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Glauber’s salt, sodium sulfate decahydrate (Na2SO4 •
Chapter 6, Problem 111P(choose chapter or problem)
Glauber's salt, sodium sulfate decahydrate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\), undergoes a phase transition (that is, melting or freezing) at a convenient temperature of about \(32^{\circ} \mathrm{C}\):
\(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
\(\Delta H^{\circ}=74.4\ \mathrm{kJ} / \mathrm{mol}\)
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 \(8.2^{\circ} \mathrm{C}\) at 1.0 atm. The dimensions of the room are \(2.80\ m \times 10.6\ m \times 17.2\ m\), the specific heat of air is \(1.2\ \mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and the molar mass of air may be taken as 29.0 g/mol.
Questions & Answers
QUESTION:
Glauber's salt, sodium sulfate decahydrate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\), undergoes a phase transition (that is, melting or freezing) at a convenient temperature of about \(32^{\circ} \mathrm{C}\):
\(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
\(\Delta H^{\circ}=74.4\ \mathrm{kJ} / \mathrm{mol}\)
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 \(8.2^{\circ} \mathrm{C}\) at 1.0 atm. The dimensions of the room are \(2.80\ m \times 10.6\ m \times 17.2\ m\), the specific heat of air is \(1.2\ \mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and the molar mass of air may be taken as 29.0 g/mol.
ANSWER:Step 1 of 3
Here we have to calculate the mass of Glauber’s salt in Kg.
Given:
Room temperature = 32 °C + 273 = 305 °C
Decrease of temperature = 8.2 °C
Pressure = 1.0 atm
Dimension of the room = 2.80 m × 10.6 m × 17.2 m
The specific heat of air = 1.2 J/g °C
Molar mass of air = 29.0 g/mol