Question

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 liters of water heated to 88 ∘C.

Use standard enthalpies of formation to calculate the change in enthalpy for dry ice sublimation. (The ΔH∘f for CO2(s) is - 427.4kJ/mol).

Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 27 ∘C. Assume no heat loss to the surroundings.

Answer 1

Amount of water = 15.0 L

Density of water = 1..0 g / ml

Therefore 15.0 L water = 15.0 kg of water

1 .00 kg = 1000 g

heat needed to raise one gram water, one degree celsius = 4.18J

4.18J * 15,000g = 62,700J = 62.7 kJ needed per degree C

Temperature difference= 88 ∘C - 27 ∘C

= 61∘C

62.7 kJ / ∘C * 61∘C = 3824.7 kJ needed from CO2

The ΔH∘f for CO2(s) is - 427.4kJ/mol

Molar mass CO2 = 44g/mol = 22.73 moles/kg

427 kJ/mol * 22.73 moles/kg
= 9704.5 kJ/kg

3824.7 kJ / 9704.5 kJ/kg

= 0.394 kg of CO2 needed

= 394 g of CO2 needed