A liquid cools from 50^{circ}C to 45^{circ}C | 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 between the average temperature of the liquid and the s

Vedclass · Accepted Answer

$33.3$

Vedclass · Answer

(D) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the average temperature of the liquid and the surrounding temperature:$\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight)$For the first interval ($50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes):$\frac{50 - 45}{5} = K \left( \frac{50 + 45}{2} - 	heta_0 ight) \implies 1 = K (47.5 - 	heta_0) \quad ...(i)$For the second interval ($45^{\circ}C$ to $41.5^{\circ}C$ in $5$ minutes):$\frac{45 - 41.5}{5} = K \left( \frac{45 + 41.5}{2} - 	heta_0 ight) \implies 0.7 = K (43.25 - 	heta_0) \quad ...(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{1}{0.7} = \frac{47.5 - 	heta_0}{43.25 - 	heta_0}$$43.25 - 	heta_0 = 0.7(47.5 - 	heta_0)$$43.25 - 	heta_0 = 33.25 - 0.7	heta_0$$0.3	heta_0 = 10$$	heta_0 = \frac{10}{0.3} = 33.3^{\circ}C$

$A$ liquid cools from $50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes and from $45^{\circ}C$ to $41.5^{\circ}C$ in the next $5$ minutes. The temperature of the surrounding is......... $^{\circ}C$.

Similar Questions

$A$ bowl filled with very hot soup cools from $98^{\circ}C$ to $86^{\circ}C$ in $2$ minutes when the room temperature is $22^{\circ}C$. How many minutes will it take to cool from $75^{\circ}C$ to $69^{\circ}C$?

The rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if

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 $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5 \, s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

$A$ body takes $5 \ min$ to cool from $90^{\circ}C$ to $60^{\circ}C$. If the temperature of the surroundings is $20^{\circ}C$,how much time (in $min$) will it take to cool from $60^{\circ}C$ to $30^{\circ}C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module