If surrounding air is kept at 20^{circ} C and | vedclass

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

Question

(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 20^{\circ} C$.Integrating this gives $T(t) = T_s + (T_0 - T_s)e^{-

Vedclass · Accepted Answer

$54.7$

Vedclass · Answer

(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 20^{\circ} C$.Integrating this gives $T(t) = T_s + (T_0 - T_s)e^{-kt}$.Given $T_0 = 80^{\circ} C$,we have $T(t) = 20 + 60e^{-kt}$.For $t = 5$ minutes,$T = 70^{\circ} C$:$70 = 20 + 60e^{-5k} \Rightarrow 50 = 60e^{-5k} \Rightarrow e^{-5k} = \frac{5}{6}$.For $t = 15$ minutes:$T(15) = 20 + 60e^{-15k} = 20 + 60(e^{-5k})^3$.Substituting $e^{-5k} = \frac{5}{6}$:$T(15) = 20 + 60 	imes \left(\frac{5}{6}ight)^3 = 20 + 60 	imes \frac{125}{216} = 20 + \frac{125}{3.6} = 20 + 34.722 = 54.722^{\circ} C$.Rounding to one decimal place,the temperature is $54.7^{\circ} C$.

If surrounding air is kept at $20^{\circ} C$ and a body cools from $80^{\circ} C$ to $70^{\circ} C$ in $5$ minutes,then the temperature of the body after $15$ minutes will be: (in $^{\circ} C$)

Similar Questions

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200$. If $p(0) = 100$,then $p(t)$ equals:

$A$ spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop,if the constant of proportionality is $K > 0$,is

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units,find the radius of the balloon after $t$ seconds.

The principal increases continuously in a newly opened bank at the rate of $10 \%$ per year. An amount of Rs. $2000$ is deposited with this bank. How much will it become after $5$ years? $(e^{0.5} = 1.648)$

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

Vedclass Test Series

Exam Paper Generator

Online Exam Module