The initial temperature of a body is 80^{circ | 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 change of temperature is given by $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	he

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of change of temperature is given by $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$,where $	heta_0$ is the surrounding temperature.For the first interval ($t = 5$ minutes):$\frac{80 - 64}{5} = K \left[ \frac{80 + 64}{2} - 	heta_0 ight] \implies 3.2 = K(72 - 	heta_0)$ ... $(i)$For the second interval ($t = 10$ minutes,from $64^{\circ}C$ to $52^{\circ}C$):$\frac{64 - 52}{10} = K \left[ \frac{64 + 52}{2} - 	heta_0 ight] \implies 1.2 = K(58 - 	heta_0)$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{3.2}{1.2} = \frac{72 - 	heta_0}{58 - 	heta_0} \implies \frac{8}{3} = \frac{72 - 	heta_0}{58 - 	heta_0}$$8(58 - 	heta_0) = 3(72 - 	heta_0) \implies 464 - 8	heta_0 = 216 - 3	heta_0$$5	heta_0 = 248 \implies 	heta_0 = 49.6^{\circ}C$. Given the options,the closest value is $49^{\circ}C$.

The initial temperature of a body is $80^{\circ}C$. If its temperature falls to $64^{\circ}C$ in $5$ minutes and in $10$ minutes to $52^{\circ}C$,then the temperature of the surrounding will be ...... $^{\circ}C$.

Similar Questions

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

Newton's law of cooling is based upon which of the following principles?

The temperature of a liquid drops from $365 \ K$ to $361 \ K$ in $2$ minutes. Find the time during which the temperature of the liquid drops from $344 \ K$ to $342 \ K$. The temperature of the room is $293 \ K$.

$A$ body cools from a temperature $60\,^{\circ}\text{C}$ to $40\,^{\circ}\text{C}$ in $10$ minutes. The room temperature is $20\,^{\circ}\text{C}$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be......... $^{\circ}\text{C}$.

$A$ solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then,choose the $CORRECT$ statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module