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 83 ∘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). Express your answer using three significant figures.

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 11 ∘C. Assume no heat loss to the surroundings. Express your answer using two significant figures.

Answer 1

We know that density of water (d) = 1kg/L

Using the formula

d = mass/volume

mass of water (m)= d×volume = (1kg/L)(15L) = 15kg = 15000g

Heat capacity of water = 4.186J/g°C

Heat given by the water = mc∆T = 15000g(4.186J/g°C)(11-83)°C

Heat given by the water = -4520880J

Number of moles of dry ice =

(heat given by the water)/(sublimation heat of dry ice)

(Sublimation heat of dry ice = -427.4kJ/mol = -427400J/mol)

Number of moles of dry ice = -4520880J/(-427400J/mol) =

10.5776mol

Mass of dry ice = number of moles of dry ice × molar mass of dry ice

Mass of dry ice = (10.5776mol)(44g/mol) = 465.4158g