A cup of coffee cools from 90^{circ} C to 80^ | vedclass

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

Question

(A) Using the average form of Newton's law of cooling: $\frac{dT}{dt} = k(T_{avg} - T_{room})$.For the first interval ($90^{\circ} C$ to $80^{\circ} C

Vedclass · Accepted Answer

$\frac{13}{5} t$

Vedclass · Answer

(A) Using the average form of Newton's law of cooling: $\frac{dT}{dt} = k(T_{avg} - T_{room})$.For the first interval ($90^{\circ} C$ to $80^{\circ} C$):$\frac{90-80}{t} = k\left(\frac{90+80}{2} - 20ight) \implies \frac{10}{t} = k(85 - 20) = 65k \dots (i)$For the second interval ($80^{\circ} C$ to $60^{\circ} C$):$\frac{80-60}{t'} = k\left(\frac{80+60}{2} - 20ight) \implies \frac{20}{t'} = k(70 - 20) = 50k \dots (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{10/t}{20/t'} = \frac{65k}{50k}$$\frac{10}{t} 	imes \frac{t'}{20} = \frac{65}{50}$$\frac{t'}{2t} = \frac{13}{10}$$t' = \frac{13}{10} 	imes 2t = \frac{13}{5} t$.

$A$ cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes when the room temperature is $20^{\circ} C$. The time taken by the similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at the same room temperature is $:$

Similar Questions

$A$ bucket full of hot water is kept in a room. If it cools from $75^{\circ} C$ to $70^{\circ} C$ in $t_1$ minutes,from $70^{\circ} C$ to $65^{\circ} C$ in $t_2$ minutes and $65^{\circ} C$ to $60^{\circ} C$ in $t_3$ minutes,then

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.

$A$ sphere and a cube made of the same material and having the same surface area are heated to the same temperature and kept in the same surroundings. The ratio of their initial rates of cooling will be

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and assuming Newton's law of cooling to hold good,the temperature of the body at the end of the next $10$ minutes will be ......... $^{\circ}C$.

$A$ body cools from a temperature of $60^{\circ} C$ to $50^{\circ} C$ in $10 \text{ minutes}$ and $50^{\circ} C$ to $40^{\circ} C$ in $15 \text{ minutes}$. The time taken in minutes for the body to cool from $40^{\circ} C$ to $30^{\circ} C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module