A body cools in 7 minutes from 60^{circ},C to | 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 proportional to the difference in temperature between the body and its surroundings:$\

Vedclass · Accepted Answer

$34^{\circ}\,C$

Vedclass · Answer

(D) According to Newton's law of cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings:$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s ight)$Given:Initial temperature $T_1 = 60^{\circ}\,C$,final temperature $T_2 = 40^{\circ}\,C$,time $t = 7$ minutes,and surrounding temperature $T_s = 10^{\circ}\,C$.Substituting these values into the formula:$\frac{60 - 40}{7} = K \left( \frac{60 + 40}{2} - 10 ight)$$\frac{20}{7} = K(50 - 10) = 40K$$K = \frac{20}{7 	imes 40} = \frac{1}{14} \ldots (i)$Now,for the next $7$ minutes,let the final temperature be $T$:$\frac{40 - T}{7} = K \left( \frac{40 + T}{2} - 10 ight)$Substitute $K = \frac{1}{14}$:$\frac{40 - T}{7} = \frac{1}{14} \left( \frac{40 + T - 20}{2} ight)$$\frac{40 - T}{7} = \frac{1}{14} \left( \frac{20 + T}{2} ight)$$2(40 - T) = \frac{20 + T}{4}$$8(40 - T) = 20 + T$$320 - 8T = 20 + T$$9T = 300$$T = \frac{300}{9} = 33.33^{\circ}\,C \approx 34^{\circ}\,C$ (rounding to the nearest option provided).

$A$ body cools in $7$ minutes from $60^{\circ}\,C$ to $40^{\circ}\,C$. The temperature of the surrounding is $10^{\circ}\,C$. The temperature of the body after the next $7$ minutes will be

Similar Questions

$A$ 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 $^{\circ} C$)

$A$ 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 $^{\circ}C$?

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. If the temperature of the surroundings is $10^{\circ} C$,then after the next $6$ minutes,its temperature will be $.........{ }^{\circ} C$.

$A$ 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:

$A$ body takes $10 \ minutes$ to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the same body after the next $10 \ minutes$ will be (if the temperature of the surroundings is $25^{\circ}C$): (in $^{\circ}C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module