I was commenting on Talmand's scenario; take "three of the world's largest ships, put them together to form one ship and then placed it in a random spot somewhere within the orbit of Pluto; how long would it take you to find it?"
Using the formula in the "Refrigeration" section of the page linked to by Avshalom, the "maximum range a ship running silent with engines shut down" (assuming diameter = 500 m and temperature = 290K is about 500 million km, or about 3.3 AU. That's well within the orbital radius of Jupiter.
That's assuming the entire asteroid is at 290K. Pluto averages about 40 AU out, given a 0.056% chance that a room-temperature asteroid would be detectable.
I'm also uncertain about the reasoning in the page linked to by Avshalom. I think that assumes a 3K background, and I think you also think there's a big heat differential. However, the average temperature of the moon is something like -25C, and if an asteroid were closer to the sun, then the difference in surface temperature between an internally heated vs. sun heated asteroid is well within the errors in measuring the physical properties of the asteroid.
You can even work it out for yourself, using the equation at http://en.wikipedia.org/wiki/Standard_asteroid_physical_char... . I worked it out to:
166C at Mercury orbit: http://www.google.com/search?&q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(0.3+*+150000000+km)**2)))+**+(1%2F4)
90C at Venus orbit: http://www.google.com/search?q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(0.6+*+150000000+km)**2)))+**+(1%2F4)
5C at Earth orbit: http://www.google.com/search?q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(1.0+*+150000000+km)**2)))+**+(1%2F4)
(The values for albedo and emissivity are not well known, so there is a wide error range in this calculation.)
Still, that's enough to show that the heat differentials inside of the Earth's orbit are not that big.