Question

Glauber's salt, sodium sulfate decahydrate (Na₂SO₄ * 10 H₂O) undergoes a phase transtion (that is, melting or freezing at a convenient temperature of about 32 degrees C:

Na₂SO₄ * 10H₂O(s) -> Na₂SO₄ * 10H₂O(l) deltaH= 74.4 kJ/mol

As a result this compound is used to regulate the temperature in homes. It is placed in plastic bags in the ceiling of a room. During the day , the endotheric melting process absorbs heat form the surroundings, cooling the room . At night, it gives off heat as it freezes. Calculate the mass of Glaubers salf in kilograms needed to lower the temperature of air in a room by 8.2 degrees C at 1.0 atm. The dimensions of the room are 2.8 m x 10.6 m x17.6 m, the specific heat of air is 1.2 J/g C, and the molar mass of air may be taken as 29.0 g/mol

Answer 1

Asuming air to be dry

Volume of air = 2.8*10.6*17.6=522.368m3

Density of air at 32 deg.c =(29/22.4)*273.15/(32+273.15) =1.16 kg/m3

Mass of air = volume* density = 522.368*1.16 kg =835.788 kg

Heat required (either removed or added) = mass* specific heat* temperature difference = 835.788*1000g*1.2*8 =1002947 Joules

Heat supplied by 1 mole of Na2SO4.10J2O= 74 Kj/mol= 74*1000j/mol

Mass of Na2SO4.10H2O= 1002947/74*1000 =13.53 moles of Na2SO4.10H2O

Molecular weight of Glaubar salt= 23*2+32+4*16+180 = 322

Mass of Na2SO4. 10H2O= 13.53*322 gm =4356.7 gms