutes,its temperature will be $.........{ }^{\circ} C$.