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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.

Respuesta :

Giangiglia

Answer:

[tex]m=8.79kg[/tex]

Explanation:

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

[tex]Q=\rho * V*Cp*\Delta T[/tex]

Where:

  • Cp is the mass heat capacity of water
  • V is the volume
  • [tex]\rho [/tex] is the density

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

[tex]Q=-5020800 J=-5020.8 kJ[/tex]

To calculate the mass of CO2 that sublimes:

[tex]-Q=\Delta H_{sub}*m[/tex]

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

[tex]5020.8 kJ=571 kJ/kg*m[/tex]

[tex]m=\frac{5020.8 kJ}{571 kJ/kg}=8.79kg[/tex]

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