In 5, text{minutes},a body cools from 75^{cir | vedclass

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

Question

(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T_{avg} - T_s)$.For the first $5\, 	ext{minutes}$:$\frac

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T_{avg} - T_s)$.For the first $5\, 	ext{minutes}$:$\frac{75 - 65}{5} = k \left( \frac{75 + 65}{2} - 25 ight)$$\frac{10}{5} = k (70 - 25)$$2 = k(45) \implies k = \frac{2}{45} \, 	ext{min}^{-1}$.For the next $5\, 	ext{minutes}$,let the final temperature be $T$:$\frac{65 - T}{5} = k \left( \frac{65 + T}{2} - 25 ight)$$\frac{65 - T}{5} = \frac{2}{45} \left( \frac{65 + T - 50}{2} ight)$$\frac{65 - T}{5} = \frac{1}{45} (T + 15)$$9(65 - T) = T + 15$$585 - 9T = T + 15$$10T = 570$$T = 57^{\circ} 	ext{C}$.

In $5\, \text{minutes}$,a body cools from $75^{\circ} \text{C}$ to $65^{\circ} \text{C}$ at a room temperature of $25^{\circ} \text{C}$. The temperature of the body at the end of the next $5\, \text{minutes}$ is $......\,^{\circ} \text{C}$.

Similar Questions

Hot water cools from $60\,^oC$ to $50\,^oC$ in the first $10\,minutes$ and to $42\,^oC$ in the next $10\,minutes.$ The temperature of the surroundings is ...... $^oC$

Two spheres have a radius ratio of $1:2$ and a density ratio of $2:1$. They have the same specific heat. If they are heated to the same temperature and placed in the same environment,what is the ratio of their rate of cooling?

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5 \text{ min}$. Calculate the time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$. The temperature of the surrounding is $20^{\circ} C$. (in $\text{ min}$)

The temperature of hot water falls from $60^oC$ to $50^oC$ in $10 \, min$. It takes another $10 \, min$ to fall from $50^oC$ to $42^oC$. Find the temperature of the surroundings in $^oC$.

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\,minutes$ and to $42\,^{\circ}C$ in the next $10\,minutes$. The temperature of the surroundings is ......... $^{\circ}C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module