My assumption is that any object with a temperature above
absolute zero emit some sort of radiation and loose heat
- is this a correct?
Yes, this is correct.
Every object with a temperature $T>0$ emits electromagnetic radiation.
According to Wien's displacement law this radiation
has its maximum at a wavelength $\lambda_\text{max}$ depending on
the temperature $T$.
$$\lambda_\text{max} = \frac{b}{T}, \quad \text{with} \quad
b = 2.9\cdot 10^{-3}\text{m}\cdot\text{K}$$
- The surface of the sun has a temperature of $5500$ °C,
i.e. $T = 5800\ \text{K}$. Therefore its radiation has its
maximum at $\lambda_\text{max}=500\ \text{nm}$,
which is visible light.
- Let's say the surface of the earth has a typical temperature of
$20$ °C, i.e. $T=290\ \text{K}$.
Then its radiation has its maximum at
$\lambda_\text{max}=10\ \mu\text{m}$, which is infra-red light.
Does earth emit infrared light during the day as well?
I mean does it loose heat in the form of infrared light
while absorbing visible light?
The earth emits infra-red light all the time, day and night.
And because the earth is warmer during day time,
it emits more infra-red light at day time than at night time.
The reason for that is the Stefan-Boltzmann law which
says that the total radiation power $P$ per area $A$ increases
with temperature $T$:
$$ \frac{P}{A} = \sigma T^4, \quad \text{with}\quad
\sigma=5.7 \cdot 10^{-8} \frac{\text W}{\text{m}^2\text{K}^4}$$
- Let's say again the surface of the earth has a typical temperature of
$20$ °C, i.e. $T=290\ \text{K}$.
Then its radiation power per area is $400\ \text{W/m}^2$.
