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 90 ∘C. 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 10 ∘C. Assume no heat loss to the surroundings.

Answer 1

Answer:

$m=8.79kg$

Explanation:

First of all we need to calculate the heat that the water in the cooler is able to release:

$Q=\rho * V*Cp*\Delta T$

Where:

• Cp is the mass heat capacity of water
• V is the volume
• $\rho$ is the density

$Q=1 g/cm^3 *15000 cm^3*4.184 \frac{J}{g*^{\circ}C}*(10-90)^{\circ}C$

$Q=-5020800 J=-5020.8 kJ$

To calculate the mass of CO2 that sublimes:

$-Q=\Delta H_{sub}*m$

Knowing that the enthalpy of sublimation for the CO2 is: $\Delta H_{sub}=571 kJ/kg$

$5020.8 kJ=571 kJ/kg*m$

$m=\frac{5020.8 kJ}{571 kJ/kg}=8.79kg$