# Dry ice is solid carbon dioxid

Dry ice is solid carbon dioxide. Instead of melting, solidcarbon dioxide sublimes according to the following equation:CO2(s)→CO2(g). When dry ice is added to warm water, heat from thewater causes the dry ice to sublime more quickly. The evaporatingcarbon dioxide produces a dense fog often used to create specialeffects. In a simple dry ice fog machine, dry ice is added to warmwater in a Styrofoam cooler. The dry ice produces fog until itevaporates away, or until the water gets too cold to sublime thedry ice quickly enough. Suppose that a small Styrofoam cooler holds15.0 liters of water heated to 81 ∘C

Part A:

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

Part B:

Calculate the mass of dry ice that should be added to the waterso that the dry ice completely sublimes away when the water reaches35 ∘C. Assume no heat loss to the surroundings.

The first thing worth noting here is that “there is no heat lossto the surroundings”, therefore all the heat transfer is from thehot water to the dry ice, which causes it to vapourise. Now thetotal amount of heat transferred during the specified scenario isreflected in the change in the temperature of the water.

The equation (that you should have already seen) that describesthis 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 = 1g; since, density of water = 1g/mL)C the specific heat capacity of water = 4.186 J/g/KT the change intemperature = (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 causethat amount of heat change. This is really simple (dimensionalanalysis should tell you right away how to do this) givenΔH(sublimation) = 33.9 kJ/mol:n(dry ice) = q / ΔH(sublimation)=  (2.9 x 103 kJ) / (427.4 kJ/mol)= 6.79 molWhere n(dry ice) is the amount of dry ice need in moles. To obtainthe final answer you clearly have to multiply that number by themolar mass of carbon dioxide (dry ice).So, 6.79 mol x 44 g/mol = 299 g

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