Heat Transfer From a Spaceship in Deep Space Space is a very low temperature environment, however it also has an extremely small number of particles per unit volume.  This leads me to believe that, contrary to popular portrayals of heat loss in space, there would be very little heat loss due to conduction on a hypothetical spaceship in deep space.  If you were put into space and provided with food, water, heat, and protected from the pressure would the spaceship cool due to head radiating away from the ship?
 A: If you are out of the sunlight, the main source of cooling will be radiation. The amount of net heat you radiate away depends on the temperature of your skin $T_{skin}$ and the ambient temperature of the surrounding environment $T_{ambient}$:
$\frac{Q}{t} = e \sigma A (T^{4}_{skin}-T^{4}_{ambient})$
where 
$Q$ = heat loss in Joules 
$t$ = time in seconds  
$e$ = emissivity of skin ($\approx$ 0.98 for human body) 
$\sigma$ = Stefan-Boltzmann constant 
$A$ = surface area of human body 
Plug $T_{skin}$ and $T_{ambient}$ into this calculator and it will calculate the heat loss in Watts using the equation above. For human body temperature ($T_{skin}$ = 34 °C = 307 K) and room temperatue ($T_{ambient}$ = 23 °C = 296 K), the heat loss is 133 Watts. For human body temperature and outer space ($T_{ambient}$ = -270 °C = 3 K), the heat loss is nearly 1,000 Watts. 
Assume a human is 70 kg of water. Since water's heat capacity is 4.18 J/gK, if the person emits 1,000 Watts for ten minutes, that is only enough to decrease its temperature by 2 K. To balance the heat radiation away from your body you would need to consume food at 1,000 Watts = 14 food Calories per minute.

One of these 50 Calorie cookies every 2-3 minutes would do the trick.
In other words, you are right that space is not as "cold" as it is made out to be. Yes the temperature is low-- but there is not enough matter out there to cool down anything quickly. A Thermos is able to keep coffee hot for long periods of time by taking advantage of exactly this phenomenon.
A: Let's start by assuming you're in the shade, so you're not receiving any radiation (apart from the cosmic microwave background, which I think we can ignore). The amount of heat per unit area that you radiate is given by Stefan's law:
$$ J = \varepsilon \sigma T^4 \tag{1} $$
The emissivity of human skin is allegedly 0.98, and the area of skin of an adult male is around 2m$^2$, so feeding in $T = $37ºC gives us a total radiated power of about a kilowatt. The power produced by an adult male is about 120W, so at body temperature you're going to lose about 880W.
To work out what temperature you would cool to we just take equation (1), feed in $J = $60W/m$^2$ and we get $T = $180K. This would be fatal.
What's interesting is to see what happens when you're in direct sunlight. At the orbit of the earth the radiation from the Sun is around 1.4kW/m$^2$. Since only half your skin would be illuminated, you would be losing a kW and gaining 1.4kW for a net gain of 400W. To work out your equilibrium temperature we just feed $J = $1400W/m$^2$ into equation (1) and we get $T = $396K, which would again be fatal.
The distance from the Sun where the heat you radiate would exactly balance the heat you receive can be worked out using the inverse square law. If $r_E$ is the radius of Earth's orbit and $r$ is the thermal balance radius we get:
$$ \frac{r^2}{r_E^2} = \frac{1400}{1000} $$
or:
$$ r = 1.18r_E $$
