A body cools from {50.0^o}C to {49.9^o}C in 5s . How long will it take to cool from {40.0^o}C to {39

English

Gujarati

Hindi

10-2.Transmission of Heat

medium

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 ....... $\sec$

$2.5 $

$10 $

$20$

$5$

Solution

(b) $\frac{{50 – 49.9}}{5} = K\,\left( {\frac{{50 + 49.9}}{2} – 30} \right)$…..$(i)$
$\frac{{40 – 39.9}}{t} = K\,\left[ {\frac{{40 + 39.9}}{2} – 30} \right]$…..$(ii)$
from equations $(i)$ and $(ii)$

we get $t ≈ 10 sec$ .

Standard 11

Physics

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

medium

There are two spherical balls $A$ and $B$ of the same material with same surface, but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool, then

medium

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.

hard

A body cools from $60\,^oC$ to $50\,^oC$ in $10\,minutes$ . If the room temperature is $25\,^oC$ and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next $10\,minutes$ will be ……… $^oC$

hard

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.

medium

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 ....... $\sec$

Solution

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

There are two spherical balls $A$ and $B$ of the same material with same surface, but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool, then

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 body cools from $60\,^oC$ to $50\,^oC$ in $10\,minutes$ . If the room temperature is $25\,^oC$ and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next $10\,minutes$ will be ……… $^oC$

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.

Start a Free Trial Now