Dry ice has a sublimation temperature of T = -78.5 degree C and an enthalpy of s

Question

Dry ice has a sublimation temperature of T = -78.5 degree C and an enthalpy of sublimation of Delta h = 571kJ/kg. It changes from solid to gas at T = -78 degree C. Assume that the specific heat of dry ice is Cp = 0 8kJ/(kg k) and is constant from T = -100 degree C to T = +30 degree C. It's not, but It'll make problem qo faster. You place 20kg of dry ice (CO_2) which is at T = -90 degree C in a meat freezer with a volume of 0.6m^3. Assume that the dry ice has negligible volume compared to the volume of the meat freezer, thus you have V = 0.6m^3 of air inside at 1 atm. The meat freezer is set to T = -20 degree C. How much energy will be absorbed by the dry ice to sublimate from T = -90 degree C to T = -20 degree C? If it takes the dry ice 120 hours to completely sublimate, what is the average power? If the door remains shut and no leaks, what is the pressure inside and what is the composition of the gas (assume initially 79% N_2 and 21% O_2 by volume - you added 20kg of CO_2 gas to this). The text examples on the last pages may help with some parts...

Explanation / Answer

Total change of enthalpy from T = -90 to T = -20 is

Since volume of freezer is much larger , Change takes place at constant pressure of 1 atmosphere.

Since Cp is given same

Energy absorbed = m*cp * (70) + sublimation energy

= 20*0.8*1000*70 + 571*1000

= 1691KJ

Average power = 1691KJ/ 120*3600 = 3.91 watts

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