Two metallic spheres S1 and S2 are made of the same material and have got identical surface finish.

Q: Why do we use $dQ = -mc \, dT$ with a negative sign?

Because the body is cooling, its final temperature is lower than its initial temperature, making $dT$ negative. The negative sign ensures that the heat lost $dQ$ is a positive quantity, matching the positive radiation output from the Stefan-Boltzmann law.

Q: Why did we ignore the density $\rho$ and specific heat $c$ when taking the ratio?

The problem states that both spheres are made of the 'same material'. This implies that their density $\rho$ and specific heat capacity $c$ are identical. When we take the ratio $R_1/R_2$, these constant terms cancel out completely.

Q: Does the 'identical surface finish' matter?

Yes, absolutely! The surface finish determines the emissivity $e$ of the body. Because they have identical finishes, $e_1 = e_2$, allowing us to cancel it out in the ratio.

Q: Why is the rate of cooling inversely proportional to mass?

Heat capacity (thermal inertia) scales with mass (volume, $r^3$), while heat loss scales with surface area ($r^2$). As an object gets larger, its volume grows faster than its surface area. Therefore, a larger mass has relatively less surface area to lose heat from, making it cool slower.

Q: What if the spheres were not solid but hollow with the same thickness?

If they were hollow with a very thin, identical thickness, their mass would scale with surface area ($m \propto r^2$). In that specific case, the ratio $A/m$ would be constant, and both spheres would cool at the exact same rate!

Question

Two metallic spheres $S_1$ and $S_2$ are made of the same material and have got identical surface finish. The mass of $S_1$ is thrice that of $S_2$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_1$ to that of $S_2$ is

Options:

(A) $\frac{1}{3}$

(B) $\frac{1}{\sqrt{3}}$

(C) $\frac{\sqrt{3}}{1}$

(D) $\left(\frac{1}{3}\right)^{1/3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine two glowing hot metallic spheres, $S_1$ and $S_2$, placed in a cooler room. We are given that $S_1$ is three times heavier than $S_2$, meaning $m_1 = 3m_2$. Both spheres are made of the exact same material and have identical surface finishes. This implies that their density $\rho$, specific heat capacity $c$, and emissivity $e$ are all identical. They are also heated to the same initial high temperature $T$ and placed in the same room at temperature $T_0$.

Our goal is to find the ratio of their initial rates of cooling. But what exactly is the rate of cooling? It is the rate at which their temperature drops, denoted mathematically as $-\frac{dT}{dt}$.

### The Master Equation
To find the rate of cooling, we must bridge two fundamental concepts: the Stefan-Boltzmann Law of radiation and the principle of calorimetry.

According to the Stefan-Boltzmann law, the net rate of heat lost by a body to its surroundings is given by:

$$ \frac{dQ}{dt} = e \sigma A (T^4 - T_0^4) $$

Where $A$ is the surface area of the body.

Now, from calorimetry, we know that when a body loses heat $dQ$, its temperature drops by $dT$. The relationship is:

$$ dQ = -mc \, dT $$

Equating the two expressions for the rate of heat loss, we get:

$$ -mc \frac{dT}{dt} = e \sigma A (T^4 - T_0^4) $$

Rearranging this to isolate the rate of cooling $R = -\frac{dT}{dt}$, we find:

$$ R = \frac{e \sigma A (T^4 - T_0^4)}{mc} $$

Since $e$, $\sigma$, $c$, $T$, and $T_0$ are all identical for both spheres, we can establish a powerful proportionality:

$$ R \propto \frac{A}{m} $$

### Relating Area to Mass
We know the ratio of their masses, but we don't directly know the ratio of their surface areas. We need to express the surface area $A$ entirely in terms of the mass $m$.

For a sphere, the mass is the product of its volume and density:

$$ m = \frac{4}{3}\pi r^3 \rho $$

From this, we can see that the radius $r$ is proportional to the cube root of the mass:

$$ r \propto m^{1/3} $$

The surface area of a sphere is $A = 4\pi r^2$. Substituting our proportionality for $r$, we get:

$$ A \propto (m^{1/3})^2 = m^{2/3} $$

### Final Calculation
Now, let's substitute this area proportionality back into our rate of cooling relation:

$$ R \propto \frac{A}{m} \propto \frac{m^{2/3}}{m} = m^{-1/3} $$

This tells us that the rate of cooling is inversely proportional to the cube root of the mass. A heavier sphere of the same material will cool down slower because its mass (which stores heat) grows faster than its surface area (which loses heat).

Finally, we can find the ratio of the initial rates of cooling for $S_1$ and $S_2$:

$$ \frac{R_1}{R_2} = \left(\frac{m_1}{m_2}\right)^{-1/3} = \left(\frac{m_2}{m_1}\right)^{1/3} $$

Since we are given that $m_1 = 3m_2$, we substitute $\frac{m_2}{m_1} = \frac{1}{3}$:

$$ \frac{R_1}{R_2} = \left(\frac{1}{3}\right)^{1/3} $$

This elegant result shows how geometry and thermodynamics intertwine perfectly!

Select Answer:

Visualized Solution

Visualizing the Setup

Stefan-Boltzmann Law

Rate of Cooling

Relating Area and Mass

Area Proportionality

Cooling Rate Proportionality

Final Ratio Calculation

The Sigma Insight: Heat Transfer

Analyzing the Setup Imagine two glowing hot metallic spheres, $S_{1}$ and $S_{2}$ , placed in a cooler room

The Master Equation

Relating Area to Mass We know the ratio of their masses, but we don't directly know the ratio of their surface areas

Final Calculation

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