Newton's law of cooling is used in the labora | vedclass

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

Question

(C) Newton's law of cooling states that the rate of loss of heat $dQ/dt$ of a body is directly proportional to the temperature difference between the

Vedclass · Accepted Answer

Specific heat of liquids

Vedclass · Answer

(C) Newton's law of cooling states that the rate of loss of heat $dQ/dt$ of a body is directly proportional to the temperature difference between the body and its surroundings,provided the difference is small.Mathematically,$\frac{dQ}{dt} = -k(	heta - 	heta_0)$.Since the rate of loss of heat is also given by $\frac{dQ}{dt} = -ms \frac{d	heta}{dt}$,where $m$ is the mass,$s$ is the specific heat,and $\frac{d	heta}{dt}$ is the rate of cooling.Equating the two,we get $-ms \frac{d	heta}{dt} = -k(	heta - 	heta_0)$,which implies $\frac{d	heta}{dt} = \frac{k}{ms}(	heta - 	heta_0)$.By measuring the rate of cooling $\frac{d	heta}{dt}$ for a liquid of known mass $m$ and comparing it with a standard liquid,the specific heat $s$ of the liquid can be determined.Thus,it is used for the determination of the specific heat of liquids.

Newton's law of cooling is used in the laboratory for the determination of the ...........

Similar Questions

$A$ 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

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 $293 \ K$.

$A$ solid sphere and a hollow sphere of the same material and same radius are heated to the same temperature. They are placed in an environment of the same temperature. If the temperature difference between both and the surroundings is $T$,then .......

Are the rate of heat emission and the rate of cooling the same? Explain this.

$A$ body takes $10 \ minutes$ to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the same body after the next $10 \ minutes$ will be (if the temperature of the surroundings is $25^{\circ}C$): (in $^{\circ}C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module