Two liquids of equal volume are cooled under | 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 heat loss is given by $\frac{dQ}{dt} = mc \frac{d	heta}{dt}$.Since the conditions are identical,

Vedclass · Accepted Answer

$8/15$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of heat loss is given by $\frac{dQ}{dt} = mc \frac{d	heta}{dt}$.Since the conditions are identical,the rate of heat loss $\frac{dQ}{dt}$ is the same for both liquids.Therefore,$m_1 s_1 \frac{d	heta}{t_1} = m_2 s_2 \frac{d	heta}{t_2}$.Since the volumes are equal $(V_1 = V_2 = V)$,we can write $m = ho V$.Substituting this,we get $ho_1 V s_1 / t_1 = ho_2 V s_2 / t_2$.This simplifies to $\frac{ho_1}{ho_2} = \frac{s_2}{s_1} 	imes \frac{t_1}{t_2}$.Given $s_1 : s_2 = 3 : 4$,so $s_2 / s_1 = 4/3$.Given $t_1 = 324 \; sec$ and $t_2 = 810 \; sec$.Ratio $\frac{ho_1}{ho_2} = \frac{4}{3} 	imes \frac{324}{810} = \frac{4}{3} 	imes \frac{2}{5} = \frac{8}{15}$.

Two liquids of equal volume are cooled under identical conditions from $60^{\circ}C$ to $50^{\circ}C$ in $324 \; sec$ and $810 \; sec$ respectively. If the ratio of their specific heats is $3:4$,find the ratio of their densities. (The water equivalent of the calorimeter is negligible.)

Similar Questions

$A$ liquid cools from $50^oC$ to $45^oC$ in $5$ minutes and from $45^oC$ to $41.5^oC$ in the next $5$ minutes. The temperature of the surrounding is ...... $^oC$.

The rate of cooling at $600 \ K$ is $R$,given the surrounding temperature is $300 \ K$. What is the rate of cooling at $900 \ K$?

The temperature of a liquid drops from $365 \ K$ to $361 \ K$ in $2$ minutes. Find the time during which the temperature of the liquid drops from $344 \ K$ to $342 \ K$. The temperature of the room is $293 \ K$.

$A$ 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

$A$ body cools from $60\,^oC$ to $50\,^oC$ in $10\, minutes$. If the room temperature is $25\,^oC$ 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 ....... $^oC$

Vedclass Test Series

Exam Paper Generator

Online Exam Module