A typical space walk uses about a half gallon of water (~5 lbs). A full tank is

Question

A typical space walk uses about a half gallon of water (~5 lbs). A full tank is ~1 gallon of water. How much heat is removed using 5 lbs of water? What might be the limitations of such a cooling system? Why do you think this method was chosen to cool the suit?

Assume the water in the tanks is 20 degrees celsius, it cools to 0 degrees celsius when it freezes in the sublimator. The Latent heat of Sublimation for water is the addition of the Latent heat of Fusion and the Latent heat of Vaporization from Table 4.4. So Ls=620 cal/g for water.

Explanation / Answer

Apply Q = mc(delta T) + mL

5 lbs = 2268 grams

Q = (2268)(1)(20) + (2268)(620)

Q = 1451520 cal which can be stated as 1452 kCal

That can also be stated as 6.073 kJ or 6.07 X 10^6 J

The limitations are that once the water is heated, there is no efficient way to cool it down and recirculate it while on the spacewalk in order to continue absorbing more heat.

This method was chosen, however, because water has the largest heat capacity of any material. Thus water, more than any other material, can absored that much heat without changing its own temperature significantly. Any other material would heat up more quickly and overall absorb much less heat. Water absorbs more heat than another other substance per degree of temperature. (Also why it is used in car radiators)

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