The temperature of a body falls from 50^oC to | vedclass

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

Question

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and the surroundings: $

Vedclass · Accepted Answer

$33.3$

Vedclass · Answer

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and the surroundings: $\frac{dT}{dt} = K(T_{avg} - T_s)$.In the first $10$ minutes,the temperature falls from $50^oC$ to $40^oC$ with surroundings at $20^oC$:$\frac{50 - 40}{10} = K \left( \frac{50 + 40}{2} - 20 ight) \implies 1 = K(45 - 20) \implies 1 = 25K \implies K = \frac{1}{25} \dots (i)$In the next $10$ minutes,let the final temperature be $	heta_2$:$\frac{40 - 	heta_2}{10} = K \left( \frac{40 + 	heta_2}{2} - 20 ight) \dots (ii)$Substituting $K = \frac{1}{25}$ into equation $(ii)$:$\frac{40 - 	heta_2}{10} = \frac{1}{25} \left( 20 + \frac{	heta_2}{2} - 20 ight) \implies \frac{40 - 	heta_2}{10} = \frac{	heta_2}{50}$$5(40 - 	heta_2) = 	heta_2 \implies 200 - 5	heta_2 = 	heta_2 \implies 6	heta_2 = 200$$	heta_2 = \frac{200}{6} = 33.33^oC$.

The temperature of a body falls from $50^oC$ to $40^oC$ in $10$ minutes. If the temperature of the surroundings is $20^oC$,then the temperature of the body after another $10$ minutes will be ........ $^oC$.

Similar Questions

$A$ can is taken out from a refrigerator at $0^{\circ}C$. The atmospheric temperature is $25^{\circ}C$. If $t_1$ is the time taken to heat from $0^{\circ}C$ to $5^{\circ}C$ and $t_2$ is the time taken to heat from $10^{\circ}C$ to $15^{\circ}C$,then:

$A$ container contains hot water at $100^{\circ}C$. If in time $T_1$ the temperature falls to $80^{\circ}C$ and in time $T_2$ the temperature falls to $60^{\circ}C$ from $80^{\circ}C$,then:

$A$ metal sphere cools at a rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. When the temperature of the sphere is $40^{\circ} C$,its rate of cooling is $0.3^{\circ} C / min$. The temperature of the surrounding $\left(\theta_0\right)$ is (in $^{\circ} C$)

There are two spherical balls $A$ and $B$ of the same material with the 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

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and Newton's law of cooling is applicable,what will be the temperature of the body after another $10$ minutes in $^{\circ}C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module