Q. It is known for a certain liquid that it takes $70s$ to cool from $55^{o}C$ to $50^{o}C$ and $30s$ to cool from $95^{o}C$ to $90^{o}C$ . Figure out the ambient temperature of the room.
NTA AbhyasNTA Abhyas 2020
Solution:
If the temperature of body decreases from $T_{1}$ and $T_{2}$ and temperature of surroundings is $T_{o}$ then according to Newton's law of cooling :
Average excess of temperature = $\left[\frac{T_{1} + T_{2}}{2} - T_{o}\right]$
$\Rightarrow \left[\frac{T_{1} - T_{2}}{t}\right]=K\left[\frac{T_{1} + T_{2}}{2} - T_{o}\right]$
$\Rightarrow \frac{95 - 90}{30}=k\left[\frac{95 + 90}{2} - T_{0}\right]$
$\Rightarrow \frac{55 - 50}{70}=k\left[\frac{55 + 50}{2} - T_{0}\right]$
On dividing both the equations ,we get
$T_{0}=22.5^\circ C$
Average excess of temperature = $\left[\frac{T_{1} + T_{2}}{2} - T_{o}\right]$
$\Rightarrow \left[\frac{T_{1} - T_{2}}{t}\right]=K\left[\frac{T_{1} + T_{2}}{2} - T_{o}\right]$
$\Rightarrow \frac{95 - 90}{30}=k\left[\frac{95 + 90}{2} - T_{0}\right]$
$\Rightarrow \frac{55 - 50}{70}=k\left[\frac{55 + 50}{2} - T_{0}\right]$
On dividing both the equations ,we get
$T_{0}=22.5^\circ C$