Water and turpentine oil (specific heat less | vedclass

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

Question

(C) The rate of cooling is given by $\frac{dT}{dt} = \frac{kA(T - T_0)}{mc}$,where $m$ is the mass and $c$ is the specific heat capacity.Since the mas

Vedclass · Accepted Answer

$B$ and $A$ will represent the cooling curves of water and oil,respectively.

Vedclass · Answer

(C) The rate of cooling is given by $\frac{dT}{dt} = \frac{kA(T - T_0)}{mc}$,where $m$ is the mass and $c$ is the specific heat capacity.Since the mass $m$ and the surface area $A$ are the same for both,the rate of cooling is inversely proportional to the specific heat capacity $c$,i.e.,$	ext{Rate of cooling} \propto \frac{1}{c}$.Given that $c_{	ext{oil}}  (	ext{Rate of cooling})_{	ext{water}}$.This means that the oil cools down faster than the water.Therefore,at any given time $t > 0$,the temperature of the oil will be lower than the temperature of the water.Looking at the graph,curve $B$ shows a faster drop in temperature compared to curve $A$.Thus,curve $B$ represents the cooling curve of oil,and curve $A$ represents the cooling curve of water.

Water and turpentine oil (specific heat less than that of water) are both heated to the same temperature. Equal amounts of these,placed in identical calorimeters,are then left to cool in the air. Which of the following statements is correct regarding their cooling curves?

Similar Questions

$A$ body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C$. The temperature of the surroundings is $20\,^{\circ} C$.

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120^{\circ}C$,then

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

$A$ calorimeter contains $10 \ g$ of water at $20^{\circ}C$. The temperature falls to $15^{\circ}C$ in $10 \ min$. When the calorimeter contains $20 \ g$ of water at $20^{\circ}C$,it takes $15 \ min$ for the temperature to become $15^{\circ}C$. The water equivalent of the calorimeter is: (in $g$)

$A$ body takes $5$ minutes to cool from $90^oC$ to $60^oC$. If the temperature of the surroundings is $20^oC$,the time taken by it to cool from $60^oC$ to $30^oC$ will be ...... $\min.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module