A solid sphere and a hollow sphere of the sam | 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 $\frac{dQ}{dt}$ depends on the surface area and temperature difference. Since both sphe

Vedclass · Accepted Answer

The rate of cooling of the hollow sphere is higher at temperature $T$.

Vedclass · Answer

(A) According to Newton's law of cooling,the rate of heat loss $\frac{dQ}{dt}$ depends on the surface area and temperature difference. Since both spheres have the same radius and material,their surface areas are equal,so $\frac{dQ}{dt}$ is the same for both.We know that $\frac{dQ}{dt} = mc \frac{dT}{dt}$,where $m$ is the mass,$c$ is the specific heat capacity,and $\frac{dT}{dt}$ is the rate of cooling.Therefore,$\frac{dT}{dt} = \frac{1}{mc} \cdot \frac{dQ}{dt}$.Since the material is the same,$c$ is constant. Because the hollow sphere has a smaller mass $(m)$ than the solid sphere,the rate of cooling $\frac{dT}{dt}$ will be higher for the hollow sphere.

$A$ 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 .......

Similar Questions

Describe the procedure to demonstrate that the rate of loss of heat from a hot body is directly dependent on the temperature difference between the body and its surroundings.

An ordinary body cools from $4 \theta$ to $3 \theta$ in $t$ minutes. The temperature of that body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature is $\theta$)

Hot water cools from $60\,^oC$ to $50\,^oC$ in the first $10\,minutes$ and to $42\,^oC$ in the next $10\,minutes.$ The temperature of the surroundings is ...... $^oC$

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$

Vedclass Test Series

Exam Paper Generator

Online Exam Module