Two circular discs A and B with equal radii a | 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 given by $\frac{d	heta}{dt} = \frac{e \sigma A_{surf}}{ms} (	heta^4 - 	heta_0^4)$.F

Vedclass · Accepted Answer

Specific heat of $A$ is less.

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{e \sigma A_{surf}}{ms} (	heta^4 - 	heta_0^4)$.For small temperature differences,this simplifies to $\frac{d	heta}{dt} \propto \frac{1}{ms}(	heta - 	heta_0)$.Since the discs have equal mass $m$,equal surface area $A_{surf}$,and are under identical conditions,the rate of cooling $\frac{d	heta}{dt}$ is inversely proportional to the specific heat $s$ for a given temperature difference $(	heta - 	heta_0)$.From the graph,the slope of the line represents $\frac{d	heta/dt}{	heta - 	heta_0} \propto \frac{1}{s}$.Since the slope of line $A$ is greater than the slope of line $B$,the specific heat of $A$ must be less than the specific heat of $B$.

Two circular discs $A$ and $B$ with equal radii and equal mass are blackened. They are heated to the same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves?

Similar Questions

$A$ 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 $:$

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5 \text{ min}$. Calculate the time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$. The temperature of the surrounding is $20^{\circ} C$. (in $\text{ min}$)

The temperature of hot water falls from $60^oC$ to $50^oC$ in $10 \, min$. It takes another $10 \, min$ to fall from $50^oC$ to $42^oC$. Find the temperature of the surroundings in $^oC$.

$A$ body takes $4\, \text{min}$ to cool from $61^{\circ} \text{C}$ to $59^{\circ} \text{C}$. If the temperature of the surroundings is $30^{\circ} \text{C}$,the time taken by the body to cool from $51^{\circ} \text{C}$ to $49^{\circ} \text{C}$ is $....\, \text{min}$.

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module