9. If 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C, how much steam at 100⁰C is produced? Show all calculations leading to an answer.
It takes 540 calories to change 1 g H2O from liquid to steam at 100 C.
540 cal/g x #g = 5400 cal.
Solve for #g
To solve this problem, we need to use the specific heat capacity and latent heat of vaporization for water.
First, we need to determine the amount of heat required to raise the temperature of 1.00 kg of water from 100°C to its boiling point at 100°C. The specific heat capacity of water is 4.18 J/g°C.
Step 1: Convert the mass of water from kilograms to grams.
1.00 kg = 1000 g
Step 2: Calculate the heat required to raise the temperature of water.
Heat = mass × specific heat capacity × temperature change
Heat = 1000 g × 4.18 J/g°C × (100°C - 100°C)
= 0 J
This means that no heat is required to raise the temperature of water from 100°C to its boiling point at 100°C.
Next, we need to determine the amount of heat required to convert the water at its boiling point into steam at the same temperature. The latent heat of vaporization for water is 2260 J/g.
Step 3: Calculate the heat required for vaporization.
Heat = mass × latent heat of vaporization
Heat = 1000 g × 2260 J/g
= 2,260,000 J
We now know that 2,260,000 J of heat is required to convert 1.00 kg of water at 100°C into steam at the same temperature.
Finally, we need to convert the heat added, which is given in kcal (kilocalories), into joules.
Step 4: Convert kcal to joules.
1 kcal = 4184 J
Heat = 5.40 kcal × 4184 J/kcal
= 22,593.6 J
Now we can determine the amount of steam produced.
Step 5: Calculate the mass of steam produced.
Heat = mass × latent heat of vaporization
22,593.6 J = mass × 2260 J/g
mass = 22,593.6 J ÷ 2260 J/g
= 10 g
Therefore, 10 grams of steam at 100°C is produced when 5.40 kcal of heat is added to 1.00 kg of water at 100°C.