# 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 Jerry Han Jia Tao
1 year, 10 months ago

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!

- 1 year, 10 months ago

Amazing solution sir , +1!

- 1 year, 10 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 .

- 1 year, 10 months ago