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 85
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 16 C. Assume no heat loss to the surroundings.
Change in enthalpy for dry ice sublimation: 33.9 kJ
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To calculate the mass of dry ice needed to completely sublime away when the water reaches 16°C, we can apply the principles of heat transfer and use the given enthalpy change for dry ice sublimation.
Here's how you can solve the problem step by step:
Step 1: Calculate the heat absorbed by the water to heat it from 85°C to 16°C.
The heat absorbed by the water can be calculated using the formula: q = m × c × ΔT
Where:
- q is the heat absorbed (in joules)
- m is the mass of water (in grams)
- c is the specific heat capacity of water (4.18 J/g·°C)
- ΔT is the change in temperature of the water (in °C)
Let's do the calculations:
q = (15,000 g) × (4.18 J/g·°C) × (16°C - 85°C)
q = (15,000 g) × (4.18 J/g·°C) × (-69°C)
q = -431,970 J
Step 2: Convert the heat absorbed by the water to kilojoules.
Since the enthalpy change for dry ice sublimation is given in kilojoules, we need to convert the heat absorbed by the water from joules to kilojoules.
1 kilojoule (kJ) = 1000 joules (J)
q = -431,970 J ÷ 1000 = -431.97 kJ
Step 3: Calculate the mass of dry ice using the enthalpy change.
The enthalpy change for dry ice sublimation is given as 33.9 kJ. We can use this value to calculate the mass of dry ice using the following formula: q = m × ΔH
Where:
- q is the heat absorbed (in kilojoules)
- m is the mass of dry ice (in grams)
- ΔH is the enthalpy change for dry ice sublimation (33.9 kJ)
Let's do the calculations:
-431.97 kJ = m × (33.9 kJ)
m = -431.97 kJ ÷ 33.9 kJ
m = -12.748 grams
The negative sign indicates that the reaction is exothermic, as heat is being released.
Therefore, the mass of dry ice that should be added to the water is approximately 12.748 grams.
To calculate the mass of dry ice that should be added to the water, we need to consider the heat transfer from the water to the dry ice, and the heat required for the dry ice to sublime.
The heat transfer (q) can be calculated using the equation:
q = mcΔT
where:
q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
The heat required for the dry ice to sublime (ΔH) is given as 33.9 kJ.
Since there is no heat loss to the surroundings, the heat lost by the water is equal to the heat gained by the dry ice. Therefore, we can set up the equation:
q (water) = q (dry ice)
mcΔT = ΔH
Rearranging this equation, we can solve for m (mass of the water):
m = ΔH / (cΔT)
Given:
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = 85°C - 16°C = 69°C
Substituting the given values into the equation:
m = 33.9 kJ / (4.18 J/g°C × 69°C)
Converting kJ to J and simplifying:
m = 33.9 × 10^3 J / (4.18 J/g°C × 69°C)
m = 734.82 g
Therefore, approximately 734.82 grams of dry ice should be added to the water.