For a system where Newton's law of cooling is | 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,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $(T - T_s) = \Delta T$ is the temperature

Vedclass · Accepted Answer

$\frac{\ln(2) \cdot \Delta T_0}{R}$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $(T - T_s) = \Delta T$ is the temperature difference.Given the initial rate of cooling is $R = k \Delta T_0$,we can find the rate constant $k = \frac{R}{\Delta T_0}$.The temperature difference at any time $t$ is given by $\Delta T(t) = \Delta T_0 e^{-kt}$.We want to find the time $t$ when $\Delta T(t) = \frac{\Delta T_0}{2}$.Substituting this into the equation: $\frac{\Delta T_0}{2} = \Delta T_0 e^{-kt}$.This simplifies to $\frac{1}{2} = e^{-kt}$,or $2 = e^{kt}$.Taking the natural logarithm on both sides: $\ln(2) = kt$.Therefore,$t = \frac{\ln(2)}{k}$.Substituting $k = \frac{R}{\Delta T_0}$,we get $t = \frac{\ln(2) \Delta T_0}{R}$.

For a system where Newton's law of cooling is applicable,the initial rate of cooling is $R \ ^\circ C/sec$. Find the time when the temperature difference $\Delta T_0$ (initial temperature difference) is reduced to half.

Similar Questions

Which of the following statements is true/correct?

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

Two spheres of the same metal have radii $r$ and $2r$. They are heated to the same temperature and placed in the same environment. The ratio of their rates of cooling will be:

According to Newton's law of cooling,the rate of cooling of a body is proportional to $(\Delta \theta )^n$,where $\Delta \theta$ is the difference of the temperature of the body and the surroundings,and $n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module