# Vacuum Question

1. Nov 18, 2014

### billbaty

What would happen under these circumstances:

A sealed container of water in a vacuum chamber. Since water doesn't expand like air, I assume that there would be no extra pressure within the sealed container once the air was evacuated around it. Since there would be no, or very little air surrounding the container, and since heat requires a medium to dissipate, would the temperature of the water inside the container stay virtually constant or close to the temperature it was when placed into the vacuum chamber? If the vacuum chamber was placed in a sub-zero environment would the water still stay around its original temperature?

Also, while on the vacuum topic, since the air in a sealed balloon will expand and blow up large in a vacuum, what would happen if that sealed balloon with a small amount of air would be placed into the above example's sealed container of water in a vacuum chamber? Would there be no extra pressure inside the container and therefore the balloon will stay its same size?

2. Nov 18, 2014

### Staff: Mentor

3. Nov 18, 2014

### Khashishi

For your second question, since water is effectively incompressible, the balloon immersed in water won't expand. The pressure will change as you change the temperature, but the volume will stay the same.

4. Nov 18, 2014

### Doug Huffman

Heat does NOT require a medium to dissipate. Heat transfer cooling by radiation alone occurs at a rate proportional to the fourth power of the difference in absolute temperature.

Be careful of how you are measuring your pressure, absolute or gauge. Water in an elastic container will still boil when the vapor pressure exceeds the applied pressure.

Water does have a bulk modulus of elasticity compressibility.

Last edited: Nov 18, 2014
5. Nov 18, 2014

### davenn

indeed ! :)

a lack of a medium only takes care of 2 of the 3 methods of heat transfer ( conduction and convection)
IR radiation doesn't require a medium

Dave

6. Nov 18, 2014

### Khashishi

I think the rate is proportional to the difference in the fourth power of absolute temperature, not the fourth root.
$P = \sigma_{SB} \epsilon_1 \epsilon_2 (T_1^4-T_2^4)$

7. Nov 18, 2014

### Doug Huffman

You are of course correct.