For a system with newtons law of cooling applicable the initial rate of cooling is $R^0\ C/sec$ find the time when temperature diff. $\Delta T_0 =$ initial temperature difference, is reduced to half.
For a system with newtons law of cooling applicable the initial rate of cooling is $R^0\ C/sec$ find the time when temperature diff. $\Delta T_0 =$ initial temperature difference, is reduced to half.
- A
$\frac{{\Delta {T_0}}}{{2R}}$
- B
$\frac{{2\Delta {T_0}}}{{R}}$
- C
$\frac{{\ln (2).\Delta {T_0}}}{R}$
- D
$\frac{{\Delta {T_0}}}{{\ln (2)R}}$
Similar Questions
A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature ${120^o}C$, then
A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature ${120^o}C$, then
A body cools from ${50.0^o}C$ to ${49.9^o}C$ in $5\;s$. How long will it take to cool from ${40.0^o}C$ to ${39.9^o}C$? Assume the temperature of surroundings to be ${30.0^o}C$ and Newton's law of cooling to be valid ....... $\sec$
A body cools from ${50.0^o}C$ to ${49.9^o}C$ in $5\;s$. How long will it take to cool from ${40.0^o}C$ to ${39.9^o}C$? Assume the temperature of surroundings to be ${30.0^o}C$ and Newton's law of cooling to be valid ....... $\sec$
A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in $\mathrm{t}$ minutes, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by a similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at a room temperature same at $20^{\circ} \mathrm{C}$ is :
A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in $\mathrm{t}$ minutes, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by a similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at a room temperature same at $20^{\circ} \mathrm{C}$ is :
- [NEET 2021]
If a liquid takes $30 \;sec$ in cooling from $80^{\circ} C$ to $70^{\circ} C$ and $70 \;sec$ in cooling from $60^{\circ} C$ to $50^{\circ} C$, then find the room temperature.
A cup of tea cools from $80\,^oC$ to $60\,^oC$ in one minute. The ambient temperature is $30\,^oC$. In cooling from $60\,^oC$ to $50\,^oC$, it will take ....... $\sec$
A cup of tea cools from $80\,^oC$ to $60\,^oC$ in one minute. The ambient temperature is $30\,^oC$. In cooling from $60\,^oC$ to $50\,^oC$, it will take ....... $\sec$