The human body obtains 1094 kJ from a candy bar. If this energy were used to vaporize water at 100°C, how much water in liters could be vaporized? (Assume that the density of water is 1.0 g/mL.)
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To calculate the amount of water that can be vaporized, we need to determine the heat required to convert water at 100°C to vapor and then divide it by the energy obtained from the candy bar.
The heat required to convert water at 100°C to vapor is given by the formula:
Q = m * ΔHv
where Q is the heat absorbed, m is the mass of water, and ΔHv is the heat of vaporization of water. The heat of vaporization of water is approximately 40.7 kJ/mol.
First, we need to convert the energy obtained from the candy bar to joules:
Energy = 1094 kJ * 1000 = 1094000 J
Now, let's calculate the mass of water that can be vaporized:
1094000 J = m * 40.7 kJ/mol
To convert kJ to J, we need to divide by 1000:
1094000 J = m * 40.7 J/mol
Simplifying the equation:
m * 40.7 J/mol = 1094000 J
m = 1094000 J / 40.7 J/mol
m ≈ 26885.1 mol
Since the density of water is 1.0 g/mL, we know that 1 mole of water has a mass of 18.02 g:
m = 26885.1 mol * 18.02 g/mol
m ≈ 484749.7 g
Finally, let's convert the mass of water to liters using the density of water:
484749.7 g = 484749.7 mL
Therefore, approximately 484749.7 liters of water could be vaporized using the energy obtained from the candy bar.
To determine how much water in liters could be vaporized using the energy obtained from the candy bar, we need to consider the heat required to vaporize the water and convert it into energy units that match the given value of 1094 kJ.
First, we need to determine the heat of vaporization for water, which is the amount of energy required to vaporize one gram of water. The heat of vaporization of water is approximately 40.7 kJ/mol.
Next, we convert the given energy value of 1094 kJ to joules by multiplying it by 1000 since 1 kJ is equal to 1000 J:
1094 kJ × 1000 = 1,094,000 J
Now, we can calculate the number of moles of water that can be vaporized using the heat of vaporization. To do this, divide the energy in joules by the heat of vaporization:
1,094,000 J ÷ 40.7 kJ/mol = 26.9 mol
Since the density of water is 1.0 g/mL, we know that 1 mole of water is equivalent to 18 grams (grams/moles) or 18 mL (milliliters/moles).
Thus, the number of milliliters of water that can be vaporized is:
26.9 mol × 18 mL/mol = 484.2 mL
Finally, we can convert the milliliters of water to liters by dividing by 1000:
484.2 mL ÷ 1000 = 0.4842 L
Therefore, approximately 0.4842 liters of water can be vaporized using the energy obtained from the candy bar.