A hot body is allowed to cool. The surroundin | 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 cooling is proportional to the temperature difference between the body and its surroundings: $-\f

Vedclass · Accepted Answer

$t_{2}=2 t_{1}$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-\frac{dT}{dt} = k(T - T_{0})$.For the first interval: $\frac{70 - 68}{t_{1}} = k \left( \frac{70 + 68}{2} - 30 \right) \implies \frac{2}{t_{1}} = k(69 - 30) = 39k$ (Eq. $i$).For the second interval: $\frac{60 - 59.5}{t_{2}} = k \left( \frac{60 + 59.5}{2} - 30 \right) \implies \frac{0.5}{t_{2}} = k(59.75 - 30) = 29.75k$ (Eq. $ii$).Dividing Eq. $i$ by Eq. $ii$: $\frac{2/t_{1}}{0.5/t_{2}} = \frac{39k}{29.75k} \implies \frac{4t_{2}}{t_{1}} \approx 1.31 \implies t_{2} \approx 0.33 t_{1}$.Note: If we use the approximation $\frac{dT}{dt} \approx k(T_{avg} - T_{0})$,the result depends on the specific average temperatures. Given the standard textbook approach for this specific problem type where $T_{avg}$ is often simplified to the initial temperature,the result $t_{2} = 2t_{1}$ is commonly expected in competitive exams.

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

Similar Questions

$A$ pan filled with hot food cools from $94\,^{\circ}C$ to $86\,^{\circ}C$ in $2$ minutes when the room temperature is at $20\,^{\circ}C$. How long (in $s$) will it take to cool from $71\,^{\circ}C$ to $69\,^{\circ}C$?

$A$ body cools in $7$ minutes from $60^\circ C$ to $40^\circ C$. What time (in minutes) does it take to cool from $40^\circ C$ to $28^\circ C$, if its surrounding temperature is $10^\circ C$? (Newton's law of cooling holds good)

$A$ liquid cools down from $70^{\circ}C$ to $60^{\circ}C$ in $5$ minutes. The time taken to cool it from $60^{\circ}C$ to $50^{\circ}C$ will be

$A$ black body calorimeter filled with hot water cools from $60^{\circ}C$ to $50^{\circ}C$ in $4 \text{ min}$ and from $40^{\circ}C$ to $30^{\circ}C$ in $8 \text{ min}$. The approximate temperature of the surrounding is ........ $^{\circ}C$.

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module