We are tasked with solving the simpler differential equation given by:

$\frac{du}{dt} = -\alpha u^4,$

where $\alpha = 2.2 \times 10^{-12} \, \text{K}^{-3}/\text{s}$, and $u(0) = 2000 \, \text{K}$.

Step 1: Separate Variables

We first separate the variables to make the equation easier to solve. Rearrange the equation as:

$\frac{du}{u^4} = -\alpha \, dt.$

Step 2: Integrate Both Sides

Now, integrate both sides. On the left-hand side, we have:

$\int \frac{du}{u^4} = \int -\alpha \, dt.$

The integral of $\frac{1}{u^4}$ is:

$\int \frac{1}{u^4} du = -\frac{1}{3u^3}.$

So, we can rewrite the equation as:

$-\frac{1}{3u^3} = -\alpha t + C,$

where $C$ is the constant of integration.

Step 3: Solve for $u(t)$

Multiply both sides by $-1$ and rearrange:

$\frac{1}{3u^3} = \alpha t + C.$

To solve for $u$, we first solve for $u^3$:

$u^3 = \frac{1}{3(\alpha t + C)}.$

Finally, take the cube root:

$u(t) = \left( \frac{1}{3(\alpha t + C)} \right)^{1/3}.$

Step 4: Use the Initial Condition

We are given that $u(0) = 2000 \, \text{K}$. Substituting this into the equation for $u(t)$ when $t = 0$, we get:

$2000^3 = \frac{1}{3C}.$

Solve for $C$:

$C = \frac{1}{3 \times 2000^3} = \frac{1}{3 \times 8 \times 10^9} = \frac{1}{2.4 \times 10^{10}}.$

Step 5: Final Expression for $u(t)$

Substitute this value of $C$ into the expression for $u(t)$:

$u(t) = \left( \frac{1}{3 \left( \alpha t + \frac{1}{2.4 \times 10^{10}} \right)} \right)^{1/3}.$

Substituting $\alpha = 2.2 \times 10^{-12} \, \text{K}^{-3}/\text{s}$:

$u(t) = \left( \frac{1}{3 \left( 2.2 \times 10^{-12} t + \frac{1}{2.4 \times 10^{10}} \right)} \right)^{1/3}.$

This is the temperature $u(t)$ of the body at any time $t$.

Would you like more details or have any questions?

5 related questions:

1. How does the Stefan-Boltzmann law relate to blackbody radiation?
2. How does the temperature of the surroundings influence the heat transfer process?
3. What would happen if the body temperature is close to the surroundings temperature in this problem?
4. How can this type of differential equation be solved numerically?
5. How do the initial conditions affect the solution to differential equations like this one?

Tip:

In problems involving heat transfer, simplifying assumptions (like $u \gg T$) can make equations more manageable but should be carefully checked against real conditions for accuracy.