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

State Newton's law of cooling and derive its mathematical equation.

Hot water cools from $80^{\circ} C$ to $60^{\circ} C$ in $1$ minute. How long will it take to cool from $60^{\circ} C$ to $50^{\circ} C$ (in $s$)? (Room temperature $= 30^{\circ} C$)

$A$ sphere,a cube,and a thin circular plate,all of the same material and same mass,are initially heated to the same high temperature and allowed to cool down under similar conditions. Then,the

An object is cooled from $75^{\circ}C$ to $65^{\circ}C$ in $2$ minutes in a room at $30^{\circ}C$. The time taken to cool another object from $55^{\circ}C$ to $45^{\circ}C$ in the same room in minutes is

The rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if

Vedclass Test Series

Exam Paper Generator

Online Exam Module