How much heat is emitted when 4.55 of water condenses at 25 ? The heat of vaporization of water at 25 is 44.0 .
44.0 WHAT?
q = mass water x deltaHvap
The 44.0 doesn't look right to me no matter what the units. Is that 25C?
To calculate the amount of heat emitted when water condenses, we can use the formula:
Q = m * ΔHv
Where:
Q is the heat emitted/released (in joules),
m is the mass of water (in grams), and
ΔHv is the heat of vaporization (in J/g).
Given:
mass of water (m) = 4.55 g
heat of vaporization (ΔHv) = 44.0 J/g
Substituting the values into the formula, we have:
Q = 4.55 g * 44.0 J/g
Q = 200.2 J
Therefore, when 4.55 g of water condenses at 25°C, approximately 200.2 J of heat is emitted.
To calculate the amount of heat emitted when water condenses, we can use the formula:
Q = m * ΔHvap
Where:
Q is the amount of heat emitted,
m is the mass of water being condensed, and
ΔHvap is the heat of vaporization of water at a given temperature.
Given:
m = 4.55 g (mass of water)
ΔHvap = 44.0 kJ/mol (heat of vaporization of water)
However, we need to convert the mass of water to moles before proceeding with the calculation. We can use the molar mass of water to do this conversion:
Molar mass of water = 18.015 g/mol
So, the number of moles of water (n) can be calculated as:
n = m / Molar mass of water
= 4.55 g / 18.015 g/mol
Now, we have the number of moles of water, and we can calculate the amount of heat emitted:
Q = n * ΔHvap
= moles of water * heat of vaporization
= (4.55 g / 18.015 g/mol) * 44.0 kJ/mol
Simplifying the above calculation:
Q = (4.55 / 18.015) * 44.0 kJ
Therefore, the amount of heat emitted when 4.55 g of water condenses at 25 °C is approximately ((4.55 / 18.015) * 44.0) kJ.