# Cooling of the Earth

How long does it take for the Earth to cool down to $3K$ if the sun was gone?

Assumptions: The Earth only loses heat to space and does not gain any heat. Heat transfer occurs only through radiation. The mass of the Earth is $5.972\cdot10^{24}\text{ kg}$. The specific heat capacity of the Earth is $1260 \text{ J/kg} K$. The current temperature of the Earth is $300K$. The Earth is a perfect blackbody. The surface area of the Earth is $5.1\cdot10^{14}m^2$. Do not take the latent heat of fusion or vaporisation into account. Assume heat is always evenly distributed around the Earth and the Earth is a uniform sphere made of the same material with the specific heat capacity as stated above. Note by A Former Brilliant Member
3 years, 5 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

From the Stephan-Boltzmann Law, $\frac{dQ}{dt} = \sigma A T^4 = \sigma A T^4$ $Q$ is heat radiated by the earth, $T$ is the temperature of the earth, $A$ is the surface area.
By the definition of specific heat,
$\frac{dQ}{dT} = -mc$ $m$ is the mass of the earth, $c$ is the specific heat.
The negative sign is because the temperature is reducing.
Dividing,
$\frac{dT}{dt} = \frac{-\sigma A}{mc} \cdot T^4$ Rearranging,
$\int_{300 K}^{3 K} \frac{-dT}{T^4} = \int_0^t \frac{\sigma A}{mc} dt$

$\frac{1}{3} \left(\frac{1}{T^3}\right)\biggl|_{300}^{3} = \frac{\sigma A t}{mc}$ so,
$t = \frac{mc}{3\sigma A} \left(\frac{1}{27} - \frac{1}{27 \cdot 10^6}\right)$

$t \approx 3.23 \cdot 10^{18} \text{ seconds } \approx 10^{11} \text{ years }$ That is a long time!

- 3 years, 4 months ago

Amazing solution sir , +1!

- 3 years, 4 months ago

Hey Jerry , its interesting to know that we can even calculate this , Can you explain how ? Anyone ? @Pranshu Gaba , @Swapnil Das , @Arjen Vreugdenhil please help .

- 3 years, 4 months ago