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10-2.Transmission of Heat
normal
It takes $10$ minutes to cool a liquid from $61^{\circ} C$ to $59^{\circ} C$. If room temperature is $30^{\circ} C$ then time taken in cooling from $51^{\circ} C$ to $49^{\circ} C$ is .......... $min$
A
$10$
B
$11$
C
$13$
D
$15$
Solution
(d)
$\frac{ dQ }{ dt }= k \left(\theta-\theta_0\right)$
$\theta_0$ is room temperature.
$\frac{61-59}{10}= k \left(\frac{61+59}{2}-30\right) \ldots \ldots .(1)$
$\frac{51-49}{ t }= k \left(\frac{51+49}{2}-30\right) \ldots \ldots \ldots .(2)$
Dividing equation $1$ and $2$ we get $t =15\, min$
Standard 11
Physics
