A body cools from 70^{circ} C to 40^{circ} C | 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 directly proportional to the temperature difference between the body and the surround

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and the surroundings.$\frac{T_i - T_f}{t} = K \left( \frac{T_i + T_f}{2} - T_0 ight)$Where $T_i$ is initial temperature, $T_f$ is final temperature, $t$ is time, $T_0$ is surrounding temperature, and $K$ is a constant.For the first case: $T_i = 70^{\circ} C$, $T_f = 40^{\circ} C$, $t = 5 \ min$, $T_0 = 20^{\circ} C$.$\frac{70 - 40}{5} = K \left( \frac{70 + 40}{2} - 20 ight)$$\frac{30}{5} = K (55 - 20)$$6 = K(35) \Rightarrow K = \frac{6}{35}$For the second case: $T_i = 60^{\circ} C$, $T_f = 30^{\circ} C$, $t = ?$, $T_0 = 20^{\circ} C$.$\frac{60 - 30}{t} = \frac{6}{35} \left( \frac{60 + 30}{2} - 20 ight)$$\frac{30}{t} = \frac{6}{35} (45 - 20)$$\frac{30}{t} = \frac{6}{35} (25)$$t = \frac{30 	imes 35}{6 	imes 25} = \frac{5 	imes 35}{25} = \frac{35}{5} = 7 \ min.$

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)

Similar Questions

$A$ certain quantity of water cools from $70\,^{\circ}C$ to $60\,^{\circ}C$ in the first $10\, minutes$ and to $54\,^{\circ}C$ in the next $10\, minutes$. The temperature of the surrounding is ......... $^{\circ}C$.

$A$ body takes $5$ minutes to cool from $90^oC$ to $60^oC$. If the temperature of the surroundings is $20^oC$,the time taken by it to cool from $60^oC$ to $30^oC$ will be ...... $\min.$

Two liquids of equal volume are cooled under identical conditions from $60^{\circ}C$ to $50^{\circ}C$ in $324 \; sec$ and $810 \; sec$ respectively. If the ratio of their specific heats is $3:4$,find the ratio of their densities. (The water equivalent of the calorimeter is negligible.)

$A$ body takes $10 \, \text{minutes}$ to cool from $60 \, ^\circ\text{C}$ to $50 \, ^\circ\text{C}$. The temperature of the surroundings is constant at $25 \, ^\circ\text{C}$. Then, the temperature of the body after the next $10 \, \text{minutes}$ will be approximately ....... $^\circ\text{C}$

$A$ body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module