Question

A solid aluminium sphere and a solid lead sphere of the same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium \[=900J/k{{g}^{0}}C\] and that of lead \[=130J/k{{g}^{0}}C\]. The density of aluminium \[={{10}^{4}}kg{{m}^{-3}}\] and that of lead \[=2.7\times {{10}^{3}}kg{{m}^{-3}}\]. Assume that the emissivity of both the spheres are the same.
1.Find the ratio of rate of heat loss of the aluminium sphere to the rate of heat loss from the lead sphere.
2.Find the ratio of rate of fall of temperature of aluminium sphere to the rate of fall of temperature of the lead sphere.

Answer 1

Hint: We need to understand the dependence of the material of the object, the heat energy supplied and the specific heat capacity with the rate of heat loss of the object to solve the problem. We can easily find it with the given information.

Complete answer:
We know that the rate of the heat loss of a material can be defined by Stefan-Boltzmann's law. According to the law, the power radiated from a body or the rate of energy loss is proportional to the emissivity of the material, the surface area of the object and the fourth power of the temperature difference between the object and the surrounding.
It is given mathematically as –
\[P=\sigma Ae{{T}^{4}}\]
We know that the two spherical balls have the same conditions of surface area (As they have the same radii), surrounding temperature, initial temperature and emissivity. So, the ratio between the heat loss can be given as –
\[\begin{align}
  & {{P}_{AL}}=\sigma {{A}_{Al}}{{e}_{Al}}{{T}^{4}} \\
 & \text{and,} \\
 & {{P}_{Ld}}=\sigma {{A}_{Ld}}{{e}_{Ld}}{{T}^{4}} \\
 & \Rightarrow \dfrac{{{P}_{AL}}}{{{P}_{Ld}}}=\dfrac{\sigma {{A}_{Al}}{{e}_{Al}}{{T}^{4}}}{\sigma {{A}_{Ld}}{{e}_{Ld}}{{T}^{4}}} \\
 & \therefore \dfrac{{{P}_{AL}}}{{{P}_{Ld}}}=1:1 \\
\end{align}\]
The ratio between the power loss is in the ratio 1:1.
Now, we can find the rate of fall of temperature during the same using the relation with the specific heat capacity and the heat energy. We know that the heat loss for the same interval is constant for the both.
For Aluminium sphere, the heat loss is given as –
\[\begin{align}
  & {{Q}_{Al}}={{m}_{Al}}{{s}_{Al}}\Delta {{T}_{Al}} \\
 & \Rightarrow Q=({{\rho }_{Al}}\dfrac{4}{3}\pi {{r}^{3}}){{s}_{Al}}\Delta {{T}_{Al}} \\
 & \Rightarrow Q={{10}^{4}}\times \dfrac{4}{3}\pi {{r}^{3}}\times 900\times \Delta {{T}_{Al}} \\
 & \therefore Q=\dfrac{4}{3}\pi {{r}^{3}}\times 9\times {{10}^{6}}\Delta {{T}_{Al}} \\
\end{align}\]
For the lead sphere, the heat loss is given by –
\[\begin{align}
  & {{Q}_{Ld}}={{m}_{Ld}}{{s}_{Ld}}\Delta {{T}_{Ld}} \\
 & \Rightarrow Q=({{\rho }_{Ld}}\dfrac{4}{3}\pi {{r}^{3}}){{s}_{Ld}}\Delta {{T}_{Ld}} \\
 & \Rightarrow Q=2.7\times {{10}^{3}}\times \dfrac{4}{3}\pi {{r}^{3}}\times 130\times \Delta {{T}_{Ld}} \\
 & \therefore Q=\dfrac{4}{3}\pi {{r}^{3}}\times 3.51\times {{10}^{4}}\Delta {{T}_{Ld}} \\
\end{align}\]
Now, we can equate the heat losses to get the ratio between the temperature drops as –
\[\begin{align}
  & {{Q}_{Al}}={{Q}_{Ld}} \\
 & \Rightarrow \dfrac{4}{3}\pi {{r}^{3}}\times 9\times {{10}^{6}}\Delta {{T}_{Al}}=\dfrac{4}{3}\pi {{r}^{3}}\times 3.51\times {{10}^{4}}\Delta {{T}_{Ld}} \\
 & \Rightarrow \Delta {{T}_{Al}}\times 9\times {{10}^{6}}=3.51\times {{10}^{4}}\Delta {{T}_{Ld}} \\
 & \Rightarrow \dfrac{\Delta {{T}_{Al}}}{\Delta {{T}_{Ld}}}=\dfrac{3.51\times {{10}^{4}}}{9\times {{10}^{6}}} \\
 & \therefore \dfrac{\Delta {{T}_{Al}}}{\Delta {{T}_{Ld}}}=39:1000 \\
\end{align}\]
So, the ratio between the rate of temperature drops of the aluminium sphere to the lead sphere is 39:1000.
This is the required solution.

Note:
In this case, we have already found from the first part that the power radiated from the two spheres are equal in a given time interval. So, we could use the same relation in the second part to find the rate of temperature drop in the two more easily.