Delta H degree = 74.4 kJ/mol As a result, this compound is used to regulate the

Question

Delta H degree = 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 endothermic melting process absorbs heat from the surroundings, cooling the room. At night, it gives off heat as it freezes. Calculate the mass of Glauber's salt in kilograms needed to lower the temperature of air in a room by 8.2 degree C at 1.0 atm. The dimensions of the room are2.8 m x 10.6 m x 16.3 m, the specific heat of air is 1.2 J/g degree C, and the molar mass of air may be taken as 29.0 g/mol. kg

Explanation / Answer

First calculate mass of air in the room

Volume of room = l x b x h = 2.8 x 10.6 x 16.3 = 483.784 m^3

which is equivalent to 483.784 x 10^3 L

mass of air = Volume x density

taking density of air = 1.2 g/L

mass = 483.784 x 10^3 x 1.2 = 5.80 x 10^5 g

heat q = mCpdT

where,

m = mass of air = 5.80 x 10^5 g

Cp = specific heat of air = 1.2 J/g.oC

dT = change in temp. = 8.2 oC

Feed values,

q = 5.80 x 10^5 x 1.2 x 8.2 = 5.71 x 10^6 J

q = delta H

we have deltaHo = 74.4 kJ/mol which is heat absorbed by 322.20 g of Galuber's salt

So how much mass of Galuber's salt will be required to absorb 5.71 x 10^3 kJ

= 5.71 x 10^3 x 322.20/74.4 = 24.74 kg

Thus, we would reuire 24.74 kg of Galuber's salt for our purpose.

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.