On what does the proportionality constant dep | vedclass

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

Question

(C) Newton's law of cooling states that the rate of loss of heat $dQ/dt$ is proportional to the temperature difference $(T - T_0)$ between the body an

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(C) Newton's law of cooling states that the rate of loss of heat $dQ/dt$ is proportional to the temperature difference $(T - T_0)$ between the body and its surroundings.Mathematically,$dQ/dt = k(T - T_0)$,where $k$ is the proportionality constant.The constant $k$ is given by $k = eA\sigma / ms$,where $e$ is the emissivity,$A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,$m$ is the mass,and $s$ is the specific heat capacity.Since $k$ depends on the surface area $A$ and the nature of the surface (which determines emissivity $e$),the constant $k$ depends on both the nature and the area of the surface of the body.Therefore,the correct option is $C$.

On what does the proportionality constant depend in Newton's law of cooling?

Similar Questions

According to $Newton's$ $Law$ $of$ $cooling$,the rate of cooling of a body is proportional to the

$A$ bowl filled with very hot soup cools from $98^{\circ}C$ to $86^{\circ}C$ in $2$ minutes when the room temperature is $22^{\circ}C$. How many minutes will it take to cool from $75^{\circ}C$ to $69^{\circ}C$?

Hot water is kept in a thermally insulated closed container. It takes $T_1 \, min$ for the temperature to drop from $75^{\circ}C$ to $70^{\circ}C$,$T_2 \, min$ to drop from $70^{\circ}C$ to $65^{\circ}C$,and $T_3 \, min$ to drop from $65^{\circ}C$ to $60^{\circ}C$. Then:

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module