A beaker full of hot water is kept in a room. | 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 directly proportional to the temperature difference between the body and its surroundi

Vedclass · Accepted Answer

$t_1 < t_2 < t_3$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundings.Mathematically,$\frac{d heta}{dt} = k \left( \frac{ heta_1 + heta_2}{2} - heta_0 ight)$,where $ heta_0$ is the room temperature.Since the temperature difference decreases as the water cools,the rate of cooling decreases.Therefore,the time taken to cool by the same amount $(5^{\circ} C)$ increases as the temperature of the water approaches the room temperature.Comparing the mean temperatures: $\left( \frac{80+75}{2} ight) > \left( \frac{75+70}{2} ight) > \left( \frac{70+65}{2} ight)$.Thus,the rate of cooling is highest for the first interval and lowest for the third interval.Consequently,$t_1 < t_2 < t_3$.

$A$ beaker full of hot water is kept in a room. If it cools from $80^{\circ} C$ to $75^{\circ} C$ in $t_1$ minutes,from $75^{\circ} C$ to $70^{\circ} C$ in $t_2$ minutes,and from $70^{\circ} C$ to $65^{\circ} C$ in $t_3$ minutes,then:

Similar Questions

Are the rate of heat emission and the rate of cooling the same? Explain this.

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:

$A$ solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is $T$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module