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 81 ∘C

Part A:

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.

Part B:

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 35 ∘C. Assume no heat loss to the surroundings.

Express your answer using two significant figures.

Answer 1

The first thing worth noting here is that "there is no heat loss to the surroundings", therefore all the heat transfer is from the hot water to the dry ice, which causes it to vapourise. Now the total amount of heat transferred during the specified scenario is reflected in the change in the temperature of the water.

The equation (that you should have already seen) that describes this situation is:

q = m C T

where q is the heat change,
m the mass of the substance water = 15000 g (as 1 mL of water = 1 g; since, density of water = 1g/mL)
C the specific heat capacity of water = 4.186 J/g/K
T the change in temperature = (81 - 35) K = 46 K

So,

q = (15000 g) (4.186 J/g/K) (46 K)
= 2888340 J
= 2888.34 kJ
= 2.9 x 10³ kJ

The next step is to calculate how much dry ice is required to cause that amount of heat change. This is really simple (dimensional analysis should tell you right away how to do this) given ΔH(sublimation) = 33.9 kJ/mol:

n(dry ice) = q / ΔH(sublimation)
=  (2.9 x 10³ kJ) / (427.4 kJ/mol)
= 6.79 mol

Where n(dry ice) is the amount of dry ice need in moles. To obtain the final answer you clearly have to multiply that number by the molar mass of carbon dioxide (dry ice).
So, 6.79 mol x 44 g/mol = 299 g