A small object is placed at the center of a large evacuated hollow spherical container. Assume that

Q: Why do we take the surrounding temperature as $0\text{ K}$?

The problem explicitly states that the container is maintained at $0\text{ K}$. According to the Stefan-Boltzmann law, the heat absorbed from the surroundings is proportional to $T_s^4$. Since $T_s = 0$, the object only loses heat and absorbs none.

Q: Why is there a negative sign in $dQ = -ms dT$?

The temperature of the object is decreasing over time, meaning $dT$ is a negative quantity. However, the heat lost $dQ$ is a positive quantity. The negative sign ensures that the equation balances correctly.

Q: Can I integrate from $100\text{ K}$ to $50\text{ K}$ to find $t_2$?

If you integrate from $100\text{ K}$ to $50\text{ K}$, the time interval you calculate will be the time taken to cool from $100\text{ K}$ to $50\text{ K}$, which is $(t_2 - t_1)$. To find $t_2$ directly, you must integrate from the initial state at $t=0$, which is $200\text{ K}$.

Q: Why did we bundle all the constants into $k$?

Bundling constants like $\sigma$, $A$, $m$, and $s$ into a single constant $k$ simplifies the algebra. Since we ultimately need to find a ratio of times, these constants will cancel out anyway, so there's no need to carry them around in their expanded form.

Q: How did the complex fraction simplify so easily at the end?

By multiplying both the numerator and the denominator by $200^3$, we eliminated the fractions inside the fraction. This turned terms like $1/50^3$ into $(200/50)^3 = 4^3$, making the final arithmetic straightforward.

Question

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at $0	ext{ K}$. At time $t = 0$, the temperature of the object is $200	ext{ K}$. The temperature of the object becomes $100	ext{ K}$ at $t = t_1$ and $50	ext{ K}$ at $t = t_2$. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio $(t_2/t_1)$ is_______.

Sigma Percentile · Accepted Answer

### The Setup: A Lonely Object in the Cold Void
Imagine a small object suspended in the center of a massive, evacuated spherical container. The container is maintained at absolute zero ($0	ext{ K}$). This means the environment is completely devoid of thermal energy—it absorbs everything and radiates nothing back.

Our small object, initially at $200	ext{ K}$, is an ideal black body. It is bleeding heat into the void purely through electromagnetic radiation. To understand how fast it cools, we must invoke the **Stefan-Boltzmann Law**.

### The Master Equation of Cooling
The Stefan-Boltzmann law states that the net rate of heat loss is given by:
$$ \frac{dQ}{dt} = \sigma A (T^4 - T_s^4) $$

Since the surrounding temperature $T_s = 0	ext{ K}$, the equation simplifies beautifully to:
$$ \frac{dQ}{dt} = \sigma A T^4 $$

But how does this heat loss affect the object's temperature? We know from calorimetry that the heat lost is proportional to the drop in temperature:
$$ dQ = -ms dT $$

Here, the negative sign is crucial—it indicates that the temperature $T$ is decreasing as time $t$ increases.

Equating these two expressions for the rate of heat transfer, we forge our master differential equation:
$$ -ms \frac{dT}{dt} = \sigma A T^4 $$

### Taming the Differential Equation
To solve this, we must separate the variables. We gather all the temperature terms on one side and the time terms on the other:
$$ \frac{dT}{T^4} = -\frac{\sigma A}{ms} dt $$

To keep our algebra clean, let's bundle all those constant physical properties into a single constant $k = \frac{\sigma A}{ms}$. Our equation becomes:
$$ \int \frac{dT}{T^4} = -k \int dt $$

### The First Interval: Cooling to 100 K
We are told that the object cools from $200	ext{ K}$ to $100	ext{ K}$ in time $t_1$. Let's set up the definite integral for this journey:
$$ \int_{200}^{100} T^{-4} dT = -k \int_0^{t_1} dt $$

Integrating $T^{-4}$ yields $\frac{T^{-3}}{-3}$. Applying the limits:
$$ \left[ \frac{T^{-3}}{-3} ight]_{200}^{100} = -k t_1 $$

$$ \frac{1}{3} \left( \frac{1}{100^3} - \frac{1}{200^3} ight) = k t_1 $$
This is our first milestone.

### The Second Interval: Cooling to 50 K
Next, we analyze the total time $t_2$ it takes for the object to cool from its initial $200	ext{ K}$ all the way down to $50	ext{ K}$. We use the exact same integral, just with a new lower limit:
$$ \int_{200}^{50} T^{-4} dT = -k \int_0^{t_2} dt $$

Evaluating this gives us our second milestone:
$$ \frac{1}{3} \left( \frac{1}{50^3} - \frac{1}{200^3} ight) = k t_2 $$

### The Grand Finale: Finding the Ratio
We are hunting for the ratio $\frac{t_2}{t_1}$. By dividing our two milestone equations, the constant $k$ and the factor of $\frac{1}{3}$ vanish entirely:
$$ \frac{t_2}{t_1} = \frac{\frac{1}{50^3} - \frac{1}{200^3}}{\frac{1}{100^3} - \frac{1}{200^3}} $$

This fraction looks intimidating, but there is an elegant algebraic trick. Multiply both the numerator and the denominator by $200^3$:
$$ \frac{t_2}{t_1} = \frac{\left(\frac{200}{50}ight)^3 - 1}{\left(\frac{200}{100}ight)^3 - 1} $$

Simplify the terms inside the parentheses:
$$ \frac{t_2}{t_1} = \frac{4^3 - 1}{2^3 - 1} $$

$$ \frac{t_2}{t_1} = \frac{64 - 1}{8 - 1} = \frac{63}{7} $$

And just like that, the math collapses into a beautiful, clean integer:
$$ \mathbf{\frac{t_2}{t_1} = 9} $$

Enter Numerical Value:

Visualized Solution

$Stefan-Boltzmann Law$

$Cooling Equation$

$Separation of Variables$

$First Interval: t_{1}$

$Second Interval: t_{2}$

$Ratio t_{2} / t_{1}$

The Sigma Insight: Heat Transfer

The Setup

The Master Equation of Cooling

Taming the Differential Equation To solve this, we must separate the variables

The First Interval

The Second Interval

The Grand Finale

Similar Questions

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

Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

The figure shows a system of two concentric spheres of radii $r_{1}$ and $r_{2}$ and kept at temperatures $T_{1}$ and $T_{2}$ , respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

A black body of temperature $T$ is inside a chamber of temperature $T_{0}$ . Now the closed chamber is slightly opened to sun such that temperature of black body ( $T$ ) and chamber ( $T_{0}$ ) remains constant

A solid sphere of copper of radius $R$ and a hollow sphere of the same material of inner radius $r$ and outer radius $R$ are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster?