In an experiment to verify Newton's law of co | 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 temperature difference $\Delta T$ at time $t$ is given by $\Delta T = \Delta T_{0} e^{-\lambda t}$,where

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) According to Newton's law of cooling,the temperature difference $\Delta T$ at time $t$ is given by $\Delta T = \Delta T_{0} e^{-\lambda t}$,where $\Delta T_{0}$ is the initial temperature difference.From the graph,at $t = 0$,$\Delta T_{0} = 60^{\circ}C$.At $t = 6 	ext{ min}$,$\Delta T = 40^{\circ}C$. Substituting these values: $40 = 60 e^{-6\lambda} \Rightarrow e^{-6\lambda} = \frac{40}{60} = \frac{2}{3} \Rightarrow 6\lambda = \ln(1.5)$.At $t = t_{2}$,$\Delta T = 20^{\circ}C$. Substituting these values: $20 = 60 e^{-t_{2}\lambda} \Rightarrow e^{-t_{2}\lambda} = \frac{20}{60} = \frac{1}{3} \Rightarrow t_{2}\lambda = \ln(3)$.Dividing the two equations: $\frac{t_{2}\lambda}{6\lambda} = \frac{\ln(3)}{\ln(1.5)} \Rightarrow \frac{t_{2}}{6} = \frac{1.0986}{0.4055} \approx 2.709$.Therefore,$t_{2} = 6 	imes 2.709 \approx 16.25 	ext{ min}$.Rounding to the nearest integer,$t_{2} \approx 16 	ext{ min}$.

Similar Questions

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

Hot water is kept in a thermally insulated closed container. It takes $T_1 \, min$ for the temperature to drop from $75^{\circ}C$ to $70^{\circ}C$,$T_2 \, min$ to drop from $70^{\circ}C$ to $65^{\circ}C$,and $T_3 \, min$ to drop from $65^{\circ}C$ to $60^{\circ}C$. Then:

$A$ cup of tea cools from $80^{\circ}C$ to $60^{\circ}C$ in $1$ minute. The ambient temperature is $30^{\circ}C$. In cooling from $60^{\circ}C$ to $50^{\circ}C$,it will take ....... $\text{sec}$.

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

If a liquid cools from $95^{\circ}C$ to $90^{\circ}C$ in $30 \, s$ and from $55^{\circ}C$ to $50^{\circ}C$ in $70 \, s$,then the temperature of the surroundings is ...... $^{\circ}C$. (in $.5$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module