It takes 10 minutes to cool a liquid from 61^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.For the first case: $\frac{61 - 59}{1

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.For the first case: $\frac{61 - 59}{10} = K\left( \frac{61 + 59}{2} - 30 ight)$.$\frac{2}{10} = K(60 - 30) \implies 0.2 = 30K \implies K = \frac{0.2}{30} = \frac{1}{150}$.For the second case,let the time be $t$ minutes: $\frac{51 - 49}{t} = K\left( \frac{51 + 49}{2} - 30 ight)$.$\frac{2}{t} = K(50 - 30) \implies \frac{2}{t} = 20K$.Substituting $K = \frac{1}{150}$: $\frac{2}{t} = 20 	imes \frac{1}{150} = \frac{2}{15}$.Therefore,$t = 15 	ext{ min}$.

It takes $10$ minutes to cool a liquid from $61^oC$ to $59^oC$. If room temperature is $30^oC$,then the time taken in cooling from $51^oC$ to $49^oC$ is ....... $\min$.

Similar Questions

In which of the following reactions will cyanide be obtained as a major product?

Which one of the following molecular hydrides acts as a Lewis acid?

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4 v^2 = 25 - x^2$,then the time period is

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module