Dry Ice (CO_2) is often used to flash-cool circuit boards once they come out of

Question

Dry Ice (CO_2) is often used to flash-cool circuit boards once they come out of a soldering bath. Typically 5 kg of (solid) dry ice is put into a box (Styrofoam) at a temperature of-100 degree C, along with a load of circuit boards. When the ice warms to -60 degree C it vaporizes(sublimes) to a gas. The latent heat of vaporization is 750 kJ/kg. Once it sublimes, it continues to absorb heat from the circuit boards until it is +20 degree C. At this time the box is opened and the circuit boards removed and the CO_2 gas allowed to escape. Overall, how much heat was absorbed by the dry ice while it was in the box? Specific heat CO_2 gas .65 kJ/kg-K solid .85 kJ/kg-K

Explanation / Answer

Mass of dry ice in the container m = 5kg

The problem states that dry ice in solid has absorbed heat, turned into vapour and then absorbed some more heat. Let us find out heat absorbed in all the processes separately and add them up to get the final result.

Given Specific heat of solid dry ice Csolid = 0.85 kJ/kg-K

Mass of ice m = 5kg

Change in temperature T = Final temperature - Initial temperature = 60o - (-100o) = 160o

Heat absorbed during the process = mCT = 5 x 0.85 x 160 = 680 kJ

Now, let us calculate heat absorbed by dry ice in the process of sublimation

The latent heat of vapourization per kg of dry ice = 750 kJ

The total heat absorbed = 750 x 5 = 3750 kJ

It is given that dry ice continues to absorb heat for another 20oc

Hence T = 20

Given specific heat Cgas=0.65 kJ/kg-k

m = 5kg

Hence heat absorbed in this stage = mCT = 5 x 0.65 x 20 = 65 kJ

Hence the overall heat absorption = 65 + 3750 + 680 = 4495 kJ

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