How much heat is absorbed when 52.3 g H2O(l) at 100°C and 101.3 kPa is converted to steam at 100°C? (The molar heat of vaporization of water is 40.7 kJ/mol.)
To find the answer to this question, we can use the following equation:
heat absorbed = (mass of water)(heat of vaporization of water)
heat absorbed = (52.3 g)(2260 J/g)
heat absorbed = 118198 J
heat absorbed = 118.198 kJ or 1.18 × 10^2 kJ
Answer: About 1.18 × 10^2 kJ of heat is absorbed when 52.3 g of H2O at 100°C and 101.3 kPa is converted to steam at 100°C. The answer is choice "B".
To calculate the heat absorbed, we need to use the equation:
Q = m × ΔHvap
Where:
Q is the heat absorbed (in Joules)
m is the mass of water (in grams)
ΔHvap is the molar heat of vaporization (in J/g)
First, we need to calculate the number of moles of water:
n = m / M
Where:
n is the number of moles
m is the mass of water (in grams)
M is the molar mass of water (18.015 g/mol)
Next, we can determine the heat absorbed:
Q = n × ΔHvap
Given:
m = 52.3 g
ΔHvap = 40.7 kJ/mol = 40.7 × 10^3 J/mol
M = 18.015 g/mol
First, calculate the number of moles:
n = 52.3 g / 18.015 g/mol = 2.9014 mol
Next, calculate the heat absorbed:
Q = 2.9014 mol × 40.7 × 10^3 J/mol = 118,102.798 J
Therefore, the heat absorbed when 52.3 g of H2O(l) is converted to steam at 100°C is approximately 118,102.798 J.
To calculate the amount of heat absorbed when converting water to steam, we need to know the mass of water being converted, the molar heat of vaporization of water, and the boiling point of water.
Given:
Mass of water (m) = 52.3 g
Boiling point of water (T) = 100°C
Molar heat of vaporization of water (ΔHvap) = 40.7 kJ/mol
First, we need to calculate the moles of water being converted. We can do this by dividing the mass of water by its molar mass.
Molar mass of water (H2O) = 18.015 g/mol
moles = mass / molar mass
moles = 52.3 g / 18.015 g/mol
Next, we calculate the amount of heat absorbed using the following formula:
Heat absorbed = moles * ΔHvap
Heat absorbed = (52.3 g / 18.015 g/mol) * 40.7 kJ/mol
Now we can calculate it:
Heat absorbed = 119.58 kJ
Therefore, when 52.3 g of water at 100°C and 101.3 kPa is converted to steam at 100°C, it absorbs approximately 119.58 kJ of heat.