A solid copper sphere (density rho and specific heat c) of radius r at an initial temperature 200 K

Q: Why do we take the chamber temperature as 0 K?

The problem states the walls are at almost 0 K. This means the radiation absorbed by the sphere from the chamber is negligible ($\sigma A T_0^4 \approx 0$), so we only consider the heat radiated outwards.

Q: Why is there a negative sign in the $dQ = -mc dT$ equation?

Since the sphere is cooling, its change in temperature $dT$ is negative. The negative sign ensures that the heat lost $dQ$ remains a positive quantity when equated to the radiation rate.

Q: Does the cooling time depend on the size of the sphere?

Yes! The final expression has $r$ in the numerator. A larger sphere has more volume (mass) relative to its surface area, meaning it stores more heat but has proportionally less area to radiate it away, so it takes longer to cool.

Q: Why didn't we use Newton's Law of Cooling here?

Newton's Law of Cooling is an approximation valid only for small temperature differences between the body and its surroundings. Here, the temperature drops from 200 K to 100 K in a 0 K chamber, which is a massive difference, requiring the exact Stefan-Boltzmann Law.

Q: Where did the $10^{-6}$ come from in the calculation?

It comes from evaluating the integral limits: $1/100^3 - 1/200^3$. $100^3$ is $10^6$, so factoring it out makes the arithmetic much simpler.

Question

A solid copper sphere (density $ho$ and specific heat $c$) of radius $r$ at an initial temperature $200	ext{ K}$ is suspended inside a chamber whose walls are at almost $0	ext{ K}$. The time required for the temperature of the sphere to drop to $100	ext{ K}$ is ......

Sigma Percentile · Accepted Answer

### The Absolute Zero Void

Imagine a glowing hot solid copper sphere suspended in the absolute freezing void of a $0	ext{ K}$ chamber. Because the surroundings are at absolute zero, the sphere doesn't absorb any heat back from the walls. It only loses heat by radiating it away into the emptiness.

This is a classic scenario of pure radiative cooling. The sphere starts at an initial temperature of $200	ext{ K}$, and we want to find out exactly how long it takes for its temperature to drop to $100	ext{ K}$.

### The Master Equation of Cooling

To find out how fast it cools, we need to invoke the **Stefan-Boltzmann Law**. The rate at which the sphere radiates heat energy is given by:

$$ \frac{dQ}{dt} = \sigma A T^4 $$

At the same time, from the principles of calorimetry, we know that the heat lost by the sphere is directly related to its drop in temperature:

$$ dQ = -mc dT $$

Equating these two powerful ideas, we get our master differential equation for cooling:

$$ -mc \frac{dT}{dt} = \sigma A T^4 $$

The negative sign perfectly captures the physical reality that the temperature $T$ is decreasing as time $t$ marches forward.

### The Calculus of Heat

Now, let's separate the variables to prepare for integration. We'll move all the temperature terms to one side and the time term to the other:

$$ dt = -\frac{mc}{\sigma A} T^{-4} dT $$

It's time to integrate. The clock starts at $t = 0$ and ticks up to our unknown time $t$. Meanwhile, the temperature drops from an initial $200	ext{ K}$ down to $100	ext{ K}$.

$$ \int_0^t dt = -\frac{mc}{\sigma A} \int_{200}^{100} T^{-4} dT $$

The integral of $T^{-4}$ is $\frac{T^{-3}}{-3}$. The negative signs beautifully cancel out. Plugging in the upper and lower limits, we get:

$$ t = \frac{mc}{3\sigma A} \left[ \frac{1}{100^3} - \frac{1}{200^3} ight] $$

Let's crunch the numbers in the bracket. $100^3$ is $10^6$, and $200^3$ is $8 	imes 10^6$.

$$ \frac{1}{10^6} - \frac{1}{8 	imes 10^6} = \frac{7}{8 	imes 10^6} $$

Multiplying by the $\frac{1}{3}$ outside, we get:

$$ t = \frac{7mc}{24\sigma A} 	imes 10^{-6} $$

### Unveiling the Geometry

We aren't given the mass $m$ or surface area $A$ directly, but we know it's a solid sphere of radius $r$ and density $ho$.

The mass is the volume times the density:

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

And the surface area of a sphere is:

$$ A = 4\pi r^2 $$

Let's substitute these geometric facts into our time equation:

$$ t = \frac{7 \left(\frac{4}{3}\pi r^3 hoight) c}{24\sigma (4\pi r^2)} 	imes 10^{-6} $$

Notice how the $\pi$ and most of the $r$'s cancel out, leaving just a single $r$ in the numerator:

$$ t = \frac{7 ho r c}{72 \sigma} 	imes 10^{-6} $$

### The Final Countdown

Now, we plug in the standard value for the Stefan-Boltzmann constant, $\sigma = 5.67 	imes 10^{-8} 	ext{ W m}^{-2}	ext{K}^{-4}$.

$$ t = \frac{7 	imes 10^{-6}}{72 	imes 5.67 	imes 10^{-8}} ho r c $$

$$ t = \frac{700}{408.24} ho r c \approx 1.71 ho r c $$

And there we have it, our final expression!

This result is fascinating. Because the cooling rate depends on the fourth power of temperature, the sphere cools very quickly at first, but as it gets colder, the process slows down drastically. Think about how long it would take to reach exactly zero Kelvin!

Visualized Solution

$Visualizing the Cooling Sphere$

$Stefan-Boltzmann Law$

$Forming the Differential Equation$

$Separating Variables$

$Setting up the Integral$

$Executing the Integration$

$Simplifying the Bracket$

$Substituting Mass and Area$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Heat Transfer

The Absolute Zero Void

The Master Equation of Cooling

The Calculus of Heat

Unveiling the Geometry

The Final Countdown

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Cooling Sphere

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stefan-Boltzmann Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forming the Differential Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separating Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Executing the Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Bracket

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Mass and Area

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward