The rate at which the metal cools in moving a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$(A)$ According to Newton's law of cooling,the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_m)$,where $T_m = 290 \ K$ is the su

Vedclass · Accepted Answer

$40 \ 	ext{min}$

Vedclass · Answer

$(A)$ According to Newton's law of cooling,the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_m)$,where $T_m = 290 \ K$ is the surrounding temperature.Integrating the equation: $\int \frac{dT}{T - 290} = \int -k \ dt \Rightarrow \ln(T - 290) = -kt + C$.At $t = 0$,$T = 370 \ K$: $\ln(370 - 290) = C \Rightarrow C = \ln(80)$.So,$\ln(T - 290) = -kt + \ln(80) \Rightarrow \ln\left(\frac{T - 290}{80}ight) = -kt$.At $t = 10 \ 	ext{min}$,$T = 330 \ K$: $\ln\left(\frac{330 - 290}{80}ight) = -10k \Rightarrow \ln\left(\frac{40}{80}ight) = -10k \Rightarrow \ln\left(\frac{1}{2}ight) = -10k \Rightarrow -\ln(2) = -10k \Rightarrow k = \frac{\ln(2)}{10}$.Now,for $T = 295 \ K$: $\ln\left(\frac{295 - 290}{80}ight) = -kt \Rightarrow \ln\left(\frac{5}{80}ight) = -\left(\frac{\ln(2)}{10}ight)t$.$\ln\left(\frac{1}{16}ight) = -\frac{\ln(2)}{10}t \Rightarrow -\ln(16) = -\frac{\ln(2)}{10}t$.$-4 \ln(2) = -\frac{\ln(2)}{10}t \Rightarrow t = 40 \ 	ext{minutes}$.

Similar Questions

The solution of the differential equation $y \frac{dy}{dx} + x = k$ represents . . . . . . .

The orthogonal trajectories of the family of curves $x^{2/3} + y^{2/3} = a^{2/3}$,where $a$ is an arbitrary constant,are given by:

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module