If 5.40 kcal of heat is added to 1.00 kg of water at 100 degrees celsius, how much steam at 100 degrees celsius is produced?
It takes 540 cal to vaporize 1 grams H2O at 100 C to steam at 100 C.
540 cal/g x #g = 540,000 cal.
Solve for #g.
To determine the amount of steam produced, we need to use the equation:
Q = mL
Where:
Q is the heat added or removed (in joules)
m is the mass of the substance (in kilograms)
L is the latent heat of vaporization of the substance (in joules per kilogram)
In this case, we have the heat added (Q = 5.40 kcal) and the mass of water (m = 1.00 kg).
However, we need the latent heat of vaporization of water (L) to calculate the amount of steam produced. The latent heat of vaporization of water is 2,260 kJ/kg.
1 kilocalorie (kcal) is equal to 4.18 kilojoules (kJ). So, we need to convert 5.40 kcal to kilojoules:
5.40 kcal × 4.18 kJ/kcal = 22.572 kJ
Now we can calculate the amount of steam produced:
Q = mL
22.572 kJ = m × 2260 kJ/kg
Solving for m:
m = (22.572 kJ) / (2260 kJ/kg)
m ≈ 0.01 kg
Therefore, approximately 0.01 kg (or 10 grams) of steam at 100 degrees celsius is produced when 5.40 kcal of heat is added to 1.00 kg of water at 100 degrees celsius.