Two identical beakers A and B contain equal v | vedclass

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

Question

(A) According to Newton's law of cooling,the rate of heat loss is given by: $-ms\frac{dT}{dt} = e\sigma A(T^4 - T_0^4)$For small temperature differenc

Vedclass · Accepted Answer

Vedclass · Answer

(A) According to Newton's law of cooling,the rate of heat loss is given by: $-ms\frac{dT}{dt} = e\sigma A(T^4 - T_0^4)$For small temperature differences,this simplifies to:$-\frac{dT}{dt} = \frac{4e\sigma AT_0^3}{ms}(T - T_0)$Let $k = \frac{4e\sigma AT_0^3}{ms}$. Since $m = ho V$,we have $k = \frac{4e\sigma AT_0^3}{ho Vs}$.Thus,the rate of cooling $|\frac{dT}{dt}|$ is inversely proportional to the product of density and specific heat $(ho s)$:$|\frac{dT}{dt}| \propto \frac{1}{ho s}$Calculating $(ho s)$ for both liquids:$(ho s)_A = (8 	imes 10^2) 	imes 2000 = 16 	imes 10^5 = 1.6 	imes 10^6\, Jm^{-3}K^{-1}$$(ho s)_B = 10^3 	imes 4000 = 4 	imes 10^6\, Jm^{-3}K^{-1}$Since $(ho s)_A  |\frac{dT}{dt}|_B)$.Therefore,liquid $A$ cools down faster than liquid $B$,which corresponds to graph $A$.

Similar Questions

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 cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes when kept in air at $30^{\circ}C$. In the next $10$ minutes,its temperature will be:

The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ will be $:$

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

$A$ cup of tea cools from $80^{\circ}C$ to $60^{\circ}C$ in $1$ minute. The ambient temperature is $30^{\circ}C$. In cooling from $60^{\circ}C$ to $50^{\circ}C$,it will take ....... $\text{sec}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module