Glauber's salt undergoes a phase transition according to the following equation:

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

Calculate the mass (g) of Glauber's salt (use the masses to two decimal places found in the front of your text) needed to lower the temperature of air in a room by 5.6oC. The mass of air in the room is 551.8 kg; the specific heat of air is 1.2J/g-C.

q = mass air x specific heat x delta T.

Calculate q. Watch the units.
From the kJ q needed, convert to moles of the salt and from there to grams. Post your work if you get stuck.

To solve this problem, we can use the equation:

q = m × c × ΔT

where:
- q is the heat transferred (in Joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g-°C)
- ΔT is the change in temperature (in °C)

First, let's calculate the heat transferred by the air in the room:

q_air = m_air × c_air × ΔT_air

Given:
m_air = 551.8 kg = 551,800 g
c_air = 1.2 J/g-°C
ΔT_air = -5.6 °C (negative because the temperature is lowered)

Plugging in the values:

q_air = 551,800 g × 1.2 J/g-°C × (-5.6 °C)
q_air = -3,296,256 J

Next, we can use the enthalpy change (∆H) of Glauber's salt to calculate the mass needed to release the equivalent amount of heat.

q_salt = m_salt × ∆H

Given:
∆H = 74.4 kJ/mol = 74,400 J/g

Plugging in the values:

-3,296,256 J = m_salt × 74,400 J/g
m_salt = -3,296,256 J / 74,400 J/g
m_salt ≈ -44.28 g

Since mass cannot be negative, we take the absolute value:

m_salt ≈ 44.28 g

Therefore, approximately 44.28 grams of Glauber's salt are needed to lower the temperature of the air in the room by 5.6 °C.